Cartesian Vectors

Drawing vectors and using the sine and cosine laws works, but it gets messy with three or more vectors. Putting vectors on a coordinate grid turns every vector into a pair of numbers, its components, and then adding, subtracting, and scaling vectors is just arithmetic. This is also the form computers, GPS units, and game engines use. All angles are in degrees.

Key ideas

Position vectors and components

Put the tail of a vector at the origin $O$. If its head lands at the point $P(x, y)$, then $\overrightarrow{OP}$ is the position vector of $P$, and we write it in Cartesian form (component form) as

$$\overrightarrow{OP} = [x, y]$$

$x$ is the horizontal component and $y$ is the vertical component. Since equal vectors can be drawn anywhere, any vector that moves $x$ units right and $y$ units up is $[x, y]$, wherever its tail is.

This site uses square brackets, as Ontario textbooks do. Some books write $(x, y)$ or $\langle x, y\rangle$ for a vector instead. Square brackets help keep the vector $[x, y]$ separate from the point $(x, y)$. IB courses write the same vector as a column, $\begin{pmatrix} x \\ y \end{pmatrix}$; it means exactly the same thing.

Magnitude

By the Pythagorean theorem, the magnitude of $\vec{v} = [x, y]$ is

$$\lvert\vec{v}\rvert = \sqrt{x^2 + y^2}$$

From magnitude and direction to components

If $\vec{v}$ has magnitude $r$ and direction angle $\theta$ (measured counterclockwise from the positive $x$-axis), then

$$x = r\cos\theta, \qquad y = r\sin\theta, \qquad \text{so} \quad \vec{v} = [r\cos\theta,\ r\sin\theta]$$

These come from the trig ratios for any angle, and the signs take care of themselves: in the figure, $\cos\theta$ is negative, so $x$ is negative. If the direction is given as a bearing, convert it first with $\theta = 90^\circ - \text{bearing}$ (taking north as the positive $y$-axis and east as the positive $x$-axis).

The components of $\overrightarrow{OP}$ are $x = r\cos\theta = -4$ and $y = r\sin\theta = 3$.

From components to magnitude and direction

Given $\vec{v} = [x, y]$, find $r = \sqrt{x^2 + y^2}$, then find $\theta$ from $\tan\theta = \dfrac{y}{x}$. Watch the quadrant: a calculator’s $\tan^{-1}$ only gives angles between $-90^\circ$ and $90^\circ$.

Where $\vec{v}$ points	Direction angle
Quadrant I ($x \gt 0$, $y \gt 0$)	$\theta = \tan^{-1}\left(\dfrac{y}{x}\right)$
Quadrant II or III ($x \lt 0$)	$\theta = \tan^{-1}\left(\dfrac{y}{x}\right) + 180^\circ$
Quadrant IV ($x \gt 0$, $y \lt 0$)	$\theta = \tan^{-1}\left(\dfrac{y}{x}\right) + 360^\circ$

A sketch is the best check. Many students prefer to find the acute reference angle $\tan^{-1}\left\lvert\dfrac{y}{x}\right\rvert$ and then place it in the right quadrant.

Operations in component form

For $\vec{u} = [u_1, u_2]$, $\vec{v} = [v_1, v_2]$ and a scalar $k$, work component by component:

$$\vec{u} + \vec{v} = [u_1 + v_1,\ u_2 + v_2], \qquad \vec{u} - \vec{v} = [u_1 - v_1,\ u_2 - v_2], \qquad k\vec{u} = [ku_1,\ ku_2]$$

Two vectors are equal exactly when their components are equal. All the properties from vector addition and scalar multiplication still hold. Two non-zero vectors are collinear when their components are proportional: $[u_1, u_2] = k[v_1, v_2]$.

The vector between two points

For points $A(x_1, y_1)$ and $B(x_2, y_2)$, $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$, so

$$\overrightarrow{AB} = [x_2 - x_1,\ y_2 - y_1]$$

It’s “head minus tail”. Its magnitude is the distance from $A$ to $B$: $\lvert\overrightarrow{AB}\rvert = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

The unit vectors i and j

The standard unit vectors are $\vec{i} = [1, 0]$ (one unit along the positive $x$-axis) and $\vec{j} = [0, 1]$ (one unit along the positive $y$-axis). Every vector is a linear combination of them:

$$[x, y] = x\vec{i} + y\vec{j}$$

For example, $[5, -2] = 5\vec{i} - 2\vec{j}$. The unit vector in the direction of $\vec{v}$ is $\dfrac{1}{\lvert\vec{v}\rvert}\vec{v}$, as before.

Worked examples

Example 1: Magnitude and direction to components

• (a) $\vec{a}$ has magnitude $12$ and direction $\theta = 150^\circ$. Write $\vec{a}$ in Cartesian form, exactly.
• (b) A car’s velocity $\vec{v}$ is $50$ km/h on a bearing of $230^\circ$. Write $\vec{v}$ in Cartesian form, with east as the positive $x$-axis and north as the positive $y$-axis, to two decimal places.

Solution.

(a) Use the special angles: $\cos 150^\circ = -\dfrac{\sqrt{3}}{2}$ and $\sin 150^\circ = \dfrac{1}{2}$.

$$\vec{a} = [12\cos 150^\circ,\ 12\sin 150^\circ] = [-6\sqrt{3},\ 6]$$

Check: $\sqrt{(-6\sqrt{3})^2 + 6^2} = \sqrt{108 + 36} = \sqrt{144} = 12$. ✓

(b) Convert the bearing: $\theta = 90^\circ - 230^\circ = -140^\circ$, or $220^\circ$.

$$\vec{v} = [50\cos 220^\circ,\ 50\sin 220^\circ] \approx [-38.30,\ -32.14]$$

Both components are negative, which makes sense: a bearing of $230^\circ$ is S $50^\circ$ W, so the car is moving south and west.

Example 2: Components to magnitude and direction

Find the magnitude and direction angle of each vector, to one decimal place.

• (a) $\vec{v} = [-5, 12]$
• (b) $\vec{w} = [3, -4]$

Solution.

(a) $\lvert\vec{v}\rvert = \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13$.

A calculator gives $\tan^{-1}\left(\dfrac{12}{-5}\right) \approx -67.4^\circ$, which points into quadrant IV. But $\vec{v}$ has $x \lt 0$ and $y \gt 0$, so it’s in quadrant II. Add $180^\circ$:

$$\theta \approx -67.4^\circ + 180^\circ = 112.6^\circ$$

(b) $\lvert\vec{w}\rvert = \sqrt{3^2 + (-4)^2} = 5$. $\tan^{-1}\left(\dfrac{-4}{3}\right) \approx -53.1^\circ$. $\vec{w}$ is in quadrant IV, so add $360^\circ$: $\theta \approx 306.9^\circ$.

Check (a): $13\cos 112.6^\circ \approx -5.0$ and $13\sin 112.6^\circ \approx 12.0$. ✓

Example 3: Vectors between points

Given $A(-2, 1)$, $B(3, 4)$ and $C(5, -2)$:

• (a) Find $\overrightarrow{AB}$, $\lvert\overrightarrow{AB}\rvert$ and $\overrightarrow{AC}$.
• (b) Find $2\overrightarrow{AB} - \overrightarrow{AC}$.
• (c) Find the unit vector in the direction of $\overrightarrow{AB}$.

$\overrightarrow{AB}$ moves $5$ right and $3$ up, so $\overrightarrow{AB} = [5, 3]$.

Solution.

(a) Head minus tail:

$$\overrightarrow{AB} = [3 - (-2),\ 4 - 1] = [5, 3], \qquad \lvert\overrightarrow{AB}\rvert = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83$$ $$\overrightarrow{AC} = [5 - (-2),\ -2 - 1] = [7, -3]$$

(b)

$$2\overrightarrow{AB} - \overrightarrow{AC} = [10, 6] - [7, -3] = [3, 9]$$

(c) Divide by the magnitude:

$$\frac{1}{\sqrt{34}}[5, 3] = \left[\frac{5}{\sqrt{34}},\ \frac{3}{\sqrt{34}}\right] \approx [0.857,\ 0.514]$$

Check: $0.857^2 + 0.514^2 \approx 1.000$. ✓

Example 4: Adding forces with components

Two forces act on a point: $\vec{F}_1$ is $50$ N at $\theta = 20^\circ$ and $\vec{F}_2$ is $80$ N at $\theta = 130^\circ$. Find the resultant in $\vec{i}, \vec{j}$ form, and its magnitude and direction, to two decimal places.

Solution. Break each force into components (rounded here to two decimal places; keep more in your calculator):

$$\begin{aligned} \vec{F}_1 &= [50\cos 20^\circ,\ 50\sin 20^\circ] \approx [46.98,\ 17.10] \\ \vec{F}_2 &= [80\cos 130^\circ,\ 80\sin 130^\circ] \approx [-51.42,\ 61.28] \end{aligned}$$

Add the components:

$$\vec{R} = \vec{F}_1 + \vec{F}_2 \approx [-4.44,\ 78.38] = -4.44\vec{i} + 78.38\vec{j}$$

Magnitude: $\lvert\vec{R}\rvert \approx \sqrt{(-4.44)^2 + 78.38^2} \approx 78.51$ N.

Direction: $\vec{R}$ is in quadrant II, so $\theta \approx \tan^{-1}\left(\dfrac{78.38}{-4.44}\right) + 180^\circ \approx -86.76^\circ + 180^\circ = 93.24^\circ$.

Check with the cosine law. The angle between the forces is $130^\circ - 20^\circ = 110^\circ$, so the angle inside the tip-to-tail triangle is $70^\circ$:

$$\lvert\vec{R}\rvert^2 = 50^2 + 80^2 - 2(50)(80)\cos 70^\circ \approx 6163.84, \qquad \lvert\vec{R}\rvert \approx 78.51 \text{ N} \checkmark$$

Components make problems like those in applications of vectors routine, especially with three or more vectors.

Common mistakes

Forgetting the quadrant when finding the direction. $\tan^{-1}\left(\dfrac{12}{-5}\right)$ gives $-67.4^\circ$, but $[-5, 12]$ points up and to the left, at $112.6^\circ$. Always check the signs of the components (or a quick sketch) before giving the angle.

Subtracting in the wrong order. $\overrightarrow{AB} = B - A$ (head minus tail), not $A - B$. Getting it backwards gives $\overrightarrow{BA} = -\overrightarrow{AB}$.

Using a bearing as if it were $\theta$. $x = r\cos\theta$ needs the angle from the positive $x$-axis, counterclockwise. A bearing of $230^\circ$ means $\theta = 220^\circ$, not $230^\circ$.

Squaring negatives on a calculator. $(-5)^2 = 25$, but typing $-5^2$ gives $-25$. Use brackets when finding magnitudes.

Adding magnitudes instead of components. $\lvert\vec{u} + \vec{v}\rvert$ is not $\lvert\vec{u}\rvert + \lvert\vec{v}\rvert$. Add the components first, then find the magnitude.

Calculator in radians. The vector pages in this course use degrees. If $50\cos 20^\circ$ comes out negative, your calculator is in radian mode.

Practice

1. (Warm-up) Find the magnitude of each vector.

• (a) $[6, -8]$
• (b) $[-2, -5]$

Solution

(a) $\sqrt{6^2 + (-8)^2} = \sqrt{100} = 10$.

(b) $\sqrt{(-2)^2 + (-5)^2} = \sqrt{29} \approx 5.39$.

2. (Warm-up) Let $\vec{u} = [3, -1]$ and $\vec{v} = [-2, 4]$. Find:

• (a) $\vec{u} + \vec{v}$
• (b) $2\vec{u} - 3\vec{v}$
• (c) $\lvert\vec{u} - \vec{v}\rvert$

Solution

(a) $[3 + (-2),\ -1 + 4] = [1, 3]$.

(b) $[6, -2] - [-6, 12] = [12, -14]$.

(c) $\vec{u} - \vec{v} = [5, -5]$, so $\lvert\vec{u} - \vec{v}\rvert = \sqrt{25 + 25} = 5\sqrt{2} \approx 7.07$.

3. (Core) Write each vector in Cartesian form. Give exact values in (a) and (b), and two decimal places in (c).

• (a) magnitude $10$, $\theta = 225^\circ$
• (b) magnitude $6$, $\theta = 300^\circ$
• (c) $20$ km/h on a bearing of $070^\circ$ (east is the positive $x$-axis, north is the positive $y$-axis)

Solution

(a) $[10\cos 225^\circ,\ 10\sin 225^\circ] = \left[10\left(-\dfrac{\sqrt{2}}{2}\right),\ 10\left(-\dfrac{\sqrt{2}}{2}\right)\right] = [-5\sqrt{2},\ -5\sqrt{2}]$.

(b) $[6\cos 300^\circ,\ 6\sin 300^\circ] = \left[6 \cdot \dfrac{1}{2},\ 6\left(-\dfrac{\sqrt{3}}{2}\right)\right] = [3,\ -3\sqrt{3}]$.

(c) $\theta = 90^\circ - 70^\circ = 20^\circ$, so $[20\cos 20^\circ,\ 20\sin 20^\circ] \approx [18.79,\ 6.84]$.

4. (Core) Find the magnitude and direction angle $\theta$ of each vector. Give exact magnitudes, and angles to one decimal place where needed.

• (a) $[-7, -7]$
• (b) $[4, -9]$
• (c) $[-\sqrt{3}, 1]$

Solution

(a) $\sqrt{49 + 49} = 7\sqrt{2}$. Both components are negative (quadrant III) and equal in size, so the reference angle is $45^\circ$ and $\theta = 180^\circ + 45^\circ = 225^\circ$.

(b) $\sqrt{16 + 81} = \sqrt{97} \approx 9.85$. $\tan^{-1}\left(\dfrac{-9}{4}\right) \approx -66.0^\circ$; quadrant IV, so $\theta \approx 294.0^\circ$.

(c) $\sqrt{3 + 1} = 2$. The reference angle is $\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = 30^\circ$, in quadrant II, so $\theta = 180^\circ - 30^\circ = 150^\circ$.

5. (Core) For $P(4, -1)$ and $Q(-2, 7)$, find $\overrightarrow{PQ}$, its magnitude, and the unit vector in its direction. Write $\overrightarrow{PQ}$ in $\vec{i}, \vec{j}$ form.

Solution

$\overrightarrow{PQ} = [-2 - 4,\ 7 - (-1)] = [-6, 8] = -6\vec{i} + 8\vec{j}$.

$\lvert\overrightarrow{PQ}\rvert = \sqrt{36 + 64} = 10$.

Unit vector: $\dfrac{1}{10}[-6, 8] = [-0.6,\ 0.8]$.

6. (Core)

• (a) Find $k$ so that $[k, 6]$ is collinear with $[2, -3]$.
• (b) Find the vector of magnitude $15$ in the same direction as $[3, 4]$.

Solution

(a) We need $[k, 6] = m[2, -3]$. From the second components, $6 = -3m$, so $m = -2$. Then $k = 2m = -4$.

(b) $\lvert[3, 4]\rvert = 5$, so the unit vector is $\left[\dfrac{3}{5}, \dfrac{4}{5}\right]$. Multiply by $15$: $[9, 12]$.

7. (Core) A plane has an air speed of $400$ km/h on a heading of $050^\circ$. A $60$ km/h wind blows from the north. Use components (east is the positive $x$-axis, north is the positive $y$-axis) to find the ground speed and the track, to one decimal place.

Solution

Air velocity: $\theta = 90^\circ - 50^\circ = 40^\circ$, so $\vec{v}_{\text{air}} = [400\cos 40^\circ,\ 400\sin 40^\circ] \approx [306.42,\ 257.12]$.

A wind from the north blows south: $\vec{w} = [0, -60]$.

$$\vec{v}_{\text{ground}} \approx [306.42,\ 197.12]$$

Ground speed: $\sqrt{306.42^2 + 197.12^2} \approx 364.3$ km/h.

Direction: quadrant I, $\theta = \tan^{-1}\left(\dfrac{197.12}{306.42}\right) \approx 32.753^\circ$. As a bearing, the track is $90^\circ - 32.753^\circ \approx 057.2^\circ$.

8. (Challenge) The points $A(1, 1)$, $B(5, 2)$, $C(6, 6)$ and $D(2, 5)$ are the vertices of quadrilateral $ABCD$.

• (a) Use vectors to show that $ABCD$ is a parallelogram.
• (b) Show that it’s actually a rhombus.

Solution

(a) $\overrightarrow{AB} = [4, 1]$ and $\overrightarrow{DC} = [6 - 2,\ 6 - 5] = [4, 1]$. Since $\overrightarrow{AB} = \overrightarrow{DC}$, the sides $AB$ and $DC$ are parallel and equal in length, so $ABCD$ is a parallelogram.

(b) $\overrightarrow{AD} = [1, 4]$. Then $\lvert\overrightarrow{AB}\rvert = \sqrt{16 + 1} = \sqrt{17}$ and $\lvert\overrightarrow{AD}\rvert = \sqrt{1 + 16} = \sqrt{17}$. A parallelogram with two adjacent sides equal is a rhombus.

9. (Challenge) Let $\vec{u} = [1, 2]$ and $\vec{v} = [3, -1]$. Write $\vec{w} = [-1, 12]$ as a linear combination $a\vec{u} + b\vec{v}$.

Solution

$a[1, 2] + b[3, -1] = [a + 3b,\ 2a - b]$. Match the components with $[-1, 12]$:

$$a + 3b = -1 \qquad\text{and}\qquad 2a - b = 12$$

From the first, $a = -1 - 3b$. Substitute: $2(-1 - 3b) - b = 12$, so $-2 - 7b = 12$ and $b = -2$. Then $a = -1 + 6 = 5$.

$$\vec{w} = 5\vec{u} - 2\vec{v}$$

Check: $5[1, 2] - 2[3, -1] = [5 - 6,\ 10 + 2] = [-1, 12]$. ✓