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Components of a Vector

In a two-dimensional coordinate system, any vector can be broken into $x$ -component and $y$ -component.

$\vec{v} = ⟨ v_{x}, v_{y} ⟩$

For example, in the figure shown below, the vector $\vec{v}$ is broken into two components, $v_{x}$ and $v_{y}$ . Let the angle between the vector and its $x$ -component be $θ$ .

Math diagram

The vector and its components form a right angled triangle as shown below.

Math diagram

In the above figure, the components can be quickly read. The vector in the component form is $\vec{v} = ⟨ 4, 5 ⟩$ .

The trigonometric ratios give the relation between magnitude of the vector and the components of the vector.

$\cos θ = \frac{Adjacent Side}{Hypotenuse} = \frac{v_{x}}{v}$

$\sin θ = \frac{Opposite Side}{Hypotenuse} = \frac{v_{y}}{v}$

$v_{x} = v \cos θ$

$v_{y} = v \sin θ$

Using the Pythagorean Theorem in the right triangle with lengths $v_{x}$ and $v_{y}$ :

$| v | = \sqrt{v_{x}^{2} + v_{y}^{2}}$

Here, the numbers shown are the magnitudes of the vectors.

Case 1: Given components of a vector, find the magnitude and direction of the vector.

Use the following formulas in this case.

Magnitude of the vector is $| v | = \sqrt{v_{x}^{2} + v_{y}^{2}}$ .

To find direction of the vector, solve $\tan θ = \frac{v_{y}}{v_{x}}$ for $θ$ .

Case 2: Given the magnitude and direction of a vector, find the components of the vector.

Use the following formulas in this case.

$v_{x} = v \cos θ$

$v_{y} = v \sin θ$

Example:

The magnitude of a vector $\vec{F}$ is $10$ units and the direction of the vector is $60 °$ with the horizontal. Find the components of the vector.

Math diagram

$\begin{array}{l} F_{x} = F \cos 60 ° \\ = 10 \cdot \frac{1}{2} \\ = 5 \end{array}$

$\begin{array}{l} F_{y} = F \sin 60 ° \\ = 10 \cdot \frac{\sqrt{3}}{2} \\ = 5 \sqrt{3} \end{array}$

So, the vector $\vec{F}$ is $⟨ 5, 5 \sqrt{3} ⟩$ .