Storage Devices

Magnetic tape is the most popular storage medium for long time storage medium for long time storage. This note contains more information on storage devices.

Summary

Magnetic tape is the most popular storage medium for long time storage medium for long time storage. This note contains more information on storage devices.

Things to Remember

  • Magnetic tape is the most popular storage medium for long time storage medium for long time storage.
  • Magnetic disk is most popular storage medium for direct access backup storage.
  • Floppy disks are very thin and flexible, so called floppies.
  • Hard disk are made up with set of rigid metal diskettes (mostly aluminum) which is permanently sealed in metal case for air-tight.
  • Due to the use of laser beam technology for recording/reading of data on the disk, these devices are called optical storage devices.

MCQs

No MCQs found.

Subjective Questions

Q1:

If \(\vec a\) = \(\begin {pmatrix} 6\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} -1\\ 6\\ \end {pmatrix}\), then prove that:

\(\vec a\) and \(\vec b\) are perpendiculsr to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = \(\begin {pmatrix} 6\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} -1\\ 6\\ \end {pmatrix}\)</p> <p>\(\vec a\) . \(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 6&times; -1 + 1&times; 6 = 6 - 6 = 0</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(6)^2 + (1)^2}\) = \(\sqrt {36 + 1}\) = \(\sqrt {37}\)</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(-1)^2 + (6)^2}\) = \(\sqrt {1 + 36}\) = \(\sqrt {37}\)</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)</p> <p>or, cos\(\theta\) = \(\frac 0{\sqrt {37} \sqrt {37}}\)</p> <p>or, cos\(\theta\) = 0</p> <p>or, \(\theta\) = cos<sup>-1</sup>(0) = 90&deg;</p> <p>The angle between two vectors is 90&deg; so, the \(\vec a\) and \(\vec b\) are perpendicular each other. <sub>Proved</sub></p>

Q2:

If \(\vec a\) = \(\begin {pmatrix} 2\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} 0\\ -2\\ \end {pmatrix}\), find the angle between \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = \(\begin {pmatrix} 2\\ 1\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} 0\\ -2\\ \end {pmatrix}\)</p> <p>\(\vec a\) . \(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 2 &times; 0 + 1 &times; -2 = 0 + -2 = -2</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(2)^2 + (1)^2}\) = \(\sqrt {4 + 1}\) = \(\sqrt 5\)</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(0)^2 + (-2)^2}\) = \(\sqrt {0 + 4}\) = 2</p> <p>cos\(\theta\) = \(\frac {-2}{2\sqrt 5}\) = -\(\frac 1{\sqrt 5}\)</p> <p>&there4; \(\theta\) = cos<sup>-1</sup> (-\(\frac 1{\sqrt 5}\)) <sub>Ans</sub></p> <p></p>

Q3:

If \(\vec p\) = \(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\), find the angle between \(\vec p\) and \(\vec q\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec p\) = \(\begin {pmatrix} 3\\ 4\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 1\\ 1\\ \end {pmatrix}\)</p> <p>\(\vec p\) . \(\vec q\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 3 &times; 1 + 4 &times; 1 = 3 + 4 = 7</p> <p>\(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt 25\) = 5</p> <p>\(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(1)^2 + (1)^2}\) = \(\sqrt {1 + 1}\) = \(\sqrt 2\)</p> <p>cos\(\theta\) = \(\frac {7}{5\sqrt 2}\) = \(\frac 7{7.07}\) = 0.99</p> <p>&there4; \(\theta\) = cos<sup>-1</sup> (0.99) = 8.11<sup>0</sup><sub>Ans</sub></p>

Q4:

If \(\vec a\) = \(\begin {pmatrix} 2\\ -3\\ \end {pmatrix}\) and \(\vec c\) = \(\begin {pmatrix} 4\\ -2\\ \end {pmatrix}\), find the angle between \(\vec a\) and \(\vec c\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = \(\begin {pmatrix} 2\\ -3\\ \end {pmatrix}\) and \(\vec c\) = \(\begin {pmatrix} 4\\ -2\\ \end {pmatrix}\)</p> <p>\(\vec a\) . \(\vec c\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 2&times; 4+ (-3) &times; (-2) = 8+ 6=14</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(2)^2 + (-3)^2}\) = \(\sqrt {4 + 9}\) = \(\sqrt {13}\)</p> <p>\(\begin {vmatrix} \vec c\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {(4)^2 + (-2)^2}\) = \(\sqrt {16 + 4}\) = \(\sqrt {20}\)</p> <p>cos\(\theta\) = \(\frac {14}{\sqrt {13} . \sqrt {20}}\) = \(\frac {14}{16.13}\) = 0.87</p> <p>&there4; \(\theta\) = cos<sup>-1</sup> (0.87) = 29.54<sup>0</sup><sub>Ans</sub></p>

Q5:

If \(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = 3, \(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = 3\(\sqrt 2\) and \(\vec p\).\(\vec q\) = 9, find the angle between \(\vec p\) and \(\vec q\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\begin {vmatrix} \vec p\\ \end {vmatrix}\) = 3, \(\begin {vmatrix} \vec q\\ \end {vmatrix}\) = 3\(\sqrt 2\) and \(\vec p\).\(\vec q\) = 9</p> <p>We know,</p> <p>cos\(\theta\) = \(\frac {\vec p . \vec q}{\begin {vmatrix} \vec p\\ \end {vmatrix}\begin {vmatrix} \vec q\\ \end {vmatrix}}\) = \(\frac 9{3 &times; 3\sqrt 2}\) = \(\frac 1{\sqrt 2}\)</p> <p>cos\(\theta\) = \(\frac 1{\sqrt 2}\)</p> <p>\(\theta\) = cos<sup>-1</sup>(\(\frac 1{\sqrt 2}\)) = 45&deg;</p> <p>&there4; The angle between \(\vec p\) and \(\vec q\) is 45&deg;. <sub>Ans</sub></p>

Q6:

If \(\vec a\) = 4 \(\vec i\) + \(\vec j\) and \(\vec b\) = 3 \(\vec i\) + 4\(\vec j\):

  1. find the value of \(\vec a\).\(\vec b\)
  2. find the angle between \(\vec a\) and \(\vec b\)

 


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = 4 \(\vec i\) + \(\vec j\) and \(\vec b\) = 3 \(\vec i\) + 4\(\vec j\)</p> <p>(i)</p> <p>\(\vec a\).\(\vec b\) = (4 \(\vec i\) + \(\vec j\)) . (3\(\vec i\) + 4\(\vec j\)) = 4&times; 3 + 1&times; 4 = 12 + 4 = 16 <sub>Ans</sub></p> <p>(ii)</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(4)^2 + (1)^2}\) = \(\sqrt {16 + 1}\) = \(\sqrt {17}\)</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5</p> <p>Let: \(\theta\) be the angle between \(\vec a\) and \(\vec b\) then,</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac {16}{5\sqrt {17}}\)</p> <p>\(\theta\) = cos<sup>-1</sup>(\(\frac {16}{5\sqrt {17}}\)) = 37.47&deg;</p> <p>&there4; The angle between \(\vec a\) and \(\vec b\) is 37.47&deg;. <sub>Ans</sub></p>

Q7:

Prove that: \(\vec a\) = -4\(\vec i\) + 5\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 10\(\vec j\) are parallel to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = -4\(\vec i\) + 5\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 10\(\vec j\)</p> <p>(x<sub>1</sub> , y<sub>1</sub>) = (-4 , 5) and (x<sub>2</sub> , y<sub>2</sub>) = (8 , -10)</p> <p>\(\vec a\).\(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = -4 &times; 8 + 5 &times; 10 = -32 - 50 = -82</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(-4)^2 + (5)^2}\) = \(\sqrt {16 + 25}\) = \(\sqrt {41}\)</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-10)^2}\) = \(\sqrt {64 + 100}\) = \(\sqrt {164}\)</p> <p>We know,</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac {-82}{\sqrt {41} \sqrt {164}}\) = - 1 (Approx.)</p> <p>\(\theta\) = cos<sup>-1</sup> (-1) = 180&deg;</p> <p>&there4; The \(\vec a\) and \(\vec b\) are parallel in opposite direction. <sub>Proved</sub></p>

Q8:

If \(\vec a\) = 3 + k and \(\vec b\) = -7 + 3 are perpendicular each other, find the value of k.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = 3 + k and \(\vec b\) = -7 + 3 are perpendiculat each other.</p> <p>Then:</p> <p>\(\vec a\) . \(\vec b\) = 0</p> <p>or, (3 + k) (-7 + 3) = 0</p> <p>or, 3&times; -7 + k&times; 3 = 0</p> <p>or, -21 + 3k = 0</p> <p>or, 3k = 21</p> <p>or, k = \(\frac {21}3\)</p> <p>&there4; k = 7 <sub>Ans</sub></p>

Q9:

Prove that: \(\vec a\) and \(\vec b\) are perpendicular, if \(\vec a\) = 4 + 2 and \(\vec b\) = -1 + 2.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = 4 + 2 and \(\vec b\) = -1 + 2</p> <p>If \(\vec a\) and \(\vec b\) are perpendicular than \(\vec a\) . \(\vec b\) = 0</p> <p>or, (4 + 2) (-1 + 2) = 0</p> <p>or, 4&times; -1 + 2&times; 2 = 0</p> <p>or, -4 + 4 = 0</p> <p>&there4; 0 = 0</p> <p>&there4; \(\vec a\) and \(\vec b\) are perpendicular to each other. <sub>Proved</sub></p>

Q10:

If \(\vec p\) = 10\(\vec i\) + 2k \(\vec j\) and \(\vec q\) = 2\(\vec i\) - 5\(\vec j\) are perpendicular to each other, find the value of k.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec p\) = 10\(\vec i\) + 2k \(\vec j\) and \(\vec q\) = 2\(\vec i\) - 5\(\vec j\)</p> <p>(x<sub>1</sub>, y<sub>1</sub>) = (10, 2k) and (x<sub>2</sub>, y<sub>2</sub>) = (2. -5)</p> <p>We know,</p> <p>\(\vec p\) and \(\vec q\) are perpendicular.</p> <p>Then: \(\vec p\) . \(\vec q\) = 0</p> <p>or, (10, 2k) (2, -5) = 0</p> <p>or, 10&times; 2 + 2k&times; -5 = 0</p> <p>or, 20 - 10k = 0</p> <p>or, 10k = 20</p> <p>or, k = \(\frac {20}{10}\)</p> <p>&there4; k = 2 <sub>Ans</sub></p>

Q11:

If \(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\), find the value of \(\angle\)AOB.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\)</p> <p>\(\vec a\) . \(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 7 &times; 5 + (-5) &times; (-7) = 35 + 35 = 70</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(7)^2 + (-5)^2}\) = \(\sqrt {49 + 25}\) = \(\sqrt {74}\)</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(5)^2 + (-7)^2}\) = \(\sqrt {25 + 49}\) = \(\sqrt {74}\)</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) =\(\frac {70}{\sqrt {74} \sqrt {74}}\) = \(\frac {70}{74}\) = 0.95</p> <p>\(\theta\) = cos<sup>-1</sup>(0.95) = 18.93&deg;</p> <p>&there4;\(\angle\)AOB = 18.93&deg; <sub>Ans</sub></p>

Q12:

In what condition \(\vec a\) and \(\vec b\) are parallel? If \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\), find the angle between the vectors \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>Let: the vector \(\vec a\) and \(\vec b\) are parallel if \(\vec a\) = m\(\vec b\) where m is scalar quantity.</p> <p>\(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and b = 8\(\vec i\) - 6\(\vec j\)</p> <p>\(\vec a\) . \(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 3 &times; 8 + 4 &times; (-6) = 24 - 24 = 0</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-6)^2}\) = \(\sqrt {64 + 36}\) = \(\sqrt {100}\) = 10</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac 0{5 &times; 10}\) = 0</p> <p>&there4; \(\theta\) = cos<sup>-1</sup> (0) = 90&deg; <sub>Ans</sub></p>

Q13:

 

vfs

In the given figure, \(\vec {OA}\) = \(\vec a\) and \(\vec {OB}\) = \(\vec b\). If \(\vec {AC} \) = 3 \(\vec {AB}\), find \(\vec {OC}\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and \(\vec {AC}\) = 3\(\vec {AB}\) then, \(\vec {OC}\) = ?</p> <p>Using triangle law in vector addition,</p> <p>\(\vec {OA}\) + \(\vec {AB}\) = \(\vec {OB}\)</p> <p>\(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) = \(\vec b\) - \(\vec a\)</p> <p>We know,</p> <p>\(\vec {AC}\) = 2\(\vec {AB}\) = 3 (\(\vec b\) - \(\vec a\)) = 3\(\vec b\) - 3\(\vec a\)</p> <p>Again,</p> <p>\(\vec {OA}\) + \(\vec {AC}\) = \(\vec {OC}\)</p> <p>&there4; \(\vec {OC}\) = \(\vec a\) + 3\(\vec b\) - 3\(\vec a\) = 3\(\vec b\) - 2\(\vec a\) <sub>Ans</sub></p>

Q14:

If A(1, 2) and B(3, 0) are two points. Express in the form x + y.


Type: Short Difficulty: Easy

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Answer: <p>Given points are: A(1, 2) and B(3, 0)</p> <p>x-component = x<sub>2</sub> - x<sub>1</sub> = 3 - 1 = 2</p> <p>y-component = y<sub>2</sub> - y<sub>1</sub> = 0 - 2 = - 2</p> <p>\(\vec {AB}\) = (2, -2)</p> <p>\(\vec {AB}\) = 2\(\vec i\) - 2\(\vec j\) <sub>Ans</sub></p>

Q15:

If C(1, 1) and D(-2, -4) are two points. Express in the form of x + y.


Type: Short Difficulty: Easy

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Answer: <p>Given points are: C(1, 1) and D(-2, -4)</p> <p>x-component = x<sub>2</sub> - x<sub>1</sub> = -2 - 1 = -3</p> <p>y-component = y<sub>2</sub> - y<sub>1</sub> = 0 - 2 = - 2</p> <p>\(\vec {CD}\) = (-3, -5)</p> <p>\(\vec {CD}\) = -3\(\vec i\) - 5\(\vec j\) <sub>Ans</sub></p>

Q16:

Find the position vector of the mid-point M of the line segment PQ. If the co-ordinates of the points P and Q are (3, 5) and (-7, 3) respectively.


Type: Short Difficulty: Easy

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Answer: <p>Given points are: P(3, 5) and Q(-7, 3).</p> <p>Let, O be the origin.</p> <p>\(\vec {OP}\) = \(\begin {pmatrix} 3\\ 5\\ \end {pmatrix}\) and \(\vec {OQ}\) = \(\begin {pmatrix} -7\\ 3\\ \end {pmatrix}\)</p> <p>Using midpoint formula,</p> <p>\(\vec {OM}\) = \(\frac {\vec {OP} + \vec {OQ}}2\)</p> <p>or, \(\vec {OM}\) = \(\frac {\begin {pmatrix}3\\ 5\\ \end{pmatrix} + \begin {pmatrix} -7\\ 3\\ \end {pmatrix}}2\)</p> <p>or, \(\vec {OM}\) = \(\frac {\begin {pmatrix}3 - 7\\ 5 + 3\\ \end {pmatrix}}2\)</p> <p>or, \(\vec {OM}\) = \(\frac {\begin {pmatrix} -4\\ 3\\ \end {pmatrix}}2\)</p> <p>or, \(\vec {OM}\) = \(\begin {pmatrix} \frac {-4}2\\ \frac 82\\ \end {pmatrix}\)</p> <p>&there4; \(\vec {OM}\) = \(\begin {pmatrix} -2\\ 4\\ \end {pmatrix}\)</p> <p>&there4; \(\vec {OM}\) = -2\(\vec i\) + 4\(\vec j\) <sub>Ans</sub></p>

Q17:

 

dfs

In \(\triangle\)ABC , if D is the midpoint of BC, then prove by vector method that: \(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>D is the midpoint of the line BC of the \(\triangle\)ABC.</p> <p>In \(\triangle\)ABD,</p> <p>\(\vec {AB}\) = \(\vec {AD}\) + \(\vec {DB}\)....................(1) [\(\because\) triangle law of vector addition]</p> <p>In \(\triangle\)ACD,</p> <p>\(\vec {AC}\) = \(\vec {AD}\) + \(\vec {DC}\)...................(2)[\(\because\) triangle law of vector addition]</p> <p>Adding (1) and (2)</p> <p>\(\vec {AB}\) + \(\vec {AC}\) = \(\vec {AD}\) + \(\vec {BD}\) + \(\vec {AD}\) + \(\vec {DC}\)</p> <p>or,\(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) + (\(\vec {DB}\) + \(\vec {DC}\))</p> <p>or,\(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) + \(\vec {DB}\) - \(\vec {DB}\) [\(\because\) -\(\vec {DB}\) = \(\vec {DC}\)]</p> <p>&there4; \(\vec {AB}\) + \(\vec {AC}\) = 2\(\vec {AD}\) <sub>Ans</sub></p>

Q18:

 

asdf

The position vector of the point A and B are 3 + 2 and 5 - 6 respectively. Find the position vectors of the midpoint M of the line AB.


Type: Short Difficulty: Easy

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Answer: <p>Let, O be the origin.</p> <p>\(\vec {OA}\) = 3\(\vec i\) + 2\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 6\(\vec j\)</p> <p>We know,</p> <p>\(\vec {OM}\) = \(\frac {\vec {OA} + \vec {OB}}2\)</p> <p>or, \(\vec {OM}\) = \(\frac {3\vec i + 2\vec j + 5\vec i - 6\vec j}2\)</p> <p>or, \(\vec {OM}\) = \(\frac {8\vec i - 4\vec j}2\)</p> <p>&there4; \(\vec {OM}\) = 4\(\vec i\) - 2\(\vec j\)</p> <p>&there4; The position vector of M = 4\(\vec i\) - 2\(\vec j\) <sub>Ans</sub></p>

Q19:

The position vectors of the points A and B are 5 + 2 and 3 + 6 respectively. Find the position vector of the point P which divides AB internally in the ratio 2 : 3.


Type: Short Difficulty: Easy

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Answer: <p>Let: O be the origin.</p> <p>\(\vec {OA}\) = 5\(\vec i\) + 2\(\vec j\) and \(\vec {OB}\) = 3\(\vec i\) + 6\(\vec j\)</p> <p>m : n = 2 : 3</p> <p>Using section internal division formula.</p> <p>\(\vec {OP}\) = \(\frac {m \vec{OB} + n\vec {OA}}{m + n}\)</p> <p>or, \(\vec {OP}\) = \(\frac {2(3\vec i + 6\vec j) + 3(5\vec i + 2\vec j)}{2 + 3}\)</p> <p>or, \(\vec {OP}\) = \(\frac {6\vec i + 12\vec j + 15\vec i + 6\vec j}5\)</p> <p>or, \(\vec {OP}\) = \(\frac {21\vec i + 18\vec j}5\)</p> <p>&there4; \(\vec {OP}\) = \(\frac {21}5\)\(\vec i\) + \(\frac {18}5\)\(\vec j\) <sub>Ans</sub></p>

Q20:

If the co-ordinates of the points A and B are (-4, 8) and (3, 7) respectively. Find the position vector of the point which divides the line segment AB externally in the ratio 4 : 3.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>A(-4, 8) and B(3, 7)</p> <p>Let, O be the origin.</p> <p>\(\vec {OA}\) = \(\begin {pmatrix} -4\\ 8\\ \end {pmatrix}\)</p> <p>\(\vec {OB}\) = \(\begin {pmatrix} 3\\ 7\\ \end {pmatrix}\)</p> <p>Using section external division formula,</p> <p>\(\vec {AB}\) = \(\frac {4\begin {pmatrix} 3\\ 7\\ \end {pmatrix} - 3\begin {pmatrix} -4\\ 8\\ \end {pmatrix}}{4 - 3}\)</p> <p>or, \(\vec {AB}\) = \(\frac {\begin {pmatrix} 12\\ 28\\ \end {pmatrix} - \begin {pmatrix} -12\\ 24\\ \end {pmatrix}}1\)</p> <p>or, \(\vec {AB}\) = \(\begin {pmatrix} 12 + 12\\ 28 - 24\\ \end {pmatrix}\)</p> <p>or, \(\vec {AB}\) = \(\begin {pmatrix} 24\\ 4\\ \end {pmatrix}\)</p> <p>&there4; \(\vec {AB}\) = 24\(\vec i\) + 4\(\vec j\) <sub>Ans</sub></p>

Q21:

If \(\vec {OA}\) = \(\begin {pmatrix} \sqrt 3\\ 1\\ \end {pmatrix}\) and \(\vec {OB}\) = \(\begin {pmatrix} \sqrt 3\\ 3\sqrt 3\\ \end {pmatrix}\), calculate the \(\angle\)AOB.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec {OA}\) = \(\begin {pmatrix} \sqrt 3\\ 1\\ \end {pmatrix}\) and \(\vec {OB}\) = \(\begin {pmatrix} \sqrt 3\\ 3\sqrt 3\\ \end {pmatrix}\)</p> <p>Let: x<sub>1</sub> = \(\sqrt 3\), y<sub>1</sub> = 1, x<sub>2</sub> = \(\sqrt 3\) and y<sub>2</sub> = 3\(\sqrt 3\)</p> <p>Let: \(\theta\) be the angle between \(\vec {OA}\) and \(\vec {OB}\)</p> <p>cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}}\)</p> <p>or, cos \(\angle\)AOB = \(\frac {\sqrt 3 &times; \sqrt 3 + 1 &times; 3\sqrt 3}{\sqrt {(\sqrt 3)^2 + 1^2} \sqrt {(\sqrt 3)^2 + (3\sqrt 3)^2}}\)</p> <p>or, cos \(\angle\)AOB = \(\frac {3 + 3\sqrt 3}{\sqrt {3 + 1}. \sqrt {3 + 27}}\)</p> <p>or, cos \(\angle\)AOB = \(\frac {3 + 3\sqrt 3}{2\sqrt {30}}\)</p> <p>or, cos \(\angle\)AOB = \(\frac {8.197}{10.95}\)</p> <p>or, cos \(\angle\)AOB = 0.75</p> <p>or, \(\angle\)AOB = cos<sup>-1</sup> (0.75)</p> <p>&there4;\(\angle\)AOB = 41.58&deg; <sub>Ans</sub></p>

Q22:

State parallelogram law of vector addition.


Type: Short Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 285px;"><img src="/uploads/0411.jpg" alt="DF" width="285" height="201"><figcaption><br></figcaption></figure><p>If the vector is \(\vec u\) and \(\vec v\) are represented by the two adjacent sides \(\vec {AB}\) and \(\vec {AD}\) of a parallelogram ABCD, then the sum \(\vec u\) + \(\vec v\) is represented by the diagonal \(\vec {AC}\).</p> <p>\(\vec {AB}\) + \(\vec {AD}\) = \(\vec {AC}\)</p> <p>\(\vec u\) + \(\vec v\) = \(\vec {AC}\)</p>

Q23:

Define unit vector and prove that: \(\vec a\) = 4\(\vec i\) - 6\(\vec j\) and \(\vec b\) = 3\(\vec i\) + 2\(\vec j\) are perpendicular.


Type: Short Difficulty: Easy

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Answer: <p>Unit vector: A vector whose modulus is 1 is called a unit vector.</p> <p>\(\vec a\) = (\(\frac 1{\sqrt 2}\), \(\frac 1{\sqrt 2}\)) is unit vector.</p> <p>\(\vec a\) . \(\vec b\)</p> <p>= (4\(\vec i\) - 6\(\vec j\)) . (3\(\vec i\) + 2\(\vec j\))</p> <p>= 12\(\vec {i^2}\) + 8\(\vec i\) \(\vec j\) - 18\(\vec i\) \(\vec j\) - 12\(\vec {j^2}\)</p> <p>= 12&times; 1 + 8&times; 0 - 18&times; 0 - 12&times; 1 [\(\vec {i^2}\) = \(\vec {j^2}\) = 1, \(\vec i\) . \(\vec j\) = 0]</p> <p>= 12 - 12</p> <p>= 0</p> <p>\(\vec a\) . \(\vec b\) = 0</p> <p>&there4; \(\vec a\) . \(\vec b\) are perpendicular each other. <sub>Proved</sub></p>

Q24:

If \(\vec a\) = (4\(\vec i\) + 5\(\vec j\)) and \(\vec b\) = (5\(\vec i\) - 4\(\vec j\)), find the angle between \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = (4\(\vec i\) + 5\(\vec j\)) and \(\vec b\) = (5\(\vec i\) - 4\(\vec j\))</p> <p>\(\vec a\) = (4, 5) and \(\vec b\) = (5, -4)</p> <p>If \(\theta\) be the angle between \(\vec a\) and \(\vec b\) then:</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)</p> <p>or, cos\(\theta\) = \(\frac {(4, 5) . (5, -4)}{\sqrt {4^2 + 5^2} \sqrt {5^2 + (-4)^2}}\)</p> <p>or, cos\(\theta\) = \(\frac {20 - 20}{\sqrt {16 + 25} \sqrt {25 + 16}}\)</p> <p>or, cos\(\theta\) = \(\frac 0{25 + 16}\)</p> <p>or, cos\(\theta\) = 0</p> <p>or, \(\theta\) = cos<sup>-1</sup> (0)</p> <p>&there4; \(\theta\) = 90&deg; <sub>Ans</sub></p>

Q25:

If \(\vec a\) = 10\(\vec i\) - 7\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 10\(\vec j\). Find the angle between \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = 10\(\vec i\) - 7\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 10\(\vec j\)</p> <p>If \(\theta\) be the angle between the vectors \(\vec a\) and \(\vec b\).</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\)</p> <p>or, cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}}\)</p> <p>or, cos\(\theta\) = \(\frac {10 &times; 7 + 10 &times; (-7)}{\sqrt {(10)^2 + (-7)^2} \sqrt {7^2 + {10}^2}}\)</p> <p>or, cos\(\theta\) = \(\frac {70 - 70}{\sqrt {149} \sqrt {149}}\)</p> <p>or, cos\(\theta\) = 0&deg;</p> <p>&there4; \(\theta\) = cos<sup>-1</sup>(0&deg;) = 90&deg; <sub>Ans</sub></p>

Q26:

If the position vectors of the vertices of \(\triangle\)ABC are A(-1, -1), B(5, -1) and C(2, 5). Find the position vector of the centroid G of the \(\triangle\)ABC.


Type: Short Difficulty: Easy

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Answer: <p>In the \(\triangle\)ABC,</p> <p>Let O be the origin.</p> <p>Position vector of the point A is = \(\vec {OA}\) = (-1, -1)</p> <p>Position vector of the point B is = \(\vec {OB}\) = (5, -1)</p> <p>Position vector of the point C is = \(\vec {OC}\) = (2, 5)</p> <p>If G be the centroid of the triangle,</p> <p>= \(\frac 13\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\))</p> <p>= \(\frac 13\) (- 1 - 1 + 5 - 1 + 2 + 5)</p> <p>= \(\frac 13\) (6 + 3)</p> <p>= 2 + 1</p> <p>Hence, the position vector of the centroid is (2, 1). <sub>Ans</sub></p>

Q27:

If \(\vec a\) = 6\(\vec i\) - 8\(\vec j\) and \(\vec b\) = 4\(\vec i\) + 3\(\vec j\) prove that the vector \(\vec a\) and \(\vec b\) are perpendicular to each other.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec a\) = 6\(\vec i\) - 8\(\vec j\) and \(\vec b\) = 4\(\vec i\) + 3\(\vec j\)</p> <p>\(\vec a\) . \(\vec b\)</p> <p>= (6\(\vec i\) - 8\(\vec j\)) (4\(\vec i\) + 3\(\vec j\))</p> <p>= 6 . 4\(\vec {i^2}\) + 18\(\vec i\) \(\vec j\) - 32 \(\vec i\) \(\vec j\) - 24\(\vec {j^2}\)</p> <p>= 24 . 1 + 18 . 0 - 32 . 0 * 24 . 1 [\(\because\) \(\vec {i^2}\) = \(\vec {j^2}\) = 1 , \(\vec j\) = \(\vec i\) \(\vec j\) = 0]</p> <p>= 24 - 24</p> <p>= 0</p> <p>&there4; The dot product of two vector a and b is equal to zero so these are perpendicular to each other. <sub>Proved</sub></p>

Q28:

If \(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\), find the value of \(\angle\)AOB.


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec {OA}\) = 7\(\vec i\) - 5\(\vec j\) and \(\vec {OB}\) = 5\(\vec i\) - 7\(\vec j\)</p> <p>\(\vec a\) . \(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 7&times; 5 + (-5) (-7) = 35 + 35 = 70</p> <p>\(\begin {vmatrix} \vec a \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {7^2 + (-5)^2}\) = \(\sqrt {49 + 25}\) = \(\sqrt {74}\)</p> <p>\(\begin {vmatrix} \vec b\end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {5^2 + (-7)^2}\) = \(\sqrt {25 + 49}\) = \(\sqrt {74}\)</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a \end {vmatrix} \begin {vmatrix} \vec b\end {vmatrix}}\)</p> <p>or, cos\(\theta\) = \(\frac {70}{\sqrt {74} . \sqrt {74}}\)</p> <p>or, cos\(\theta\) = \(\frac {70}{74}\)</p> <p>or, cos\(\theta\) = 0.95</p> <p>or, \(\theta\) = cos<sup>-1</sup> (0.95)</p> <p>&there4; \(\theta\) = 18.93&deg;</p> <p>&there4; \(\angle\)AOB = 18.93&deg; <sub>Ans</sub></p>

Q29:

What do you mean by a unit vector? If the position vectors of A and B are 3\(\vec i\) + 4\(\vec j\) and 7\(\vec i\) + 8\(\vec j\) respectively, find the position vector of the mid-point of the line joining A and B.


Type: Short Difficulty: Easy

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Answer: <p>Unit Vector: If the magnitude of the vector is 1 such type of vector is known as unit vector.</p> <p>i.e., \(\vec {OP}\) = \(\begin {pmatrix} 1\\ 0\\ \end {pmatrix}\)</p> <p>Let, \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 7\(\vec i\) + 8\(\vec j\) are the position vectors of A and B,</p> <p>Position vector of mid-point \(\vec M\) = \(\frac 12\) (\(\vec a\) + \(\vec b\))</p> <p>\(\vec M\) = \(\frac 12\) (3\(\vec i\) + 4\(\vec j\) + 7\(\vec i\) + 8\(\vec j\)) = \(\frac 12\) (10\(\vec i\) + 12\(\vec j\)) = 5\(\vec i\) + 6\(\vec j\) <sub>Ans</sub></p>

Q30:

 

sdgv

In the given figure, \(\vec {OA}\) = \(\vec a\) and \(\vec {OB}\) = \(\vec b\). If \(\vec {AC}\) = 3\(\vec {AB}\), find \(\vec {OC}\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>\(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and\(\vec {AC}\) = 3\(\vec {AB}\) then, \(\vec {OC}\) = ?</p> <p>Using triangle law of vector addition:</p> <p>\(\vec {OA}\) + \(\vec {AB}\) \(\vec {OB}\)</p> <p>or, \(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) = \(\vec b\) - \(\vec a\)</p> <p>We know,</p> <p>\(\vec {AC}\) = 3\(\vec {AB}\) = 3(\(\vec b\) - \(\vec a\)) = 3\(\vec b\) - 3\(\vec a\)</p> <p>Again,</p> <p>\(\vec {OA}\) + \(\vec {AC}\) = \(\vec {OC}\)</p> <p>or, \(\vec {OC}\) = \(\vec a\) + 3\(\vec b\) - 3\(\vec a\)</p> <p>&there4; \(\vec {OC}\) = 3\(\vec b\) - 2\(\vec a\) <sub>Ans</sub></p>

Q31:

On what condition \(\vec a\) and \(\vec b\) are parallel? If \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\), find the angle between the vectors \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>The two vectors are \(\vec a\) and \(\vec b\) are parallel if \(\vec a\) = m\(\vec b\) where m is scalar product.</p> <p>\(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = 8\(\vec i\) - 6\(\vec j\)</p> <p>\(\vec a\).\(\vec b\) = x<sub>1</sub>x<sub>2</sub> + y<sub>1</sub>y<sub>2</sub> = 3&times; 8 + (-4) &times; 6= 24 - 24 = 0</p> <p>\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(3)^2 + (4)^2}\) = \(\sqrt {9 + 16}\) = \(\sqrt {25}\) = 5</p> <p>\(\begin {vmatrix} \vec b\\ \end {vmatrix}\) = \(\sqrt {x_1^2 + y_1^2}\) = \(\sqrt {(8)^2 + (-6)^2}\) = \(\sqrt {64 + 36}\) = \(\sqrt {100}\) = 10</p> <p>We know,</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a\\ \end {vmatrix} \begin {vmatrix} \vec b\\ \end {vmatrix}}\) = \(\frac 0{5 &times; 10}\) = 0</p> <p>&there4; \(\theta\) = cos<sup>-1</sup> (0) = 90&deg; <sub>Ans</sub></p>

Q32:

Point C divides the line AB internally in the ratio of 3 : 1. If the position vectors of A and B are \(\vec i\) - 3\(\vec j\) and 2\(\vec i\) + 5\(\vec j\) respectively, find \(\begin {vmatrix} \vec {AB} \end {vmatrix}\) and the position vector of point C.


Type: Short Difficulty: Easy

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Answer: <p>Let: \(\vec a\) = \(\vec i\) - 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) - 5\(\vec j\) , m:n = 3:1</p> <p>Position vector of C (\(\vec {OC}\)) = \(\frac {n\vec a + m \vec b}{m + n}\)</p> <p>\(\vec c\) = \(\vec {1 (\vec i - 3\vec j)}{3 + 1}\)</p> <p>\(\vec c\) = \(\frac {\vec i - 3\vec j + 6\vec i + 15\vec j}{4}\)</p> <p>\(\vec c\) = \(\frac {7\vec i + 12\vec j}4\) <sub>Ans</sub></p> <p>\(\vec {AB}\) = \(\begin {pmatrix} x_2 - x_1\\ y_2 - y_1\\ \end {pmatrix}\) =\(\begin {pmatrix} 2 - 1\\ 5 + 3\\ \end {pmatrix}\) = \(\begin {pmatrix} 1\\ 8\\ \end {pmatrix}\)</p> <p>\(\begin {vmatrix} \vec {AB}\\ \end {vmatrix}\) = \(\sqrt {x^2 + y^2}\) = \(\sqrt {1^2 + 8^2}\) = \(\sqrt {1 + 64}\) = \(\sqrt {65}\) <sub>Ans</sub></p>

Q33:

 

VG

In the figure, \(\vec {OA}\) = \(\vec a\) . \(\vec {OB}\) = \(\vec b\) and \(\vec {AC}\) = 5\(\vec AB\), find \(\vec {OC}\) in terms of \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Using triangle law of vector addition,</p> <p>\(\vec {OC}\) = \(\vec {OA}\) + \(\vec {AC}\)</p> <p>or, \(\vec {OC}\) = \(\vec a\) + 5\(\vec {AB}\)</p> <p>or, \(\vec {OC}\) = \(\vec a\) + 5 (\(\vec {OB}\) - \(\vec {OA}\))</p> <p>or, \(\vec {OC}\) = \(\vec a\) + 5 (\(\vec b\) - \(\vec a\))</p> <p>or, \(\vec {OC}\) = \(\vec a\) + 5\(\vec b\) - 5\(\vec a\)</p> <p>&there4; \(\vec {OC}\) = 5\(\vec b\) - 4\(\vec a\) <sub>Ans</sub></p>

Q34:

What do you mean by orthogonal vectors? If \(\vec p\) = \(\begin {pmatrix} 3\\ -2\\ \end {pmatrix}\) and \(\vec q\) = \(\begin {pmatrix} 2\\ 3\\ \end {pmatrix}\), show that \(\vec p\) and \(\vec q\) are orthogonal vectors.


Type: Short Difficulty: Easy

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Answer: <p>Orthogonal Vectors: If two vectors are perpendicular each other then vector are known as orthogonal vector.</p> <p>\(\vec p\) . \(\vec q\) = 3&times; 2 - 2&times; 3 = 0</p> <p>&there4; The dot product od \(\vec p\) and \(\vec q\) is equal to zero, so these vectors are orthogonal. <sub>Proved</sub></p>

Q35:

If the position vectors of the points A and B are 3\(\vec i\) - 2\(\vec j\) and 1\(\vec i\) + 8\(\vec j\) respectively, find the position vector of the mid-point of the line AB.


Type: Short Difficulty: Easy

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Answer: <p>Let: \(\vec {OA}\) = 3\(\vec i\) - 2\(\vec j\) and \(\vec {OB}\) = 1\(\vec i\) + 8\(\vec j\)</p> <p>Mid-point of \(\vec {AB}\)</p> <p>= \(\frac {\vec {OA} + \vec {OB}}2\)</p> <p>= \(\frac {3\vec i - 2\vec j + 1\vec i + 8\vec j}2\)</p> <p>= \(\frac {4\vec i + 6\vec j}2\)</p> <p>= 2\(\vec i\) + 3\(\vec j\)</p> <p>&there4; Position vector of the mid-point of \(\vec {AB}\) = 2\(\vec i\) + 3\(\vec j\) <sub>Ans</sub></p>

Q36:

Define position vector. If P is the mid-point of the straight line joining the points A(3, 4) and B(-1, 2), find the position vector of P.


Type: Short Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 248px;"><img src="/uploads/074.jpg" alt="czv" width="248" height="212"><figcaption><br></figcaption></figure><p>Position Vectors: Let P(x, y) be any point on the co-ordinates axes, then (x, y) is called the position vector of the point P with referred to the origin O. OP is a position vector.</p> <p>Let: \(\vec a\) = 3\(\vec i\) + 4\(\vec j\) and \(\vec b\) = -\(\vec i\) + 2\(\vec j\)</p> <p>\(\vec m\) = ?</p> <p>\(\vec m\) = \(\frac {\vec a + \vec b}2\) = \(\frac {3\vec i + 4\vec j - \vec i + 2\vec j}2\) = \(\frac {2\vec i + 6\vec j}2\) = \(\vec i\) + 3\(\vec j\) <sub>Ans</sub></p>

Q37:

If PQRS is a quadrilateral, prove that:

\(\vec {QR}\) + \(\vec {RS}\) + \(\vec {SP}\) = \(\vec {QP}\)


Type: Short Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 321px;"><img src="/uploads/083.jpg" alt="ASD" width="321" height="228"><figcaption><br></figcaption></figure><p>Using triangle law of vector addition,</p> <p>In \(\triangle\)QRS,</p> <p>\(\vec {QS}\) = \(\vec {QR}\) + \(\vec {RS}\) ....................................(1)</p> <p>In \(\triangle\)PQS,</p> <p>\(\vec {QS}\) = \(\vec {QP}\) + \(\vec {PS}\) ...................................(2)</p> <p>From (1) and (2),</p> <p>\(\vec {QR}\) + \(\vec {RS}\) = \(\vec {QP}\) + \(\vec {PS}\)</p> <p>or, \(\vec {QR}\) + \(\vec {RS}\) - \(\vec {PS}\) = \(\vec {QP}\)</p> <p>&there4; \(\vec {QR}\) + \(\vec {RS}\) +\(\vec {SP}\) = \(\vec {QP}\) <sub>Proved</sub></p>

Q38:

If \(\vec a\) = \(\vec i\) + 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) + \(\vec j\), find the angle between \(\vec a\) and \(\vec b\).


Type: Short Difficulty: Easy

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Answer: <p>Given,</p> <p>\(\vec a\) = \(\vec i\) + 3\(\vec j\) and \(\vec b\) = 2\(\vec i\) + \(\vec j\)</p> <p>Let an anglebetween \(\vec a\) and \(\vec b\) be \(\theta\)</p> <p>So,</p> <p>cos\(\theta\) = \(\frac {\vec a . \vec b}{\begin {vmatrix} \vec a \end {vmatrix} \begin {vmatrix} \vec b \end {vmatrix}}\)</p> <p>or, cos\(\theta\) = \(\frac {x_1x_2 + y_1y_2}{\sqrt {{x_1}^2 +{y_1}^2} \sqrt {{x_2}^2 + {y_2}^2}}\) [where x<sub>1</sub> = 1, y<sub>1</sub> = 3, x<sub>2</sub> = 2 and y<sub>2</sub> = 1]</p> <p>or, cos\(\theta\) = \(\frac {1 &times; 2 + 3 &times; 1}{\sqrt {1^2 + 3^2} \sqrt {2^2 + 1^2}}\)</p> <p>or, cos\(\theta\) = \(\frac {2 + 3}{\sqrt {10} \sqrt 5}\)</p> <p>or, cos\(\theta\) = \(\frac 5{\sqrt {50}}\)</p> <p>or, cos\(\theta\) = \(\frac 5{5\sqrt 2}\)</p> <p>or, cos\(\theta\) = \(\frac 1{\sqrt 2}\)</p> <p>or, cos\(\theta\) = cos 45&deg;</p> <p>&there4; \(\theta\) = 45&deg; <sub>Ans</sub></p>

Q39:

define a null vector. If \(\vec a\) = \(\begin {pmatrix} -5\\ 3\\ \end {pmatrix}\) and \(\vec b\) = \(\begin {pmatrix} p\\ p + 2\\ \end {pmatrix}\) are perpendicular to each other, find the value of p.


Type: Short Difficulty: Easy

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Answer: <p>Null Vector: If the magnitude of vector is equal to zero such type of vector is known as zero or null vector. \(\vec {AA}\) = 0</p> <p>If \(\vec a\) and \(\vec b\) are perpendicular then,</p> <p>\(\vec a\) . \(\vec b) = 0</p> <p>or, (-5, 3) (p, p+2) = 0</p> <p>or, -5&times; p + 3&times; (p + 2) = 0</p> <p>or, -5p + 3p + 6 = 0</p> <p>or, -2p = - 6</p> <p>or, p = \(\frac 62\)</p> <p>&there4; p = 3 <sub>Ans</sub></p> <p></p>

Q40:

 

ASD

Prove that: \(\vec {PQ}\) + \(\vec {QR}\) + \(\vec {RS}\) + \(\vec {SP}\) = 0 in the given quadrilateral PQRS.


Type: Short Difficulty: Easy

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Answer: <p>Joining point P and R,</p> <p>In \(\triangle\)RSP,</p> <p>\(\vec {RP}\) = \(\vec {RS}\) + \(\vec {SP}\).................................(1)</p> <p>In \(\triangle\)PQR,</p> <p>\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)..............................(2)</p> <p>Adding equation (1) and (2),</p> <p>\(\vec {RP}\) + \(\vec {PR}\) = \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\)</p> <p>or, -\(\vec {PR}\)+ \(\vec {PR}\) = \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\) [\(\because\) \(\vec {RP}\) = -\(\vec {PR}\)]</p> <p>or, 0= \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\)</p> <p>&there4; \(\vec {RS}\) + \(\vec {SP}\) + \(\vec {PQ}\) + \(\vec {QR}\) = 0 <sub>Proved</sub></p>

Q41:

 

d

Prove that: \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) = 0 in given triangle ABC.


Type: Short Difficulty: Easy

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Answer: <p>\(\vec {AB}\), \(\vec {BC}\) and \(\vec {AC}\) are the vertices which represent the side of \(\triangle\)ABC using triangle law of addition.</p> <p>\(\vec {BC}\) + \(\vec {CA}\) = \(\vec {BA}\)</p> <p>or, \(\vec {BC}\) + \(\vec {CA}\) = -\(\vec {AB}\) [\(\because\) \(\vec {BA}\) = - \(\vec {AB}\)]</p> <p>&there4; \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) = 0 <sub>Proved</sub></p>

Q42:

 

sf

In the given diagram, prove that:

\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) + \(\vec {EA}\) = 0


Type: Short Difficulty: Easy

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Answer: <p>In \(\triangle\)ABC,</p> <p>\(\vec {AB}\) + \(\vec {BC}\) = \(\vec {AC}\)............................(1)</p> <p>In \(\triangle\)ACD,</p> <p>\(\vec {AC}\) + \(\vec {CD}\) = \(\vec {AD}\)...........................(2)</p> <p>In \(\triangle\)ADE,</p> <p>\(\vec {AD}\) + \(\vec {DE}\) = \(\vec {AE}\).............................(3)</p> <p>or, \(\vec {AC}\) + \(\vec {CD}\) + \(\vec {DE}\) = \(\vec {AE}\)</p> <p>or, \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) = \(\vec {AE}\)</p> <p>&there4;\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CD}\) + \(\vec {DE}\) + \(\vec {EA}\) = 0 <sub>Proved</sub></p>

Q43:

 

sdf

In the given figure, ABCD is a parallelogram. If \(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\), \(\vec {OC}\) = \(\vec c\), find \(\vec {OD}\).


Type: Short Difficulty: Easy

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Answer: <p>Here,</p> <p>Using triangle law of vector addition,</p> <p>In \(\triangle\)OBC,</p> <p>\(\vec {BC}\) = \(\vec {BO}\) + \(\vec {OC}\)</p> <p>&there4; \(\vec {BC}\) = -\(\vec b\) + \(\vec c\)</p> <p>In \(\triangle\)OAD,</p> <p>\(\vec {OD}\) = \(\vec {OA}\) + \(\vec {AD}\) = \(\vec {OA}\) + \(\vec {BC}\) [\(\vec {AD}\) = \(\vec {BC}\)]</p> <p>&there4; \(\vec {OD}\) = \(\vec a\) - \(\vec b\) + \(\vec c\) <sub>Ans</sub></p>

Q44:

Prove by a vector method that the parallelogram with equal diagonals is a rectangle.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 281px;"><img src="/uploads/1121.jpg" alt="FG" width="281" height="164"><figcaption><br></figcaption></figure><p>Let, PQRS be a parallelogram which PR and SQ are equal diagonals.</p> <p>Let: \(\vec {SR}\) = \(\vec a\) and \(\vec {SP}\) = \(\vec b\)</p> <p>To prove: PQRS is a parallelogram.</p> <p>\(\vec {SQ}\) = \(\vec {SR}\) + \(\vec {RQ}\) [\(\because\) Triangle law of vector addition]</p> <p>\(vec {SR}\) = \(\vec {SR} \) + \(\vec {SP}\) [\(\because\) \(\vec {RQ}\) = \(\vec {SP}\)]</p> <p>\(\vec {SQ}\) = \(\vec a\) + \(\vec b\)</p> <p>\(\vec {RP}\) =\(\vec {RS}\) + \(\vec {SP}\) = -\(\vec {SR}\) + \(\vec {SP}\) = -\(\vec a\) \(\vec b\) = \(\vec b\) - \(\vec a\)</p> <p>\(\begin {vmatrix} \vec {SQ}\\ \end {vmatrix}\) = \(\begin {vmatrix} \vec {RP}\\ \end {vmatrix}\)</p> <p>or, \(\vec {SQ}\)<sup>2</sup> = \(\vec {RP}\)<sup>2</sup></p> <p>or, (\(\vec a\) + \(\vec b\))<sup>2</sup> = (\(\vec b\) - \(\vec a\))<sup>2</sup></p> <p>or, \(\vec a\)<sup>2</sup> + 2\(\vec a\) \(\vec b\) + \(\vec b\)<sup>2</sup> = \(\vec b\)<sup>2</sup>- 2\(\vec a\) \(\vec b\) + \(\vec a\)<sup>2</sup></p> <p>or, 4\(\vec a\) \(\vec b\) = 0</p> <p>or, \(\vec a\) . \(\vec b\) = 0</p> <p>&there4; \(\angle\)PSR = 90&deg;</p> <p>Hence, PQRS is a rectangle. <sub>Proved</sub></p>

Q45:

Prove by vector method that diagonals of a rhombus bisect each at right angle.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/1215.jpg" alt="XCV" width="250" height="181"><figcaption><br></figcaption></figure><p>Let: ABCD is a rhombus.</p> <p>Let: \(\vec {AB}\) = \(\vec a\) and \(\vec {AD}\) \ \(\vec b\)</p> <p>To prove: AC &perp;BD and AO = \(\frac 12\) AC, BO = \(\frac {BD}\).</p> <p>In \(\triangle\)ABC,</p> <p>\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {BC}\) [\(\because\) Triangle law of vector addition]</p> <p>\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {AD}\) [\(\because\) BC = AD]</p> <p>\(\vec {AC}\) = \(\vec a\) + \(\vec b\)</p> <p>\(\vec {BD}\) = \(\vec {BA}\) + \(\vec {AD}\) = -\(\vec a\) + \(\vec b\) = \(\vec b\) - \(\vec a\)</p> <p>\(\vec {AC}\) . \(\vec {BD}\)</p> <p>= (\(\vec a\) + \(\vec b\)) (\(\vec b\) - \(\vec a\))</p> <p>= \(\vec b\)<sup>2</sup> - \(\vec a\)<sup>2 </sup></p> <p>=\(\vec {AD}\)<sup>2</sup> - \(\vec {AB}\)<sup>2</sup></p> <p>= \(\vec {AB}\)<sup>2</sup> - \(\vec {AB}\)<sup>2</sup> [\(\because\) AB = AD]</p> <p>= 0</p> <p>&there4; AC &perp; BD</p> <p>\(\vec {AO}\)= \(\vec {AB}\) + \(\vec {BO}\) = \(\vec {AB}\) + \(\frac 12\) \(\vec {BD}\) = a + \(\frac 12\)(\(\vec b\) - \(\vec a\))</p> <p>\(\vec {AO}\) = \(\frac {2a + \vec b - \vec a}2\) = \(\frac 12\)(\(\vec a\) + \(\vec b\)) = \(\frac 12\)\(\vec {AC}\)..........................(1)</p> <p>\(\vec {BO}\) = \(\vec {BA}\) + \(\vec {AO}\) = -\(\vec {AB}\) + \(\vec {AO}\) = - \(\vec a\) + \(\frac 12\)(\(\vec a\) + \(\vec b\)</p> <p>\(\vec {BO}\) = \(\frac {-2\vec a + \vec a + \vec b}2\) = \(\frac 12\)(\(\vec b\) - \(\vec a\)) = \(\frac 12\)\(\vec {BD}\)............(2)</p> <p>From relation (1) and (2):</p> <p>Diagonal bisect each other.</p> <p>Hence, diagonals of a rhombus bisects each at right angle. Proved</p>

Q46:

Prove  by vector method that opposite sides of a parallelogram are equal.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/1313.jpg" alt="FV" width="250" height="169"><figcaption><br></figcaption></figure><p>Let: PQRS is a parallelogram, in which PQ//SR and PS//QR.</p> <p>To prove: PQ = SR and PS = QR</p> <p>Let \(\vec {PQ}\) = m\(\vec {SR}\) and \(\vec {PS}\) = n\(\vec {QR}\) where m and n are scalars.</p> <p>In \(\triangle\)PQR,</p> <p>\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)</p> <p>\(\vec {PR}\) = m\(\vec {SR}\) + n\(\vec {QR}\)......................(1)</p> <p>In \(\triangle\)PSR,</p> <p>\(\vec {PR}\) = \(\vec {PS}\) + \(\vec {SR}\)...............................(2)</p> <p>From (1) and (2),</p> <p>m\(\vec {SR}\) + n\(\vec {QR}\) = \(\vec {PS}\) + \(\vec {SR}\)</p> <p>or, m\(\vec {SR}\) - \(\vec {SR}\) = \(\vec {QR}\) - n\(\vec {QR}\)</p> <p>or, \(\vec {SR}\) (m - 1) = \(\vec {QR}\) (1 - n)</p> <p>\(\vec {QR}\) and \(\vec {SR}\) are not parallel so;</p> <p>m - 1 = 0&there4; m = 1</p> <p>1 - n = 0&there4; n = 1</p> <p>Putting the value of m and n in</p> <p>\(\vec {PQ}\) = m\(\vec {SR}\) and \(\vec {PS}\) = n\(\vec {QR}\)</p> <p>\(\vec {PQ}\) = \(\vec {SR}\) and \(\vec {PS}\) = \(\vec {QR}\)</p> <p>Hence, opposite sides of a parallelogram are equal. <sub>Proved</sub></p>

Q47:

If diagonal of parallelogram are at right angles to each other prove by vector method that it is a rhombus.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 268px;"><img src="/uploads/1411.jpg" alt="DS" width="268" height="181"><figcaption><br></figcaption></figure><p>ABCD is a parallelogram, diagonals AC and BD are perpendicular to each other.</p> <p>To prove: ABCD is a rhombus.</p> <p>Let: \(\vec {AB}\) = \(\vec {DC}\) = \(\vec a\) and \(\vec {BC}\) = \(\vec {AD}\) = \(\vec b\)</p> <p>Using triangle law of vector addition:</p> <p>In \(\triangle\)ABC,</p> <p>\(\vec {AC}\) = \(\vec {AB}\) + \(\vec {BC}\)</p> <p>\(\vec {AC}\) = \(\vec a\) + \(\vec b\)</p> <p>In \(\triangle\)BCD,</p> <p>\(\vec {BD}\) = \(\vec {BC}\) + \(\vec {CD}\)</p> <p>\(\vec {BD}\) = \(\vec {BC}\) - \(\vec {DC}\) = \(\vec b\) - \(\vec a\)</p> <p>AC and BD are perpendicular to each other.</p> <p>\(\vec {AC}\) . \(\vec {BD}\) = 0</p> <p>(\(\vec a\) + \(\vec b\)) + (\(\vec b\) - \(\vec a\)) = 0</p> <p>or, \(\vec {b^2}\) - \(\vec {a^2}\) = 0</p> <p>or, \(\vec {b^2}\) = \(\vec {a^2}\)</p> <p>or, \(\begin {vmatrix} \vec {b^2} \end {vmatrix}\) =\(\begin {vmatrix} \vec {a^2} \end {vmatrix}\)</p> <p>&there4;\(\begin {vmatrix} \vec b\end {vmatrix}\) =\(\begin {vmatrix} \vec a\end {vmatrix}\)</p> <p>AB = BC</p> <p>Hence, ABCD is a rhombus. <sub>Proved</sub></p>

Q48:

Prove by vector method that the sum of the squares of the diagonals of any parallelogram is equal to the sum of the squares of the four sides.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/1514.jpg" alt="FDG" width="250" height="169"><figcaption><br></figcaption></figure><p>PQRS is a parallelogram, PR and QS are diagonals.</p> <p>To prove: PR<sup>2</sup>+ QS<sup>2</sup> = PQ<sup>2</sup> + QR<sup>2</sup> + RS<sup>2</sup> + PS<sup>2</sup></p> <p>In \(\triangle\)PQR,</p> <p>\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\) [\(\because\) Triangle law of vector addition]</p> <p>Squaring on both sides,</p> <p>\(\vec {PR}^2\) = (\(\vec {PQ}\) + \(\vec {QR}\))<sup>2</sup></p> <p>\(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 . \(\vec {PQ}\) . \(\vec {QR}\) + \(\vec {QR}^2\)</p> <p>PR<sup>2</sup> = PQ<sup>2</sup> + QR<sup>2</sup> + 2 PQ. QR.....................(1)</p> <p>In \(\triangle\)QRS,</p> <p>\(\vec {QS}\) = \(\vec {QR}\) + \(\vec {RS}\) [\(\because\) Triangle law of vector addition]</p> <p>Squaring on both sides,</p> <p>\(\vec {QS}^2\) = (\(\vec {QR}\) + \(\vec {RS}\))<sup>2</sup> = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . \(\vec {RS}\) + \(\vec {RS}^2\)</p> <p>\(\vec {QS}^2\) = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . \(\vec {QP}\) + \(\vec {RS}^2\) [\(\because\) \(\vec {RS}\) = \(\vec {QP}\)]</p> <p>\(\vec {QS}^2\) = \(\vec {QR}^2\) + 2 . \(\vec {QR}\) . (-\(\vec {PQ}\)) + \(\vec {RS}^2\)</p> <p>\(\vec {QS}^2\) = \(\vec {QR}^2\) -2 . \(\vec {QR}\) . \(\vec {PQ}\) + \(\vec {RS}^2\)</p> <p>QS<sup>2</sup> = QR<sup>2</sup> + RS<sup>2</sup> - 2 . PQ . QR............................(2)</p> <p>Adding (1) and (2)</p> <p>PR<sup>2</sup> + QS<sup>2</sup> = PQ<sup>2</sup> + QR<sup>2</sup> + QR<sup>2</sup> + RS<sup>2</sup> + 2 PQ . QR - 2 PQ . QR</p> <p>&there4;PR<sup>2</sup> + QS<sup>2</sup> = PQ<sup>2</sup> + QR<sup>2</sup> + PS<sup>2</sup> + RS<sup>2</sup><sub>Hence, Proved</sub></p>

Q49:

 

sxc

In the given figure, PQRS is a trapezium where PQ\\SR. A and B are the mid-points of PR and QS respectively. Prove by vector method that

  1.  AB//PQ//SR
  2.  AB = \(\frac 12\)(PQ - SR)

 


Type: Long Difficulty: Easy

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Answer: <p>PQRS is a trapezium in which PQ\\SR. A and B are the mid-points of PR and QS. Let, P be the origin.</p> <p>To prove: i. AB//PQ//SR ii. AB = \(\frac 12\)(PQ - SR)</p> <p>Construction: join PB</p> <p>Let, P be the origin.</p> <p>\(\vec {SR}\) = m\(\vec {PQ}\) [\(\because\) PQ//SR]</p> <p>\(\vec {PR}\) = \(\vec {PS}\) + \(\vec {SR}\) [Using triangle law of vector addition]</p> <p>\(\vec {PR}\) = \(\vec {PS}\) + m\(\vec {PQ}\)</p> <p>\(\vec {PB}\) = \(\frac {\vec {PS} + \vec {PQ}}2\) [\(\because\) B is the mid-point of QS]</p> <p>\(\vec {PA}\) = \(\frac 12\)\(\vec {PR}\) = \(\frac 12\)(\(\vec {PS}\) + m\(\vec {PQ}\))</p> <p>\(\vec {AB}\) = \(\vec {PB}\) - \(\vec {PA}\) = \(\frac 12\)(1 - m)\(\vec {PQ}\)</p> <p>\(\vec {AB}\)//\(\vec {PQ}\)</p> <p>&there4; AB//PQ//SR</p> <p>Again,</p> <p>\(\vec {AB}\) = \(\frac 12\)\(\vec {PQ}\) - \(\frac 12\)m\(\vec {PQ}\) = \(\frac 12\)(\(\vec {PQ}\) - m\(\vec {PQ}\)) = \(\frac 12\)(\(\vec {PQ}\) - \(\vec {SR}\))</p> <p>&there4; AB = \(\frac 12\)(PQ - SR) <sub>Proved</sub></p>

Q50:

 

ds

In the given figure, PQRS is a parallelogram. M and N are two points on the diagonal SQ. If SM = NQ, prove by vector method that PMRN is a parallelogram.


Type: Long Difficulty: Easy

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Answer: <p>Given:</p> <p>PQRS is a parallelogram.</p> <p>In which: PQ =SR, PQ//SR and PS = QR, PS//QR</p> <p>To prove: PMRN is a parallelogram.</p> <p>In \(\triangle\)PQN,</p> <p>\(\vec {PQ}\) = \(\vec {PN}\) + \(\vec {NQ}\)....................(1) [Using triangle law of vector addition]</p> <p>In \(\triangle\)SMR,</p> <p>\(\vec {SR}\) = \(\vec {SM}\) + \(\vec {MR}\).....................(2)</p> <p>From equation (1) and (2)</p> <p>\(\vec {PQ}\) = \(\vec {SR}\)</p> <p>or, \(\vec {PN}\) + \(\vec {NQ}\) = \(\vec {SM}\) + \(\vec {MR}\)</p> <p>\(\vec {PN}\) = \(\vec {MR}\) [\(\because\) \(\vec {NQ}\) = \(\vec {SM}\)]</p> <p>PN and MR are equal and parallel.</p> <p>Similarly, PM and NR are equal and parallel.</p> <p>&there4; PMRN is a parallelogram. <sub>Proved</sub></p>

Q51:

If D, E and F are the mid-point of the sides of the triangle ABC respectively and O is any point, prove that:

\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) = \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\)


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/248.jpg" alt="AS" width="250" height="143"><figcaption><br></figcaption></figure><p>In the figure, ABC is a triangle with medians. AE, BF and CD of sides BC, AC and AB respectively. BE = \(\frac 12\)BC, CF = \(\frac 12\)CA, AD = \(\frac 12\)AB.</p> <p>To prove:\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) = \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\)</p> <p>Using triangle law of vector addition,</p> <p>\(\vec {AE}\) = \(\vec {AB}\) + \(\vec {BE}\) = \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\)........................(1)</p> <p>\(\vec {BF}\) = \(\vec {BC}\) + \(\vec {CF}\) = \(\vec {BC}\) + \(\frac 12\)\(\vec {CA}\)....................(2)</p> <p>\(\vec {CD}\) = \(\vec {CA}\) + \(\vec {AD}\) = \(\vec {CA}\) + \(\frac 12\)\(\vec {AB}\)......................(3)</p> <p>Adding (1), (2) and (3),</p> <p>\(\vec {AE}\) + \(\vec {BF}\) + \(\vec {CD}\)</p> <p>= \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\) +\(\vec {BC}\) + \(\frac 12\)\(\vec {CA}\) +\(\vec {CA}\) + \(\frac 12\)\(\vec {AB}\)</p> <p>= \(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\) + \(\frac 12\)(\(\vec {AB}\) + \(\vec {BC}\) + \(\vec {CA}\))</p> <p>= \(\vec {AC}\) + \(\vec {CA}\) + \(\frac 12\)(\(\vec {AC}\) + \(\vec {CA}\)) [\(\because\) \(\vec {AB}\) + \(\vec{BC}\) = \(\vec {AC}\)]</p> <p>\(\vec {AE}\) + \(\vec {BF}\) + \(\vec {CD}\) = -\(\vec {CA}\) + \(\vec {CA}\) + \(\frac 12\)(-\(\vec {CA}\) + \(\vec {CA}\)) = 0</p> <p>or, \(\vec {AO}\) + \(\vec {OE}\) + \(\vec {BO}\) + \(\vec {OF}\) + \(\vec {CO}\) + \(\vec {OD}\) = 0 [\(\because\) \(\vec {AE}\) = \(\vec {AO}\) + \(\vec {OE}\)]</p> <p>or, \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) - \(\vec {OA}\) - \(\vec {OB}\) - \(\vec {OC}\) = 0</p> <p>or, \(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\) = \(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\)</p> <p>&there4;\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) =\(\vec {OD}\) + \(\vec {OE}\) + \(\vec {OF}\)<sub> Hence, Proved</sub></p>

Q52:

 

FDG

In the given trapezium ABCD, AD//BC. M and N are the mid-point of AB and DC respectively. Prove by vector method:

MN = \(\frac 12\)(AD + BC)


Type: Long Difficulty: Easy

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Answer: <p>Given:</p> <p>ABCD is a trapezium.</p> <p>AD//BC//MN and M and N are the mid-point of AB and DC.</p> <p>To prove:MN = \(\frac 12\)(AD + BC)</p> <p>Construction: join point A and N, join point B and N</p> <p>Using triangle law of vector addition:</p> <p>In \(\triangle\)BMN,</p> <p>\(\vec {MN}\) = \(\vec {MB}\) + \(\vec {BN}\) = \(\vec {MB}\) + (\(\vec {BC}\) + \(\vec {CN}\))........................(1)</p> <p>In \(\triangle\)MAN,</p> <p>\(\vec {MN}\) = \(\vec {MA}\) + \(\vec {AN}\) = \(\vec {MB}\) + (\(\vec {AD}\) + \(\vec {DN}\))........................(2)</p> <p>Adding (1) and (2)</p> <p>2\(\vec {MN}\) = \(\vec {MB}\) + \(\vec {BC}\) + \(\vec {CN}\) + \(\vec {MA}\) + \(\vec {AD}\) + \(\vec {DN}\)</p> <p>or, 2\(\vec {MN}\) = (\(\vec {MB}\) + \(\vec {BM}\)) + \(\vec {BC}\) + \(\vec {AD}\) + \(\vec {CN}\) + \(\vec {DN}\)</p> <p>or, 2\(\vec {MN}\) = \(\vec {MB}\) - \(\vec {MB}\) + \(\vec {BC}\) + \(\vec {AD}\) + \(\vec {CN}\) + \(\vec {DN}\)</p> <p>or, 2\(\vec {MN}\) = \(\vec {BC}\) + \(\vec {AD}\)</p> <p>&there4;MN = \(\frac 12\)(AD + BC) <sub>Hence, Proved</sub></p>

Q53:

Prove by vector method that a line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/1715.jpg" alt="dsf" width="250" height="169"><figcaption><br></figcaption></figure><p>Given:</p> <p>In \(\triangle\)ABC, M and N are the mid-point of Ab and AC. M and N are joined.</p> <p>To prove: \(\vec {MN}\) //\(\vec {BC}\) and \(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\)</p> <p>Using triangle law of vector addition,</p> <p>In \(\triangle\)AMN,</p> <p>\(\vec {MN}\) = \(\vec {MA}\) + \(\vec {AN}\).........................(1)</p> <p>In \(\triangle\)ABC,</p> <p>\(\vec {BC}\) = \(\vec {BA}\) + \(\vec {AC}\)...........................(2)</p> <p>Since, M and N are mid-point of BA and AC.</p> <p>\(\vec {BC}\) = 2\(\vec {MA}\) + 2\(\vec {AN}\)</p> <p>\(\vec {BC}\) = 2 (\(\vec {MA}\) + \(\vec {AN}\))</p> <p>\(\frac {\vec {BC}}2\) = \(\vec {MN}\) [From equation (1)]</p> <p>\(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\)</p> <p>\(\vec {MN}\)//\(\vec {BC}\) [\(\because\) \(\vec a\) = m\(\vec b\) where, m is a scalar quantity then \(\vec a\)//\(\vec b\)]</p> <p>&there4; \(\vec {MN}\) = \(\frac 12\)\(\vec {BC}\) and \(\vec {MN}\)//\(\vec {BC}\) <sub>Hence, Proved</sub></p>

Q54:

 

vz

In the given figure, P and Q are the middle points of AB and AC respectively of the triangle ABC. Prove vectorially that PQ//BC and BC = 2PQ.


Type: Long Difficulty: Easy

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Answer: <p>Given:</p> <p>In \(\triangle\)ABC, P and Q are the mid points of AB and AC.</p> <p>To Prove: \(\vec {PQ}\)// \(\vec {BC}\) and \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\)</p> <p>Using triangle law of vector addition:</p> <p>In \(\triangle\)APQ,</p> <p>\(\vec {PQ}\) = \(\vec {PA}\) + \(\vec {AQ}\)....................................(1)</p> <p>In \(\triangle\)ABC,</p> <p>\(\vec {BC}\) = \(\vec {BA}\) + \(\vec {AC}\).....................................(2)</p> <p>Dividing by 2 on both sides of equation (2)</p> <p>\(\frac 12\)\(\vec {BC}\) =\(\frac 12\)(\(\vec {BA}\) + \(\vec {AC}\))</p> <p>\(\frac 12\)\(\vec {BC}\) = \(\vec {PA}\) + \(\vec {AQ}\) [\(\because\) P and Q are the mid-point of AB and AC]</p> <p>\(\frac 12\)\(\vec {BC}\) = \(\vec {PQ}\) [From equation (1)]</p> <p>&there4; \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\)</p> <p>\(\vec {PQ}\)//\(\vec {BC}\) [\(\because\) \(\vec a\) = m\(\vec b\) then \(\vec a\)//\(\vec b\)]</p> <p>Hence, \(\vec {PQ}\)//\(\vec {BC}\) and \(\vec {PQ}\) = \(\frac 12\)\(\vec {BC}\) <sub>Hence, Proved</sub></p>

Q55:

ABCD is a parallelogram and G is the point of intersection of its diagonals. If O is any point. Prove that \(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\) = 4\(\vec {OG}\).


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 250px;"><img src="/uploads/206.jpg" alt="sdf" width="250" height="204"><figcaption><br></figcaption></figure><p>AC and BD are the diagonals of a parallelogram of ABCD which intersect at a point G.GA = GC and GB = GD. If O be any point, then join O and A, O and B, O and C, O and D and O and G.</p> <p>Using triangle law in vector addition,</p> <p>In \(\triangle\)OAC,</p> <p>\(\vec {OA}\) + \(\vec {OC}\) = \(\vec {AC}\) = 2\(\vec {OG}\).............................(1)</p> <p>[\(\because\) G is a mid-point of AC]</p> <p>In \(\triangle\)OBD,</p> <p>\(\vec {OB}\) + \(\vec {OD}\) = 2\(\vec {OG}\)..........................................(2)</p> <p>Adding equation (1) and (2)</p> <p>\(\vec {OA}\) + \(\vec {OC}\) + \(\vec {OB}\) + \(\vec {OD}\) = 2\(\vec {OG}\) + 2\(\vec {OG}\)</p> <p>&there4;\(\vec {OA}\) + \(\vec {OC}\) + \(\vec {OB}\) + \(\vec {OD}\) = 4\(\vec {OG}\) <sub>Hence, Proved</sub></p>

Q56:

 

as

In the given figure, P, Q, R and S are the mid-points of the sides AB, AD, CD and BC respectively of the quadrilateral ABCD. Prove by vector method that PR and QS bisect each other.


Type: Long Difficulty: Easy

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Answer: <p>Position vector of P, \(\vec {OP}\) = \(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\))</p> <p>Position vector of R, \(\vec {OR}\) = \(\frac 12\)(\(\vec {OD}\) + \(\vec {OC}\))</p> <p>Position vector of mid-point of \(\vec {PR}\)</p> <p>= \(\frac 12\)[\(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\)) + \(\frac 12\)(\(\vec {OD}\) + \(\vec {OC}\))]</p> <p>= \(\frac 12\) [\(\frac 12\)(\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\))]</p> <p>= \(\frac 14\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {OC}\) + \(\vec {OD}\))</p> <p>Position vector of Q, \(\vec {OQ}\) = \(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\))</p> <p>Position vector of S, \(\vec {OS}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OC}\))</p> <p>Position vector of mid-point of \(\vec {QS}\)</p> <p>= \(\frac 12\)[\(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\)) + \(\frac 12\)(\(\vec {OB}\) + \(\vec {OC}\))]</p> <p>= \(\frac 12\) [\(\frac 12\)(\(\vec {OA}\) + \(\vec {OD}\) + \(\vec {OB}\) + \(\vec {OC}\))]</p> <p>= \(\frac 14\)(\(\vec {OA}\) + \(\vec {OD}\) + \(\vec {OB}\) + \(\vec {OC}\))</p> <p>&there4; Position vector of mid-point of \(\vec {PR}\) and \(\vec {QS}\) is same. So, \(\vec {PR}\) and \(\vec {QS}\) bisect each other. <sub>Proved</sub></p>

Q57:

Prove that the middle point of the hypotenuse of a right-angled triangle is equidistance from its vertices by vector method.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 268px;"><img src="/uploads/1910.jpg" alt="sd" width="268" height="181"><figcaption><br></figcaption></figure><p>Let, \(\triangle\)ABC be the right angled triangle whose right angle is at A. Let, P be the middle point of the hypotenuse BC.</p> <p>Take A is the origin and let;</p> <p>\(\vec {AC}\) = \(\vec c\), \(\vec {AB}\) = \(\vec b\)</p> <p>\(\vec {AP}\) = \(\vec d\) then \(\vec {AB}\) = \(\vec {AP}\) + \(\vec {PB}\)</p> <p>[Using triangle law of vector addition]</p> <p>\(\vec b\) = \(\vec d\) + \(\vec {PB}\) and \(\vec {AC}\) = \(\vec {AP}\) + \(\vec {PC}\)</p> <p>or, \(\vec c\) = \(\vec d\) - \(\vec {CP}\) i.e. \(\vec c\) = \(\vec d\) - \(\vec {PB}\) [\(\because\) PB = PC]</p> <p>Now,</p> <p>It is given that: AB&perp; AC</p> <p>&there4; \(\vec b\) . \(\vec c\) = 0</p> <p>or, (\(\vec d\) + \(\vec {PB}\)) (\(\vec d\) - \(\vec {PB}\)) = 0</p> <p>or, d<sup>2</sup> - PB<sup>2</sup> = 0</p> <p>or, d<sup>2</sup> = PB<sup>2</sup></p> <p>&there4; d = PB = PC = AP</p> <p>&there4; The middle point of the hypotenuse of a right-angled triangle is equidistance from the vertices. <sub>Proved</sub></p>

Q58:

 

zd

In the given figure, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively of the quadrilateral ABCD. Prove by vector method that PQRS is a parallelogram.


Type: Long Difficulty: Easy

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Answer: <p>Let: ABCD be quadrilateral.</p> <p>Take A as origin and suppose the position vectors of the vertices B, C, D are \(\vec a\), \(\vec b\) and \(\vec c\) respectively.</p> <p>Let, P, Q, R and S be the mid-points od AB, BC, CD and DA. PQ, QR, RS and SP are joined.</p> <p>By using the mid-point formula. We get the position vectors of P, Q, R and S as follows:</p> <p>\(\vec {AP}\) = \(\frac {\vec o + \vec a}2\) = \(\frac 12\)\(\vec a\)</p> <p>\(\vec {AQ}\) = \(\frac {\vec a + \vec b}2\) = \(\frac 12\)\(\vec a\) + \(\frac 12\)\(\vec b\)</p> <p>\(\vec {AR}\) = \(\frac {\vec b + \vec c}2\) = \(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\)</p> <p>\(\vec {AS}\) = \(\frac {\vec o + \vec c}2\) = \(\frac 12\)\(\vec c\)</p> <p>\(\vec {QR}\)</p> <p>= (position vector of R) - (position vector of Q)</p> <p>= (\(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\)) - (\(\frac 12\)\(\vec a\) + \(\frac 12\) \(\vec b\))</p> <p>= \(\frac 12\)\(\vec b\) + \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\) - \(\frac 12\)\(\vec b\)</p> <p>= \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\)..................................(1)</p> <p>Also,</p> <p>\(\vec {PS}\)</p> <p>= (position vector of S) - (position vector of P)</p> <p>= \(\frac 12\)\(\vec c\) - \(\frac 12\)\(\vec a\)..................................(2)</p> <p>From relation (1) and (2)</p> <p>\(\vec {QR}\) = \(\vec {PS}\)</p> <p>&there4; QR = PS and QR//PS</p> <p>&there4; PQRS is a parallelogram. <sub>Proved</sub></p>

Q59:

 

dfv

In the given figure, AD, BE and CF are the medians of the triangle ABC, prove that:

\(\vec {AD}\) + \(\vec {BE}\) + \(\vec {CF}\) = 0


Type: Long Difficulty: Easy

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Answer: <p>D, E and F are the mid-points of BC, CA and AB.</p> <p>By using triangle law of vector addition,</p> <p>\(\vec {AB}\) + \(\vec {BC}\) = \(\vec {AC}\)</p> <p>&there4; \(\vec {BC}\) = \(\vec {AC}\) - \(\vec {AB}\)</p> <p>Again,</p> <p>\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {BD}\)..................................(1)</p> <p>\(\vec {AD}\) = \(\vec {Ac}\) + \(\vec {CD}\)...................................(2)</p> <p>Adding (1) and (2)</p> <p>2\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {AC}\) + \(\vec {BD}\) + \(\vec {CD}\)</p> <p>or, 2\(\vec {AD}\) = \(\vec {AB}\) + \(\vec {AC}\) +0 [\(\because\) \(\vec {BD}\) = -\(\vec {CD}\)]</p> <p>&there4; \(\vec {AD}\) = \(\frac 12\)(\(\vec {AB}\) + \(\vec {AC}\))</p> <p>\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\) = -\(\vec {AB}\) + \(\frac 12\)\(\vec {AC}\) = \(\frac 12\)\(\vec {AC}\) - \(\vec {AB}\)</p> <p>\(\vec {CF}\) = \(\vec {CA}\) + \(\vec {AF}\) = -\(\vec {AC}\) + \(\frac 12\)\(\vec {AB}\) = \(\frac 12\)\(\vec {AB}\) - \(\vec {AC}\)</p> <p>Now,,</p> <p>\(\vec {AD}\) +\(\vec {BE}\) +\(\vec {CF}\)</p> <p>=\(\frac 12\)(\(\vec {AB}\) + \(\vec {AC}\)) +\(\frac 12\)\(\vec {AC}\) - \(\vec {AB}\) +\(\frac 12\)\(\vec {AB}\) - \(\vec {AC}\)</p> <p>= \(\vec {AB}\) + \(\vec {AC}\) - \(\vec {Ab}\) - \(\vec {AC}\)</p> <p>= 0</p> <p>&there4;\(\vec {AD}\) + \(\vec {BE}\) + \(\vec {CF}\) = 0 <sub>Hence, Proved</sub></p>

Q60:

 

zdcf

In the following figure, M and N are the mid-points of AB and BC respectively. Use vector method to prove that: \(\vec {AC}\) = 2\(\vec {MN}\).


Type: Long Difficulty: Easy

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Answer: <p>Let: the position vector of A, B and C are \(\vec a\), \(\vec b\) and \(\vec c\) respectively. \(\vec {OA}\) = \(\vec a\), \(\vec {OB}\) = \(\vec b\) and \(\vec {OC}\) = \(\vec c\).</p> <p>M is the mid-point of AB.</p> <p>Position vector of M (\(\vec {OM}\)) = \(\frac {\vec {OA} + \vec {OB}}2\) = \(\frac {\vec a + \vec b}2\)</p> <p>Position vector of N(\(\vec {ON}\)) = \(\frac {\vec {OB} + \vec {OC}}2\) = \(\frac {\vec b + \vec c}2\)</p> <p>Position vector of MN (\(\vec {MN}\)) = \(\vec {ON}\) - \(\vec {OM}\)</p> <p>or, \(\vec {MN}\) = \(\frac {\vec b + \vec c}2\) - \(\frac {\vec a + \vec b}2\)</p> <p>or, \(\vec {MN}\) = \(\frac {\vec b + \vec c - \vec a - \vec b}2\)</p> <p>or, \(\vec {MN}\) = \(\frac {\vec c - \vec a}2\)</p> <p>&there4; 2\(\vec {MN}\) = \(\vec c\) - \(\vec a\)........................................(1)</p> <p>\(\vec {AC}\) = \(\vec {OC}\) - \(\vec {OA}\) = \(\vec c\) - \(\vec a\).........................(2)</p> <p>From (1) and (2)</p> <p>\(\vec {AC}\) = 2\(\vec {MN}\) <sub>Hence, Proved</sub></p>

Q61:

 

sdf

In the given figure, AB is the diameter of circle and O its centre. C is any point on the circumference of the circle. Prove by vector method that the \(\angle\)ACB = a right angle.


Type: Long Difficulty: Easy

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Answer: <p>Let: O be the centre of a circle and AB be the diameter of the circle.</p> <p>Let: C be any point in the semi-circle.</p> <p>Let: \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OC}\) be the position vectors of A, B and C respectively with respect to the origin O.</p> <p>Using triangle law of vector addition,</p> <p>\(\vec {AC}\) = \(\vec {OC}\) - \(\vec {OA}\) and</p> <p>\(\vec {BC}\) = \(\vec {OC}\) - \(\vec {OB}\)</p> <p>&there4; \(\vec {AC}\) . \(\vec {BC}\)</p> <p>= (\(\vec {OC}\) - \(\vec {OA}\)) (\(\vec {OC}\) - \(\vec {OB}\))</p> <p>= \(\vec {OC}\) . \(\vec {OC}\) - \(\vec {OC}\) . \(\vec {OB}\) - \(\vec {OA}\) . \(\vec {OC}\)+ \(\vec {OA}\) . \(\vec {OB}\) [\(\because\) \(\vec {OA}\) = - \(\vec {OB}\)]</p> <p>= \(\vec {OC}^2\) + \(\vec {OC}\) . \(\vec {OA}\) - \(\vec {OC}\) . \(\vec {OA}\) - \(\vec {OB}^2\)</p> <p>= \(\vec {OC}^2\) - \(\vec {OB}^2\)</p> <p>= 0 [\(\because\) \(\begin {vmatrix} OC\\ \end {vmatrix}\) = \(\begin {vmatrix} OB\\ \end {vmatrix}\)]</p> <p>&there4; \(\vec {AC}\)&perp; \(\vec {BC}\)</p> <p>Hence, \(\angle\)ACB = 90&deg; <sub>Proved</sub></p>

Q62:

Prove by vector method that the diagonals of a parallelogram bisect each other.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 200px;"><img src="/uploads/309.jpg" alt="dasf" width="200" height="158"><figcaption><br></figcaption></figure><p>Let: ABCD be a parallelogram and M be the mid-point of the diagonal DB. Let: \(\vec {OA}\), \(\vec {OB}\), \(\vec {OC}\) and \(\vec {OD}\) be the position vectors of the points A, B, C and D of the parallelogram with reference to origin O.</p> <p>Using mid-point theorem;</p> <p>We have,</p> <p>\(\vec {OM}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OD}\))</p> <p>or, \(\vec {OM}\) = \(\frac 12\)(\(\vec {OB}\) + \(\vec {OA}\) + \(\vec {AD}\)) [\(\because\) \(\vec {OD}\) = \(\vec {OA}\) + \(\vec {AD}\)]</p> <p>or, \(\vec {OM}\) = \(\frac 12\) (\(\vec {OA}\) + \(\vec {OB}\) + \(\vec {BC}\)) [\(\because\) \(\vec {AD}\) = \(\vec {BC}\)]</p> <p>&there4; \(\vec {OM}\) = \(\frac 12\) (\(\vec {OA}\) + \(\vec {OC}\) which is the position vector of the mid-point of \(\vec {AC}\)</p> <p>&there4; M is the mid-point of AC and BC.</p> <p>&there4; AC and BD bisect each other. <sub>Proved</sub></p>

Q63:

If the position vector of the vertices A, B and C of a \(\triangle\)ABC are: \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OC}\) respectively, prove that the position vector of the centroid G of the triangle ABC is:

\(\vec g\) = \(\frac 13\)(\(\vec a\) + \(\vec b\) + \(\vec c\))


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 180px;"><img src="/uploads/329.jpg" alt="s" width="202" height="202"><figcaption><br></figcaption></figure><p>Let: \(\vec {OA}\) = \(\vec a\) , \(\vec {OB}\) = \(\vec b\), \(\vec {OC}\) = \(\vec c\) anf \(\vec {OG}\) = \(\vec g\)</p> <p>The medians AE, BF and CD intersect at G. G is a centroid of the \(\triangle\)ABC.</p> <p>Using mid-point formula,</p> <p>In \(\triangle\)BOC,</p> <p>\(\vec {OE}\) = \(\frac {\vec {OB} + \vec {OC}}2\).......................(1)</p> <p>In any triangle, the centroid divides median in the ratio of 2 : 1.</p> <p>In \(\triangle\)AOE,</p> <p>\(\vec {OG}\) = \(\frac {1 . \vec {OA} + 2 . \vec {OE}}{1 + 2}\) [\(\because\) m : n = 2 : 1]</p> <p>or, \(\vec g\) = \(\frac {\vec a + 2 . (\frac {\vec {OB} + \vec {OC}}2)}3\) [\(\because\) From eq<sup>n</sup> (1)]</p> <p>or, \(\vec g\) = \(\frac {\vec a + \vec b + \vec c}3\)</p> <p>&there4;\(\vec g\) = \(\frac13\)(\(\vec a\) + \(\vec b\) + \(\vec c\)) <sub>Proved</sub></p>

Q64:

Prove vectorially that line joining vertex and mid-points of the base of an isosceles triangle is perpendicular to the base.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 170px;"><img src="/uploads/336.jpg" alt="sdf" width="170" height="170"><figcaption><br></figcaption></figure><p>Let: PQR is an isosceles triangle where PQ = PR. M is the mid-ppointof QR.</p> <p>Vertex P and M are joined.</p> <p>To prove: PM&perp; QR</p> <p>Using triangle law of vector addition:</p> <p>\(\vec {QR}\) = \(\vec {QP}\) + \(\vec {PR}\)</p> <p>\(\vec {QR}\) = - \(\vec {PQ}\) + \(\vec {PR}\) [\(\because\) \(\vec {QP}\) = - \(\vec {PQ}\)]</p> <p>\(\vec {PM}\) = \(\frac 12\) (\(\vec {PQ}\) + \(\vec {PR})\) [\(\because\) M is the mid-point of QR]</p> <p>Now,</p> <p>\(\vec {PM}\) . \(\vec {QR}\)</p> <p>= \(\frac 12\) (\(\vec {PQ}\) + \(\vec {PR}\)) (-\(\vec {PQ}\) + \(\vec {PR}\))</p> <p>= \(\frac 12\) (\(\vec {PR}^2\) - \(\vec {PQ}^2\))</p> <p>= \(\frac 12\) (PR<sup>2</sup> - PQ<sup>2</sup>)</p> <p>= \(\frac 12\) (PQ<sup>2</sup> - PQ<sup>2</sup>) [\(\because\) PR = PQ]</p> <p>= \(\frac 12\)&times; 0</p> <p>= 0</p> <p>Hence, PM&perp; QR.<sub>Proved</sub></p>

Q65:

Prove by vector method that the perpendicular line is drawn from the vertex to the base of an isosceles triangle is bisect of the base.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 170px;"><img src="/uploads/344.jpg" alt="sdf" width="170" height="170"><figcaption><br></figcaption></figure><p>Let: PQR is an isosceles triangle.</p> <p>PM&perp; QR and PQ = PR</p> <p>To prove: QM = MR</p> <p>Using triangle law of vector addition,</p> <p>In \(\triangle\)PQM,</p> <p>\(\vec {PM}\) = \(\vec {PQ}\) + \(\vec {QM}\)........................(1)</p> <p>In \(\triangle\)PRM,</p> <p>\(\vec {PM}\) = \(\vec {PR}\) + \(\vec {RM}\)..........................(2)</p> <p>From relation (1) and (2)</p> <p>\(\vec {PQ}\) + \(\vec {QM}\) = \(\vec {PR}\) + \(\vec {RM}\)</p> <p>or, \(\vec {PQ}\) + \(\vec {QM}\) = \(\vec {PQ}\) + \(\vec {RM}\) [\(\because\) \(\vec {PQ}\) = \(\vec {PR}\)]</p> <p>&there4;\(\vec {QM}\) =\(\vec {RM}\)</p> <p>Hence, PM bisect the line QR. <sub>Proved</sub></p>

Q66:

Prove by vector method that medians of an isosceles triangle drawn from extremes of the base are equal.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 181px;"><img src="/uploads/355.jpg" alt="wefr" width="181" height="181"><figcaption><br></figcaption></figure><p>Let: ABC is an isosceles triangle. CD and BE are medians. AB = AC</p> <p>To prove: BE = CD</p> <p>Let: \(\vec {BD}\) = \(\vec {DA}\) = \(\vec p\) and \(\vec {CE}\) = \(\vec {EA}\) = \(\vec q\)</p> <p>Using triangle law of vector addition,</p> <p>In \(\triangle\)BCE,</p> <p>\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\)</p> <p>\(\vec {BE}\) = \(\vec p\) + \(\vec p\) - \(\vec {EA}\) = 2\(\vec p\) - \(\vec q\).......................(1)</p> <p>Squaring on both sides,</p> <p>\(\vec {BE}^2\) = (2\(\vec p\) - \(\vec q\))<sup>2</sup></p> <p>\(\vec {BE}^2\) = 4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec q^2\)</p> <p>\(\vec {BE}^2\) =4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\) [\(\because\) \(\vec p\) = \(\vec q\)]</p> <p>\(\vec {BE}^2\) = 5\(\vec p^2\) - 4\(\vec p\)\(\vec q\)</p> <p>In \(\triangle\)BCD,</p> <p>\(\vec {CD}\) = \(\vec {CA}\) + \(\vec {AD}\) = \(\vec q\) + \(\vec q\) - \(\vec {DA}\) = 2\(\vec q\) - \(\vec p\)...................................(2)</p> <p>Squaring on both sides,</p> <p>\(\vec {CD}^2\) = (2\(\vec q\) - \(\vec p\))<sup>2</sup></p> <p>\(\vec {CD}^2\) = 4\(\vec q^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\)</p> <p>\(\vec {CD}^2\) = 4\(\vec p^2\) - 4\(\vec p\)\(\vec q\) + \(\vec p^2\) [\(\because\) \(\vec p\) = \(\vec q\)]</p> <p>\(\vec {CD}^2\) = 5\(\vec p^2\) - 4\(\vec p\)\(\vec q\)</p> <p>From above relation,</p> <p>\(\vec {BE}^2\) = \(\vec {CD}^2\)</p> <p>or, \(\vec {BE}\) = \(\vec {CD}\)</p> <p>&there4; BE = CD <sub>Proved</sub></p>

Q67:

Prove by vector method that the triangle having two equal medians is an isosceles triangle.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 181px;"><img src="/uploads/365.jpg" alt="sdf" width="181" height="181"><figcaption><br></figcaption></figure><p>Given: BE and CD are median of the \(\triangle\)ABC.</p> <p>Let: \(\vec {BD}\) = \(\vec {DA}\) = \(\vec a\) , \(\vec {AE}\) = \(\vec {EC}\) = \(\vec b\)</p> <p>To prove: AB = AC</p> <p>In \(\triangle\)BAE,</p> <p>\(\vec {BE}\) = \(\vec {BA}\) + \(\vec {AE}\) = 2\(\vec a\) + \(\vec b\)</p> <p>In \(\triangle\)ACD,</p> <p>\(\vec {DC}\) = \(\vec {DA}\) + \(\vec {AC}\) = \(\vec a\) + 2\(\vec b\)</p> <p>From question,</p> <p>\(\vec {BE}\) = \(\vec {DC}\)</p> <p>or, \(\vec {BE}^2\) = \(\vec {DC}^2\)</p> <p>or, (2\(\vec a\) + \(\vec b\))<sup>2</sup> = (\(\vec a\) + 2\(\vec b\))<sup>2</sup></p> <p>or, 4\(\vec a^2\) + 2\(\vec a\)\(\vec b\) + \(\vec b^2\) = \(\vec a^2\) + 4\(\vec a\)\(\vec b^2\) + 4\(\vec b^2\)</p> <p>or, 4\(\vec a^2\) - \(\vec a^2\) + 4\(\vec a\)\(\vec b\) - 4\(\vec a\)\(\vec b\) = 4\(\vec b^2\) - \(\vec b^2\)</p> <p>or, 3\(\vec a^2\) = 3\(\vec b^2\)</p> <p>or, \(\begin {vmatrix} \vec a^2\\ \end {vmatrix}\) =\(\begin {vmatrix} \vec b^2\\ \end {vmatrix}\)</p> <p>or,\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) =\(\begin {vmatrix} \vec b\\ \end {vmatrix}\)</p> <p>or, 2\(\begin {vmatrix} \vec a\\ \end {vmatrix}\) = 2\(\begin {vmatrix} \vec b\\ \end {vmatrix}\)</p> <p>&there4;AB = AC</p> <p>Hence, ABC is an isosceles triangle. <sub>Proved</sub></p>

Q68:

If a line is drawn from the centre of the circle to the mid-point of a chord prove by vector method that the line is perpendicular to the chord.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 181px;"><img src="/uploads/373.jpg" alt="dfc" width="181" height="181"><figcaption><br></figcaption></figure><p>Let: O be the centre of the circle.</p> <p>Join OA and OB.</p> <p>Using triangle law of vector addition,</p> <p>In \(\triangle\)AOC,</p> <p>\(\vec {OA}\) = \(\vec {OC}\) + \(\vec {CA}\)</p> <p>Squaring on both sides,</p> <p>\(\vec {OA}^2\) = (\(\vec {OC}\) + \(\vec {CA}\))<sup>2</sup> ................................(1)</p> <p>In \(\triangle\)BOC,</p> <p>\(\vec {OB}\) = \(\vec {OC}\) + \(\vec {CB}\)</p> <p>\(\vec {OB}\) = \(\vec {OC}\) - \(\vec {CA}\) [\(\because\) \(\vec {CB}\) = -\(\vec {CA}\)]</p> <p>Squaring on both sides,</p> <p>\(\vec {OB}^2\) = (\(\vec {OC}\) - \(\vec {CA}\))<sup>2</sup> ...............................(2)</p> <p>Subtracting eq<sup>n</sup> (2) from (1)</p> <p>\(\vec {OA}^2\) - \(\vec {OB}^2\) = (\(\vec {OC}\) + \(\vec {CA}\))<sup>2</sup>-(\(\vec {OC}\) - \(\vec {CA}\))<sup>2</sup></p> <p>or, \(\vec {OA}^2\) - \(\vec {OA}^2\) = \(\vec {OC}^2\) + \(\vec {CA}^2\) + 2\(\vec {OC}\) \(\vec {CA}\) - \(\vec {OC}^2\) - \(\vec {CA}^2\) + 2\(\vec {OC}\)\(\vec {CA}\)</p> <p>or, 0 = 4\(\vec {OC}\)\(\vec {CA}\)</p> <p>or, \(\vec {OC}\)\(\vec {CA}\) = 0</p> <p>&there4; \(\vec {OC}\)&perp; \(\vec {CA}\)</p> <p>Hence, OC is perpendicular to AB. <sub>Proved</sub></p>

Q69:

 

A

PQR is a triangle in which \(\angle\)Q = 90°, prove vectorially that:

\(\vec {PR}^2\) = \(\vec {PQ}^2\) + \(\vec {QR}^2\)


Type: Long Difficulty: Easy

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Answer: <p>Let, PQR is a right angled triangle.</p> <p>To prove:PR<sup>2</sup> = PQ<sup>2</sup> + QR<sup>2</sup></p> <p>Using triangle law of vector addition:</p> <p>\(\vec {PR}\) = \(\vec {PQ}\) + \(\vec {QR}\)</p> <p>Squaring on both sides;</p> <p>\(\vec {PR}^2\) = (\(\vec {PQ}\) + \(\vec {QR}\))<sup>2</sup></p> <p>or, \(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 . \(\vec {PQ}\) . \(\vec {QR}\) + \(\vec {QR}^2\)</p> <p>or, \(\vec {PR}^2\) = \(\vec {PQ}^2\) + 2 &times; 0 + \(\vec {QR}^2\) [\(\because\) \(\vec {PQ}\) . \(\vec {QR}\) = 0]</p> <p>&there4; \(\vec {PR}^2\) = \(\vec {PQ}^2\) +\(\vec {QR}^2\) <sub>Proved</sub></p>

Q70:

Derive \(\frac xa\) + \(\frac yb\) = 1 vectorially where the symbols have their usual meaning.


Type: Long Difficulty: Easy

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Answer: <p></p> <figure class="inline-right" style="width: 295px;"><img src="/uploads/391.jpg" alt="sad" width="295" height="200"><figcaption><br></figcaption></figure><p>Let, AB be a straight line cuts x-axis at A(a, 0) and y-axis at B(0. b).</p> <p>Let: P(x, y) be any point on the line AB.</p> <p>\(\vec {OA}\) = \(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\)</p> <p>\(\vec {OB}\) = \(\begin {pmatrix} 0\\ b\\ \end {pmatrix}\)</p> <p>\(\vec {OP}\) = \(\begin {pmatrix} x\\ y\\ \end {pmatrix}\)</p> <p>\(\vec {AB}\) = \(\vec {OB}\) - \(\vec {OA}\) =\(\begin {pmatrix} 0\\ b\\ \end {pmatrix}\) -\(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\) =\(\begin {pmatrix} -a\\ b\\ \end {pmatrix}\)</p> <p>\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) =\(\begin {pmatrix} x\\ y\\ \end {pmatrix}\) -\(\begin {pmatrix} a\\ 0\\ \end {pmatrix}\) =\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\)</p> <p>\(\vec {AB}\) and \(\vec {AP}\) are parallel so,</p> <p>\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\) = k\(\begin {pmatrix} -a\\ b\\ \end {pmatrix}\)</p> <p>\(\begin {pmatrix} x-a\\ y\\ \end {pmatrix}\) =\(\begin {pmatrix} -ka\\ kb\\ \end {pmatrix}\)</p> <p>Taking corresponding elements of the equal vectors,</p> <p>x - a = -ka</p> <p>\(\frac xa\) = 1 - k .....................(1)</p> <p>y = bk</p> <p>\(\frac yb\) = k ............................(2)</p> <p>Adding relation (1) and (2)</p> <p>\(\frac xa\) +\(\frac yb\) = 1 - k + k</p> <p>&there4;\(\frac xa\) +\(\frac yb\) = 1<sub> Hence, Proved</sub></p>

Q71:

 

f

Let, P be the point on the line determined by the point A and B such that AP : PB = m : n, then \(\vec {OP}\) = \(\frac {m.\vec {OB} + n.\vec {OA}}{m+n}\), the position vector of P in the relation to the origin O.


Type: Long Difficulty: Easy

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Answer: <p>Proof:</p> <p>Let, \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OP}\) be the position vectors of the points A, B and P relative to the origin O respectively.</p> <p>From Question,</p> <p>\(\frac {AP}{PB}\) = \(\frac mn\)</p> <p>AP = \(\frac mn\)PB..........................(1)</p> <p>Using vector of the above relation,</p> <p>\(\vec {AP}\) = \(\frac mn\)\(\vec {PB}\)</p> <p>\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) and</p> <p>\(\vec {PB}\) = \(\vec {OB}\) - \(\vec {OP}\)</p> <p>Putting the value of \(\vec {AP}\) and \(\vec {PB}\) in relation (1)</p> <p>\(\vec {OP}\) - \(\vec {OA}\) = \(\frac mn\)(\(\vec {OB}\) - \(\vec {OP})\)</p> <p>or, n \(\vec {OP}\) - n \(\vec {OA}\) = m \(\vec {OB}\) - m \(\vec {OP}\)</p> <p>or,m \(\vec {OP}\) +n \(\vec {OP}\) = m \(\vec {OB}\) +n \(\vec {OA}\)</p> <p>or, \(\vec {OP}\) (m + n) =m \(\vec {OB}\) +n \(\vec {OA}\)</p> <p>&there4; \(\vec {OP}\) = \(\frac {m.\vec {OB} + n.\vec {OA}}{m+n}\) <sub>Hence, Proved</sub></p>

Q72:

 

sc

Let, P be the external point on the line determined by the points A and B such that AP : PB = m : n such that: \(\vec {OP}\) = \(\frac {m. \vec {OB} - n. \vec {OA}}{m - n}\) the position vector of P in relation to the origin O.


Type: Long Difficulty: Easy

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Answer: <p>Proof:</p> <p>Let, \(\vec {OA}\), \(\vec {OB}\) and \(\vec {OP}\) be the position vectors of the points A, B and P respectively in relation to the origin O.</p> <p>From Question,</p> <p>\(\frac {AP}{PB}\) = \(\frac mn\)</p> <p>n. AP = m. PB</p> <p>Using vector of the above relation,</p> <p>n \(\vec {AP}\) = m \(\vec {PB}\)...........................(1)</p> <p>From triangle law of vector addition:</p> <p>\(\vec {AP}\) = \(\vec {OP}\) - \(\vec {OA}\) and</p> <p>\(\vec {BP}\) = \(\vec {OP}\) - \(\vec {OB}\)</p> <p>Putting the value of \(\vec {AP}\) and \(\vec {BP}\) in relation (1)</p> <p>n (\(\vec {OP}\) - \(\vec {OA}\)) = m (\(\vec {OP}\) - \(\vec {OB}\))</p> <p>or, n . \(\vec {OP}\) - n . \(\vec {OA}\) = m . \(\vec {OP}\) - m . \(\vec {OB}\)</p> <p>or,n . \(\vec {OP}\) -m . \(\vec {OP}\) =n . \(\vec {OA}\)- m . \(\vec {OB}\)</p> <p>or,\(\vec {OP}\) (n - m) =n . \(\vec {OA}\)- m . \(\vec {OB}\)</p> <p>or,\(\vec {OP}\) = \(\frac {n . \vec {OA} - m . \vec {OB}}{n - m}\)</p> <p>&there4;\(\vec {OP}\) = \(\frac {m. \vec {OB} - n. \vec {OA}}{m - n}\) <sub>Hence, Proved</sub></p>

Q73:

 

D

OABC is a parallelogram, P and Q divide OC and CB, \(\vec {CP}\) : \(\vec {OP}\) = \(\vec {CQ}\) : \(\vec {QB}\) = 1 : 3. If \(\vec {OA}\) = \(\vec a\) and \(\vec {OC}\) = \(\vec c\), express PQ in terms of a and c and prove that: PQ and OB are parallel.


Type: Long Difficulty: Easy

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Answer: <p>Given:</p> <p>OABC is a parallelogram.</p> <p>\(\vec {CP}\) : \(\vec {OP}\) = \(\vec {CQ}\) : \(\vec {QB}\) = 1 : 3</p> <p>\(\vec {OA}\) = \(\vec a\) and \(\vec {OC}\) = \(\vec c\)</p> <p>To prove: \(\vec {PQ}\)//\(\vec {OB}\) and find \(\vec {PQ}\) = ?</p> <p>From Question:</p> <p>\(\frac {\vec {CP}}{\vec {PO}}\) = \(\frac 13\) = \(\frac {\vec {CQ}}{\vec {QB}}\)</p> <p>&there4; \(\vec {PO}\) = 3 \(\vec {CP}\) and \(\vec {QB}\) = 3 \(\vec {CQ}\)</p> <p>Using triangle law of vector addition:</p> <p>\(\vec {OB}\) = \(\vec {OA}\) + \(\vec {AB}\)</p> <p>or, \(\vec {OB}\)= \(\vec a\) + \(\vec c\)</p> <p>or, \(\vec {OB}\) = \(\vec {OC}\) + \(\vec {CB}\)</p> <p>or, \(\vec {OB}\) = \(\vec {OP}\) + \(\vec {PC}\) + \(\vec {CQ}\) + \(\vec {QB}\)</p> <p>or, \(\vec {OB}\) = - \(\vec {PO}\) + \(\vec {PC}\) + \(\vec {CQ}\) + 3\(\vec {CQ}\)</p> <p>or, \(\vec {OB}\) = -3 \(\vec {CP}\) - \(\vec {CP}\) + 4 \(\vec {CQ}\) [\(\because\) \(\vec {PO}\) = 3 \(\vec {CP}\)]</p> <p>or, \(\vec {OB}\) = - 4 \(\vec {CP}\) + 4 \(\vec {CQ}\)</p> <p>or, \(\vec {OB}\) = 4 \(\vec {PC}\) + 4 \(\vec {CQ}\)</p> <p>or,\(\vec {OB}\) = 4 (\(\vec {PC}\) + \(\vec {CQ}\))</p> <p>&there4;\(\vec {OB}\) = 4 \(\vec {PQ}\)</p> <p>&there4;\(\vec {OB}\)//\(\vec {PQ}\) [\(\vec a\) = m\(\vec b\) then \(\vec a\)//\(\vec b\)]</p> <p>and \(\vec {PQ}\) = \(\frac 14\)\(\vec {OB}\) = \(\frac 14\)(\(\vec a\) + \(\vec c\)) <sub>Proved</sub></p>

Q74:

 

SAD

In the figure, ABCD is a parallelogram, L and M are mid-points of sides BC and CD respectively, prove that:

\(\vec {AL}\) + \(\vec {AM}\) = \(\frac 32\)\(\vec {AC}\)


Type: Long Difficulty: Easy

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Answer: <p>Given:</p> <p>ABCD is a parallelogram. L and M are mid-points of sides BC and CD respectively.</p> <p>Using triangle law of vector addition:</p> <p>\(\vec {AM}\) = \(\vec {AD}\) + \(\vec {DM}\) = \(\vec {AD}\) + \(\frac 12\)\(\vec {DC}\)...............(1)</p> <p>\(\vec {AL}\) = \(\vec {AB}\) + \(\vec {BL}\) = \(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\).....................(2)</p> <p>Adding (1) and (2)</p> <p>\(\vec {AM}\) +\(\vec {AL}\) =\(\vec {AD}\) + \(\frac 12\)\(\vec {DC}\) +\(\vec {AB}\) + \(\frac 12\)\(\vec {BC}\)</p> <p>or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {2 \vec {AD} + \vec {DC} + 2 \vec {AB} + \vec {BC}}2\)</p> <p>or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {2 \vec {BC} + \vec {AB} + 2 \vec {AB} + \vec {BC}}2\) [\(\because\) \(\vec {AB}\) = \(\vec {DC}\) and \(\vec {AD}\) = \(\vec {BC}\)]</p> <p>or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac {3 \vec {AB} + 3 \vec {BC}}2\)</p> <p>or,\(\vec {AM}\) +\(\vec {AL}\) = \(\frac 32\) (\(\vec {AB}\) + \(\vec {BC}\))</p> <p>&there4; \(\vec {AM}\) +\(\vec {AL}\) = \(\frac 32\)\(\vec {AC}\)</p> <p>Hence, \(\vec {AL}\) +\(\vec {AM}\) = \(\frac 32\)\(\vec {AC}\) <sub>Proved</sub></p>

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Storage Devices

Storage Devices

External memory where data and instructions are stored in terms of magnetic spot using magnetic coated area is magnetic storage. Basically we have two types of magnetic storage these are magnetic tape and magnetic disk.

Magnetic Tape

It is a plastic ribbon, usually 0.5 inch or 0.25 inch wide and 50 to 2500 feet long, with magnetic coating (such as iron oxide) at one side. The tape ribbon is again covered inside reels or on a small cassette. Magnetic tape is the most popular storage medium for long time storage medium for long time storage. The tapes can also be erased and reused several times. Old data on the tape are automatically erased and overwrite as new data are recorded in the same area, just like in audio or video tape.

Magnetic Disk

Magnetic disk is most popular storage medium for direct access backup storage. It is thin, circular metal plate or flat rotating disc coated with magnetic material on both sides. Like tape, disks can also be erased and reused several times. The surface of disk is divided into a number of concentric circles, called tracks. Each of the track is further divided into sectors. Each sector typically contains 512 bytes and all are assigned with unique number called sector number. Popular disks are: hard drive and floppy disk.

Floppy disks:

These are also called diskettes. These are very thin and flexible, so called floppies. It is a thin, round, flexible plastic, coated with metallic oxide, encased in a square plastic or vinyl jacket cover protection. It comes in various sizes: 3.5 inches (Micro-floppy) and 5.25 inches (Mini-floppy) in a diameter. The advantage of diskette is that it is highly portable and very cheap as compared to other secondary storage devices. But now a day people are not using this storage frequently. The capacity of floppy disk varies from 1.44 MB to 2.0 MB.

Hard Disk:

These are most popular secondary storage devices. These are made up with set of rigid metal diskettes (mostly aluminum) which is permanently sealed in metal case for air-tight. Nowadays, removable hard disks have been introduced. We can use hard disks externally. The capacity of hard disk varies from some GB to 350 GB.

Optical storage:

Secondary storage where data are stored in terms of optical spot in optical disk. Due to the use of laser beam technology for recording/reading of data on the disk, these devices are called optical storage devices. This consists of rotating disk coated with thin metal or some other material that is highly reflective. The information stored in the disk can be accessed detecting the light. Optical disks have one long rack, starting at the outer edge and growth inwards to the center. The spiral track is best for reading large blocks of sequential data such as music.

Some useful technical terms

  1. The integration of input, processing & output unit. (Computer System)
  2. Board where all the vital components of computer are connected. (Mother board)
  3. A device that gives soft output. (monitor)
  4. A storage media that stores data for future uses. (Secondary storage)
  5. A device which converts analog signal to digital & vice-versa. (Modem)
  6. A device which can enter text, graphics etc. directly on the computer as input. (Scanner)
  7. The volatile memory of computer. (RAM)
  8. The processor built on a single chip. (Microprocessor)
  9. A device which allows providing data to the computer. (Input device)
  10. The electronic pathway on motherboard. (Bus)
  11. The device which is used to produce audio output. (Speaker)
  12. The device which supplies regular power to the computer even in black out. (UPS)
  13. The high speed memory kept in between CPU and main memory. (Cache memory)
  14. The par which make computer system. (Hardware)
  15. The section of CPU which performs mathematical calculations. (ALU)
  16. The output on printed form. (Hard copy output)
  17. The software stored in ROM chip. (Firmware)
  18. A device that stores electrical charges. (Capacitor)
  19. A program that makes devices of a computer functional. (Device driver)
  20. An interface or connection point where hardware devices are joined. (Port)
  21. The process of converting analog signal to digital form. (Digitizing)

Lesson

Computer Hardware

Subject

Computer

Grade

Grade 9

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