# How to Find Vector Components?

Components of a vector help to split a given vector into parts according to different directions. Let us talk more about the components of a vector and how to find the components of a vector.

The components of a vector give a split of the vector. The vector is split according to each of the axes and we can calculate the components of a vector.

## A step-by-step guide to vector components

• Any vector in a two-dimensional coordinate system can be broken down into its $$x$$ and $$y$$-components.

$$v⃗ =(v_x,v_y)$$

For example, in the picture given below, the vector $$v⃗$$ is divided into two components, $$v_x$$ and $$v_y$$. Let the angle between the vector and its $$x$$-component be $$θ$$.

In the diagram below, the vector and its component form a right-angled triangle:

In the above figure, the components can be easily and quickly read. The vector in the component form is $$v⃗ =(4,5)$$.

• Trigonometric ratios show the relationship between the magnitude of the vector and the components of the vector.

$$\color{blue}{cos θ=\frac{v_x}{v}}$$ → $$\color{blue}{v_x= v \cos θ}$$

$$\color{blue}{sin θ=\frac{v_y}{v}}$$ → $$\color{blue}{v_y= v \sin θ}$$

• Using “Pythagoras theorem” in right triangles with lengths $$v_x$$ and $$v_y$$:

$$\color{blue}{|v|=\sqrt{v_x^2+v_y^2}}$$

Note1:

Find the magnitude and direction of the vector with respect to the components of a vector. In this case, use the following formulas:

The magnitude of the vector is $$\color{blue}{|v|=\sqrt{v_x^2+v_y^2}}$$.

To find the direction of the vector, solve $$\color{blue}{tan θ=\frac{v_y}{v_x}}$$ for $$θ$$.

Note 2:

Find the components of a vector according to the magnitude and direction of a vector. In this case, use the following formulas:

$$\color{blue}{v_x= v \cos θ}$$

$$\color{blue}{v_y= v \sin θ}$$

### Vector Components – Example 1:

The magnitude of a vector $$v⃗$$ is $$20$$ units and the direction of the vector is $$60°$$ with the horizontal. Find the components of the vector.

To find the components of a vector use these formulas:

$$\color{blue}{v_x= v \cos θ}$$

$$\color{blue}{v_y= v \sin θ}$$

$$v_x= v\cos 60°$$ → $$v_x= 20×\frac{1}{2}= \frac{20}{2}=10$$

$$v_y= v\sin 60°$$ → $$v_y= 20×\frac{\sqrt{3}}{2}= \frac{20\sqrt{3}}{2} =10\sqrt{3}$$

So, the vector $$v⃗$$ is $$(10, 10\sqrt{3})$$.

### Vector Components – Example 2:

Find the $$x$$ and $$y$$ components of a vector having a magnitude of $$10$$ and make an angle of $$45$$ degrees with the positive $$x$$-axis.

To find the components of a vector use these formulas:

$$\color{blue}{v_x= v \cos θ}$$

$$\color{blue}{v_y= v \sin θ}$$

$$v_x= v\cos 45°$$ → $$v_x= 10×\frac{\sqrt{2}}{2}= \frac{10\sqrt{2}}{2} =5\sqrt{2}$$

$$v_y= v\sin 45°$$ → $$v_y= 10×\frac{\sqrt{2}}{2}= \frac{10\sqrt{2}}{2} =5\sqrt{2}$$

So, the $$x$$-component and the $$y$$-components of the vector are both equal to $$5\sqrt{2}$$.

## Exercises for Vector Components

1. Find the value of $$θ$$, if $$v_x=15$$ and $$v_y=8.66$$.
2. Find out the magnitude of a vector $$OA=(-3,4)$$.
3. Find the components of the vector, if the magnitude of a vector $$v⃗$$ is $$6$$ units and the direction of the vector is $$30°$$ with the horizontal.
4. Find the direction of $$(-4,3)$$.
1. $$\color{blue}{θ=30^\circ}$$
2. $$\color{blue}{|OA|=5}$$
3. $$\color{blue}{v⃗=3, 3\sqrt{3}}$$
4. $$\color{blue}{θ=143.13^\circ}$$

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