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.

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.

​

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: 

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

​​

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

​

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