Vectors in Three Dimensions

A drone, a satellite, or a character in a video game doesn’t just move left-right and up-down: it moves in three dimensions. Adding a third axis lets you describe any point in space with three coordinates and any vector with three components. The good news is that everything from Cartesian vectors carries over, with one extra number.

Key ideas

The 3-D coordinate system

Three-space uses three number lines, the $x$-, $y$- and $z$-axes, that meet at the origin $O$ and are each perpendicular to the other two. On paper we draw them with the $y$-axis to the right, the $z$-axis up, and the $x$-axis coming out of the page toward you (drawn slanting down and to the left).

This arrangement is right-handed: point the fingers of your right hand along the positive $x$-axis and curl them toward the positive $y$-axis; your thumb points along the positive $z$-axis. Ontario courses, and most math and physics books, use right-handed systems. (For example, if $x$ points east and $y$ points north, then $z$ points up.)

Each pair of axes forms a coordinate plane: the $xy$-plane (where $z = 0$), the $xz$-plane ($y = 0$), and the $yz$-plane ($x = 0$). Together the three planes split space into eight octants.

Plotting points

The point $P(x, y, z)$ is reached by moving $x$ units along the $x$-axis, then $y$ units parallel to the $y$-axis, then $z$ units parallel to the $z$-axis. Drawing a box (rectangular prism) with one corner at $O$ and the opposite corner at $P$ makes the picture much easier to read.

To plot $P(2, 3, 4)$, go $2$ along $x$, $3$ along $y$, and $4$ up. The blue arrow is the position vector $\overrightarrow{OP} = [2, 3, 4]$.

Points and vectors

Just as in two dimensions, the point $P(x, y, z)$ has position vector

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

and any vector equal to it is also $[x, y, z]$. (Some books write $(x, y, z)$ or $\langle x, y, z\rangle$.) The same three numbers describe both a point and a vector; the brackets tell you which one is meant.

Unit vectors i, j and k

The standard unit vectors in three-space are

$$\vec{i} = [1, 0, 0], \qquad \vec{j} = [0, 1, 0], \qquad \vec{k} = [0, 0, 1]$$

and every vector is a linear combination of them: $[x, y, z] = x\vec{i} + y\vec{j} + z\vec{k}$. For example, $[2, -5, 3] = 2\vec{i} - 5\vec{j} + 3\vec{k}$.

$\vec{i}$, $\vec{j}$ and $\vec{k}$ each have length $1$ and point along the positive axes.

Magnitude and distance

Applying the Pythagorean theorem twice (once across the floor of the box, once up the height) gives

$$\lvert[x, y, z]\rvert = \sqrt{x^2 + y^2 + z^2}$$

For points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, the vector between them is still “head minus tail”, and its magnitude is the distance $AB$:

$$\overrightarrow{AB} = [x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1], \qquad AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Operations in component form

Add, subtract and multiply by scalars component by component, exactly as in 2-D:

$$[u_1, u_2, u_3] + [v_1, v_2, v_3] = [u_1 + v_1,\ u_2 + v_2,\ u_3 + v_3], \qquad k[u_1, u_2, u_3] = [ku_1,\ ku_2,\ ku_3]$$

All the properties of addition and scalar multiplication still hold. Two non-zero vectors are collinear when one is a scalar multiple of the other, and points $A$, $B$, $C$ are collinear when $\overrightarrow{AB} = k\overrightarrow{AC}$ for some scalar $k$. The midpoint of $AB$ is $\left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2},\ \dfrac{z_1 + z_2}{2}\right)$.

3-D graphing software (such as GeoGebra 3D) is a great way to check a sketch, but everything on this page can be done by hand.

Worked examples

Example 1: Locating points

Describe where each point is, and sketch $P(2, 3, 4)$.

• (a) $A(0, 0, -5)$
• (b) $B(3, 0, 2)$
• (c) $C(-1, 4, 0)$

Solution.

(a) Two coordinates are zero, so $A$ is on an axis: the negative $z$-axis, $5$ units below the origin.

(b) $y = 0$, so $B$ is in the $xz$-plane.

(c) $z = 0$, so $C$ is in the $xy$-plane.

To sketch $P(2, 3, 4)$: mark $2$ on the $x$-axis, move $3$ units parallel to the $y$-axis to reach $(2, 3, 0)$ in the $xy$-plane, then go up $4$ units. Completing the box, as in the figure above, shows clearly where $P$ is.

Example 2: Magnitude and unit vector

Let $\vec{v} = [2, -3, 6]$. Find $\lvert\vec{v}\rvert$, write $\vec{v}$ in $\vec{i}, \vec{j}, \vec{k}$ form, and find the unit vector in the direction of $\vec{v}$.

Solution.

$$\lvert\vec{v}\rvert = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

$\vec{v} = 2\vec{i} - 3\vec{j} + 6\vec{k}$. The unit vector is

$$\frac{1}{7}[2, -3, 6] = \left[\frac{2}{7},\ -\frac{3}{7},\ \frac{6}{7}\right]$$

Check: $\dfrac{4}{49} + \dfrac{9}{49} + \dfrac{36}{49} = 1$. ✓

Example 3: Distance and midpoint

For $A(1, -2, 4)$ and $B(4, 2, -8)$, find $\overrightarrow{AB}$, the distance $AB$, and the midpoint of $AB$.

Solution. Head minus tail:

$$\overrightarrow{AB} = [4 - 1,\ 2 - (-2),\ -8 - 4] = [3, 4, -12]$$ $$AB = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$$

Midpoint: $\left(\dfrac{1 + 4}{2},\ \dfrac{-2 + 2}{2},\ \dfrac{4 + (-8)}{2}\right) = (2.5,\ 0,\ -2)$.

Example 4: Operations and collinearity

• (a) For $\vec{u} = [1, 2, -3]$ and $\vec{v} = [-2, 0, 5]$, find $3\vec{u} - 2\vec{v}$.
• (b) Are the points $P(1, 0, 2)$, $Q(3, 4, -2)$ and $R(6, 10, -8)$ collinear?

Solution.

(a)

$$3\vec{u} - 2\vec{v} = [3, 6, -9] - [-4, 0, 10] = [7, 6, -19]$$

(b) Find two vectors that share the point $P$:

$$\overrightarrow{PQ} = [2, 4, -4], \qquad \overrightarrow{PR} = [5, 10, -10]$$

Compare components: $\dfrac{5}{2} = \dfrac{10}{4} = \dfrac{-10}{-4} = 2.5$. So $\overrightarrow{PR} = 2.5\,\overrightarrow{PQ}$. The vectors are parallel and share $P$, so $P$, $Q$ and $R$ are collinear. (In fact $R$ is $2.5$ times as far from $P$ as $Q$ is.)

Common mistakes

Drawing a left-handed system. If you put $x$ to the right and $y$ out of the page with $z$ up, the system is left-handed. Use the right-hand rule: fingers from $+x$ to $+y$, thumb along $+z$.

Mixing up “on an axis” and “in a plane”. A point with one zero coordinate is in a coordinate plane (for example $(3, 0, 2)$ is in the $xz$-plane). A point with two zero coordinates is on an axis.

Forgetting the third component. $\lvert[2, -3, 6]\rvert$ is $\sqrt{4 + 9 + 36}$, not $\sqrt{4 + 9}$. Every formula from 2-D gets one extra term.

Subtracting in the wrong order, or mixing orders between components. $\overrightarrow{AB}$ is $B - A$ in every component. Writing $x_2 - x_1$ but $y_1 - y_2$ gives a vector that’s neither $\overrightarrow{AB}$ nor $\overrightarrow{BA}$.

Checking only two components for collinearity. All three ratios must agree. $[2, 4, -4]$ and $[5, 10, 7]$ have matching first two ratios, but they aren’t collinear because $\dfrac{7}{-4} \ne 2.5$.

Confusing a point with its position vector. $P(2, 3, 4)$ is a location; $\overrightarrow{OP} = [2, 3, 4]$ is the arrow from the origin to it. You can add vectors, but adding points doesn’t mean anything on its own.

Practice

1. (Warm-up) State whether each point is on an axis (which one?) or in a coordinate plane (which one?).

• (a) $(0, 5, 0)$
• (b) $(-2, 0, 3)$
• (c) $(4, -1, 0)$

Solution

(a) On the $y$-axis (two coordinates are zero).

(b) In the $xz$-plane ($y = 0$).

(c) In the $xy$-plane ($z = 0$).

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

• (a) $[1, 2, 2]$
• (b) $[-4, 0, 3]$
• (c) $6\vec{i} - 2\vec{j} + 3\vec{k}$

Solution

(a) $\sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

(b) $\sqrt{16 + 0 + 9} = \sqrt{25} = 5$.

(c) $\sqrt{36 + 4 + 9} = \sqrt{49} = 7$.

3. (Core) Let $\vec{u} = [2, -1, 3]$ and $\vec{v} = [0, 4, -2]$. Find:

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

Solution

(a) $[2, 3, 1]$.

(b) $[4, -2, 6] - [0, 4, -2] = [4, -6, 8]$.

(c) $\vec{u} - \vec{v} = [2, -5, 5]$, so $\lvert\vec{u} - \vec{v}\rvert = \sqrt{4 + 25 + 25} = \sqrt{54} = 3\sqrt{6} \approx 7.35$.

4. (Core) Find the distance between $A(-3, 2, 5)$ and $B(1, -2, 7)$.

Solution

$\overrightarrow{AB} = [1 - (-3),\ -2 - 2,\ 7 - 5] = [4, -4, 2]$, so

$$AB = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$$

5. (Core) Let $\vec{w} = [4, -4, 7]$.

• (a) Find the unit vector in the direction of $\vec{w}$.
• (b) Find the vector of magnitude $18$ in the direction opposite to $\vec{w}$.

Solution

(a) $\lvert\vec{w}\rvert = \sqrt{16 + 16 + 49} = \sqrt{81} = 9$, so the unit vector is $\left[\dfrac{4}{9},\ -\dfrac{4}{9},\ \dfrac{7}{9}\right]$.

(b) Multiply the unit vector by $-18$: $[-8, 8, -14]$. (That’s $-2\vec{w}$.) Check: $\sqrt{64 + 64 + 196} = \sqrt{324} = 18$. ✓

6. (Core) Find $a$ and $b$ so that $[a, 6, -4]$ and $[3, b, 2]$ are collinear.

Solution

We need $[a, 6, -4] = k[3, b, 2]$. The third components give $-4 = 2k$, so $k = -2$.

Then $a = 3k = -6$, and $6 = kb = -2b$, so $b = -3$.

Check: $-2[3, -3, 2] = [-6, 6, -4]$. ✓

7. (Core) A drone’s position is measured from its launch pad, with $x$ east, $y$ north and $z$ up, in metres. It flies in a straight line from $A(30, 40, 25)$ to $B(-10, 70, 45)$. Find its displacement vector and how far it flew, to one decimal place.

Solution

$$\overrightarrow{AB} = [-10 - 30,\ 70 - 40,\ 45 - 25] = [-40, 30, 20]$$

So the drone moved $40$ m west, $30$ m north, and $20$ m up. The distance is

$$\sqrt{(-40)^2 + 30^2 + 20^2} = \sqrt{2900} = 10\sqrt{29} \approx 53.9 \text{ m}$$

8. (Challenge) Show that the triangle with vertices $A(2, 1, 0)$, $B(4, 3, 1)$ and $C(3, -1, 2)$ is a right isosceles triangle, and find its perimeter.

Solution

$$\overrightarrow{AB} = [2, 2, 1], \quad \overrightarrow{AC} = [1, -2, 2], \quad \overrightarrow{BC} = [-1, -4, 1]$$$$AB = \sqrt{4 + 4 + 1} = 3, \quad AC = \sqrt{1 + 4 + 4} = 3, \quad BC = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$

$AB = AC$, so the triangle is isosceles. Also $AB^2 + AC^2 = 9 + 9 = 18 = BC^2$, so by the converse of the Pythagorean theorem the angle at $A$ is a right angle.

Perimeter: $3 + 3 + 3\sqrt{2} = 6 + 3\sqrt{2} \approx 10.24$.

9. (Challenge) Find the point on the $z$-axis that is the same distance from $A(1, 2, 3)$ as from $B(3, 0, -1)$.

Solution

A point on the $z$-axis has the form $P(0, 0, z)$. Set $PA^2 = PB^2$ (squaring avoids square roots):

$$\begin{aligned} 1^2 + 2^2 + (z - 3)^2 &= 3^2 + 0^2 + (z + 1)^2 \\ 5 + z^2 - 6z + 9 &= 9 + z^2 + 2z + 1 \\ 14 - 6z &= 10 + 2z \\ z &= 0.5 \end{aligned}$$

The point is $(0, 0, 0.5)$. Check: $PA^2 = 1 + 4 + 6.25 = 11.25$ and $PB^2 = 9 + 0 + 2.25 = 11.25$. ✓