Resolution of Vectors - Any Learn Nepal

Table Of Contents

Resolution of vetors
Expressions for components
- Magnitude of vector
- Direction of vector
FAQ’s

Resolution of vetors

The process of splitting vectors into many components is called the resolution of vectors. It is also known as a splitting of vectors. Think of it as distributing vector total effect into smaller parts.

We can resolve vectors into infinite numbers of vectors each acting in different directions. But we can only have 2 rectangular components in a plane (components in x and y direction) and 3 rectangular components in a space(components in x, y, and z-direction).

When splitting a vector into two components, those components should be mutually perpendicular to each other. Hence, we always resolve a vector into two components so that one component acts in the x-axis direction and the other component acts in the y-axis direction. And two components can be treated independently to one another.

It is a very important topic in physics as it has many applications. This concept is used to find the expressions related to projectile motion too.

The figure above shows the resolution of vector $\overrightarrow{v}$ into two rectangular components: $v_x$ along x-axis and $v_y$ along y-axis.

Expressions for components

In right-angled triangle OAB, we have:

$\begin{align*} \cos\theta = \frac{b}{h} &= \frac{OB}{OA} \\ \cos\theta &= \frac{v_x}{v} \qquad \text{[using magnitude]} \end{align*}$

(1) $\begin{equation*}v_x = v \cos\theta \end{equation*}$

Now,

$\begin{align*} \sin\theta = \frac{p}{h} &= \frac{AB}{OA} \\ \sin\theta &= \frac{v_y}{v} \qquad \text{[using magnitude, $AB=v_y$]} \end{align*}$

(2) $\begin{equation*}v_y = v \sin\theta \end{equation*}$

Magnitude of vector

To get magnitude of vector, squaring on both equations 1 and 2 and adding:

$\begin{gather*} {v_x}^2 + {v_y}^2 = (v \cos\theta)^2 + (v \sin\theta)^2 \\ {v_x}^2 + {v_y}^2= v^2 \cos^2\theta + v^2\sin^2\theta \\ {v_x}^2 + {v_y}^2 = v^2(\cos^2\theta + \sin^2\theta) = v^2 \\{v_x}^2 + {v_y}^2 = v^2 \end{gather*}$

(3) $\begin{equation*}v = \sqrt{{v_x}^2 + {v_y}^2 }\end{equation*}$

Direction of vector

To get direction of vector, dividing equation 2 by 1:

$\begin{gather*} \frac{v_y}{v_x} = \frac{v \sin\theta}{v \cos\theta} \\ \frac{v_y}{v_x} = \frac{\sin\theta}{\cos\theta} \\ \frac{v_y}{v_x} = \tan\theta \end{gather*}$