Vectors in three dimensions

The unit vectors i, j and k, components and column vectors in three dimensions, magnitude, unit vectors, operations with components, position vectors, and parallel vectors and collinear points.

Components in 3D

Add a third unit vector $\underset{\sim}{k}$ along the positive $z$-axis. Every vector in space is $$\underset{\sim}{u} = x\underset{\sim}{i} + y\underset{\sim}{j} + z\underset{\sim}{k} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \qquad |\underset{\sim}{u}| = \sqrt{x^2 + y^2 + z^2}.$$ Everything from two dimensions carries over: operations are done component by component, and $\overrightarrow{AB} = \underset{\sim}{b} - \underset{\sim}{a}$.

Example 1●

$\underset{\sim}{u} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ and $\underset{\sim}{v} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$. Find $2\underset{\sim}{u} - \underset{\sim}{v}$, $|\underset{\sim}{u}|$ and the unit vector $\hat{v}$.

$2\underset{\sim}{u} - \underset{\sim}{v} = \begin{pmatrix} 2 - 3 \\ 4 - 0 \\ -4 - 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -8 \end{pmatrix}$.

$|\underset{\sim}{u}| = \sqrt{1 + 4 + 4} = 3$.

$|\underset{\sim}{v}| = 5$, so $\hat{v} = \tfrac15\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$.

Parallel vectors and collinear points

$\underset{\sim}{u}$ and $\underset{\sim}{v}$ are parallel when $\underset{\sim}{u} = \lambda\underset{\sim}{v}$, so all their components are in the same ratio. Points $A$, $B$ and $C$ are collinear when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.

Example 2●●

Show that $A(1, 0, 2)$, $B(2, 3, 1)$ and $C(4, 9, -1)$ are collinear.

$\overrightarrow{AB} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} 3 \\ 9 \\ -3 \end{pmatrix} = 3\overrightarrow{AB}$.

The vectors are parallel and share the point $A$, so $A$, $B$ and $C$ lie on one line.

Example 3●●

Find $p$ and $q$ so that $\begin{pmatrix} 2 \\ p \\ 6 \end{pmatrix}$ is parallel to $\begin{pmatrix} 1 \\ -2 \\ q \end{pmatrix}$.

The first $x$-component is twice the second, so $\lambda = 2$.

$p = 2 \times (-2) = -4$ and $6 = 2q$, so $q = 3$.

Flashcards

$|xunderset{sim}{i} + yunderset{sim}{j} + zunderset{sim}{k}|$

$\sqrt{x^2 + y^2 + z^2}$

Parallel vectors

$\underset{\sim}{u} = \lambda\underset{\sim}{v}$: the components are in the same ratio.

Collinear points $A$, $B$, $C$

$\overrightarrow{AC} = \lambda\overrightarrow{AB}$

Practice

Find $|\underset{\sim}{i} + 2\underset{\sim}{j} - 2\underset{\sim}{k}|$.
$\underset{\sim}{u} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ and $\underset{\sim}{v} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$. Find the components of $2\underset{\sim}{u} - \underset{\sim}{v}$. Enter them as $x, y, z$.
Find the $z$-component of the unit vector in the direction of $3\underset{\sim}{i} + 4\underset{\sim}{k}$.
Find $p$ so that $\begin{pmatrix} 2 \\ p \\ 6 \end{pmatrix}$ is parallel to $\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$.
$A = (1, 0, 2)$ and $B = (2, 3, 1)$. Find the components of $\overrightarrow{AB}$. Enter them as $x, y, z$.
Find $|\overrightarrow{AB}|$ for the same points. Give an exact answer or a decimal.
Which vector is parallel to $\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$?
$\overrightarrow{AC} = 3\overrightarrow{AB}$. Then $A$, $B$ and $C$ are:
$\underset{\sim}{k}$ is the unit vector along: