# Vector magnitude and direction review

Review your knowledge of vector magnitude and direction, and use them to solve problems.
Magnitude of $(a,b)$
$\mid\mid\!\!\!(a,b)\!\!\mid\mid=\sqrt{a^2+b^2}$
Direction of $(a,b)$
$\theta=\tan^{-1}\left(\dfrac{b}{a}\right)$
Components from magnitude $\mid\mid\!\!\!\vec u\!\!\mid\mid$ and direction $\theta$
$\bigg(\mid\mid\!\!\!\vec u\!\!\mid\mid\cos(\theta),\mid\mid\!\!\!\vec u\!\!\mid\mid\sin(\theta)\bigg)$

## What are vector magnitude and direction?

We are used to describing vectors in component form. For example, $(3,4)$. We can plot vectors in the coordinate plane by drawing a directed line segment from the origin to the point that corresponds to the vector's components:
Considered graphically, there's another way to uniquely describe vectors — their $\blueD{\text{magnitude}}$ and $\greenD{\text{direction}}$:
The $\blueD{\text{magnitude}}$ of a vector gives the length of the line segment, while the $\greenD{\text{direction}}$ gives the angle the line forms with the positive $x$-axis.
The magnitude of vector $\vec v$ is usually written as $||\vec v||$.

## Practice set 1: Magnitude from components

To find the magnitude of a vector from its components, we take the square root of the sum of the components' squares (this is a direct result of the Pythagorean theorem):
$||(a,b)||=\sqrt{a^2+b^2}$
For example, the magnitude of $(3,4)$ is $\sqrt{3^2+4^2}=\sqrt{25}=5$.
Want to try more problems like this? Check out this exercise.

## Practice set 2: Direction from components

To find the direction of a vector from its components, we take the inverse tangent of the ratio of the components:
$\theta=\tan^{-1}\left(\dfrac{b}{a}\right)$
This results from using trigonometry in the right triangle formed by the vector and the $x$-axis.

### Example 1: Quadrant $\text{I}$

Let's find the direction of $(3,4)$:
$\tan^{-1}\left(\dfrac{4}{3}\right)\approx 53^\circ$

### Example 2: Quadrant $\text{IV}$

Let's find the direction of $(3,-4)$:
$\tan^{-1}\left(\dfrac{-4}{3}\right)\approx -53^\circ$
The calculator returned a negative angle, but it's common to use positive values for the direction of a vector, so we must add $360^\circ$:
$-53^\circ+360^\circ=307^\circ$

### Example 3: Quadrant $\text{II}$

Let's find the direction of $(-3,4)$. First, notice that $(-3,4)$ is in Quadrant $\text{II}$.
$\tan^{-1}\left(\dfrac{4}{-3}\right)\approx -53^\circ$
$-53^\circ$ is in Quadrant $\text{IV}$, not $\text{II}$. We must add $180^\circ$ to obtain the opposite angle:
$-53^\circ+180^\circ=127^\circ$
Want to try more problems like this? Check out this exercise.

## Practice set 3: Components from magnitude and direction

To find the components of a vector from its magnitude and direction, we multiply the magnitude by the sine or cosine of the angle:
$\vec u=\bigg(||\vec u||\cos(\theta),||\vec u||\sin(\theta)\bigg)$
This results from using trigonometry in the right triangle formed by the vector and the $x$-axis.
For example, this is the component form of the vector with magnitude $\blueD 2$ and angle $\greenD{30^\circ}$:
$\bigg(\blueD 2\cos(\greenD{30^\circ}),\blueD 2\sin(\greenD{30^\circ})\bigg)=(\sqrt 3,1)$
Want to try more problems like this? Check out this exercise.