Magnitude and Direction of Vector

Magnitude of a vector

The magnitude of the vector is a positive number which represents the length of the vector or directed line segment. It is also known as modulus of the vector. If \(\overrightarrow {OP}\) is the position vector of P, then its magnitude is denoted by \(|\overrightarrow {OP}|\).If \(|\overrightarrow {AB}|\)is a vector, then magnitude is denoted by \(|\overrightarrow {AB}|\). If the vector is written in the single letter from \(\overrightarrow {a}\), then its magnitude is denoted by \(\overrightarrow {a}\).

If the initial point is at the origin and the terminal point at P (x,y).

\(\overrightarrow {OP}\) = \(\frac{x}{y}\) is given by \(|\overrightarrow {OP}|\) = \(\sqrt{(x)^2+(y)^2}\)

If A (x₁,y₁) and B(x₂,y₂) are two points,

We know, \(\overrightarrow {AB}\) =\(\begin{pmatrix} x_2-x_1\\y_2-y_1\end{pmatrix}\). Then \(|\overrightarrow {AB}|\) = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Thus\(|\overrightarrow {AB}|\) = \(\sqrt{(x-component)^2+ (y-component)^2}\)

If the initial point is at A(x₁,y₁) and terminal point at B(x_2,y₂).

Here, For \(\overrightarrow{AB}\)

x component of \(\overrightarrow{AB}\)= AE = x₂ - x₁

y component of \(\overrightarrow{AB}\) = BE = y₂ - y₁

Using pythagoras theorem,

(AB)² = (AE)² + (BE)²

or, AB² = (x₂ - x₁)² + (y₂ - y₁)²

or, AB = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

or, AB = \(\sqrt{(x\;component)^2 + (y\;component)^2}\)

\(\therefore\) \(|\overrightarrow {AB}|\) = AB =\(\sqrt{(x\; comp)^2 + (y\; comp)^2}\)

Direction of Vector

Direction of a vector is the value of angle made by the vector with x-axis in positive direction. It is generally measured in degree which ranges from 0º to 360º. Direction of a vector is denoted by θ. In the figure, \(\overrightarrow {OP}\) makes an angle θwith x-axis in anticlockwise direction. Draw PM perpendicular to x-axis.

When the initial point is at the origin and terminal point at P(x,y).

Then, OM = x and MP =y.

In ΔPOM, tan θ = \(\frac{Perpendicular}{base}\)

or, tan θ =\(\frac{MP}{OM}\)

or, tan θ = \(\frac{y}{x}\)

∴ θ = tan^-1 \((\frac{y}{x}\))

Which gives the direction of the vector.

Hence, direction of vector (θ) =tan^-1 \((\frac{y-component}{x-component}\))

When the initial point is at A(x₁,y₁) and the terminal point is at B(x₂,y₂)

If A(x₁,y₁) and B(x₂,y₂) are two points, then direction of \(\overrightarrow {AB}\) is given by θ = tan^-1 \(\begin{pmatrix}y_2-y_1\\x_2-x_1\\\end{pmatrix}\).