Vectors in 3D

In a nutshell

Vectors can be used to describe position and displacement in three dimensions. As in two dimensions where you have unit vectors $\bf{i}$ and $\bf{j}$ for the $x\text-$ and $y$ -axes respectively, the unit vector along the $z$ -axis is denoted by $\bf{k}$ . Vectors in 3D can be represented with column vectors or unit vector notation ( $\bf{i}$ , $\bf{j}$ and $\bf{k}$ ). Similar to vectors in 2D, you can work out the angle a vector makes with any positive axis.

Unit vectors

The unit vectors along the $x$ -, $y$ - and $z$ -axes are denoted as $\bf{i}$ , $\bf{j}$ and $\bf{k}$ , respectively.

\bf{i}=\begin{pmatrix}1\\0\\0\end{pmatrix}

\bf{j}=\begin{pmatrix}0\\1\\0\end{pmatrix}

\bf{k}=\begin{pmatrix}0\\0\\1\end{pmatrix}

For any vector in 3D:

$\boxed{p{\bf i}+q{\bf j}+r{\bf k}= \begin{pmatrix}p\\q\\r\end{pmatrix}}$

Example 1

Find the position vectors of $A(5,4,2)$ and $B(9,2,4)$ . Write the answer in component form. Find $\overrightarrow{AB}$ as a column vector.

The position vectors are:

$\overrightarrow{OA}=\underline{5{\bf i}+4{\bf j}+2{\bf k}}, \overrightarrow{OB}=\underline{9{\bf i}+2{\bf j}+4{\bf k}}$

$\begin{aligned}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\&=\begin{pmatrix}9\\2\\4\end{pmatrix}-\begin{pmatrix}5\\4\\2\end{pmatrix}=\underline{\begin{pmatrix}4\\-2\\2\end{pmatrix} } \end{aligned}$

Example 2

The vectors $\bf{a}$ and $\bf{ b}$ are given by ${\bf a}=\begin{pmatrix}18\\-3\\3\end{pmatrix}$ and ${\bf b}=\begin{pmatrix}1\\5\\3\end{pmatrix}$ . Find $\frac13{\bf a}+\bf{b}$ .

$\begin{aligned}\frac13{\bf a}+{\bf b}&=\frac13\begin{pmatrix}18\\-3\\3\end{pmatrix}+\begin{pmatrix}1\\5\\3\end{pmatrix}=\begin{pmatrix}6\\-1\\1\end{pmatrix}+\begin{pmatrix}1\\5\\3\end{pmatrix}=\underline{\begin{pmatrix}7\\4\\4\end{pmatrix}}\end{aligned}$

Example 3

Find the magnitude of ${\bf a}= {\bf i}+ 3 {\bf j} -2{\bf k}$ , and hence find the unit vector in the direction of ${\bf a}$ .

By Pythagoras' theorem:

$|{\bf a}| = \sqrt{1^2+3^2+(-2)^2}=\sqrt{1+9+4}=\underline{\sqrt{14}}$ 

For the unit vector in the direction of ${\bf a}$ :

$\hat{\bf a}=\frac{{\bf a}}{|{\bf a }|}=\underline{ \frac1{\sqrt{14}}({\bf i}+ 3 {\bf j} -2{\bf k})}$

Finding the angle

You can find the angle between a given vector and any of the coordinate axes by using the appropriate right-angled triangle.

Maths; Vectors II; KS5 Year 13; Vectors in 3D

If vector ${\bf a}=x{\bf i}+y{\bf j} +z{\bf k}$ and makes an angle $\theta_x$ with the positive $x$ -axis, then:

$\boxed{\cos\theta_x=\frac{x}{|{\bf a}|}}$

Note: This also applies for angles $\theta_y$ and $\theta_z$ .

Example 4

Find the angles that the vector ${\bf a}=2{\bf i}-1{\bf j} +7{\bf k}$ makes with each of the positive coordinates axes to $1\ d.p.$ 

First, find $|{\bf a}|$ :

$|{\bf a}| = \sqrt{2^2+(-1)^2+7^2}=\sqrt{4+1+49}=\sqrt{54}$ 

Use the respective formula for each axis:

$\cos\theta_x=\frac{x}{|{\bf a}|}=\frac{2}{\sqrt{54}}\implies \theta_x=\underline{74.2\degree}$

$\cos\theta_y=\frac{y}{|{\bf a}|}=\frac{-1}{\sqrt{54}}\implies \theta_y=\underline{97.8\degree}$

$\cos\theta_z=\frac{z}{|{\bf a}|}=\frac{7}{\sqrt{54}}\implies \theta_z=\underline{17.7\degree}$ 