A vector can be described by its components or by its length and direction. Switching between the two descriptions is the same as switching between rectangular coordinates and the unit circle, scaled up.
Key idea
For $\mathbf{v} = \langle a, b \rangle$: magnitude $|\mathbf{v}| = \sqrt{a^2 + b^2}$; direction angle $\theta$ with $\tan\theta = \frac{b}{a}$, placed in the quadrant of $(a, b)$. Going back: $$\mathbf{v} = \langle |\mathbf{v}|\cos\theta,\ |\mathbf{v}|\sin\theta \rangle.$$ The unit vector in the direction of $\mathbf{v}$ is $\frac{\mathbf{v}}{|\mathbf{v}|}$.
Why it works
The magnitude is the hypotenuse of a right triangle with legs $|a|$ and $|b|$. A vector of length $1$ at angle $\theta$ ends at the unit-circle point $(\cos\theta, \sin\theta)$; stretching it to length $|\mathbf{v}|$ multiplies both components by $|\mathbf{v}|$.
Worked examples
Worked example: Magnitude and direction
Find the magnitude and direction angle of $\langle -3, 4 \rangle$.
- $|\mathbf{v}| = \sqrt{9 + 16} = 5$.
- A calculator gives $\tan^{-1}\left(\frac{4}{-3}\right) \approx -53.1^\circ$, an angle in Quadrant IV. But $\langle -3, 4 \rangle$ points left and up, into Quadrant II.
- Reference angle $53.1^\circ$, so $\theta = 180^\circ - 53.1^\circ = 126.9^\circ$.
Answer: $5$ and about $126.9^\circ$
Worked example: From length and angle
A vector has magnitude $10$ and direction angle $150^\circ$. Find its components.
- $$\langle 10\cos 150^\circ,\ 10\sin 150^\circ \rangle = \left\langle 10 \cdot \left(-\frac{\sqrt{3}}{2}\right),\ 10 \cdot \frac{1}{2} \right\rangle.$$
Answer: $\langle -5\sqrt{3}, 5 \rangle$
Worked example: A unit vector
Find the unit vector in the direction of $\langle 5, -12 \rangle$.
- $|\mathbf{v}| = \sqrt{25 + 144} = 13$.
- Divide each component by $13$.
Answer: $\left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle$
Common mistake
Taking the direction from $\tan^{-1}\frac{b}{a}$ without checking the quadrant. $\tan^{-1}$ only gives angles between $-90^\circ$ and $90^\circ$, so for a vector pointing left (negative $a$) you must add $180^\circ$. Sketch the arrow first; the picture tells you the quadrant.