Collinear Points and Orthogonal Vectors in 2D

Lessons

Senior 3 · Mathematics

Collinear Points and Orthogonal Vectors in 2D

3 lessons

Lesson 2 of 3

Orthogonal vectors and the scalar product

25 min

Squaring a foundation

Before the walls of a new classroom go up, builders set out the corners with pegs and string, and every corner must be a true right angle. On site they check with a rope knotted into lengths 3, 4 and 5. On a coordinate plan the check is even quicker: two multiplications and one addition. Suppose the plan (in metres) has the corner at $A (0, 0)$ and pegs at $B (12, 5)$ and $D (- 5, 12)$ . Is $\angle BAD$ a right angle?

Definition

The scalar product (or dot product) of $\vec{u} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ is the number

\vec{u} \cdot \vec{v} = x_1x_2 + y_1y_2 .

Two vectors are orthogonal (perpendicular), written $\vec{u} \perp \vec{v}$ , if they meet at a right angle. The zero vector counts as orthogonal to every vector.

Example: $\begin{pmatrix} 3 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 3 \times 2 + 4 \times (-1) = 6 - 4 = 2$ .

Important

Properties. $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ ; $(k\vec{u}) \cdot \vec{v} = k(\vec{u} \cdot \vec{v})$ ; and $\vec{u} \cdot \vec{u} = x_1^2 + y_1^2 = |\vec{u}|^2$ , the square of the length.

The orthogonality test.

\vec{u} \perp \vec{v} \iff \vec{u} \cdot \vec{v} = x_1x_2 + y_1y_2 = 0 .

Why. Draw $\vec{u} = \overrightarrow{OA}$ and $\vec{v} = \overrightarrow{OB}$ from the origin, so $\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$ . Expanding,

AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) - 2(x_1x_2 + y_1y_2) = OA^2 + OB^2 - 2\,\vec{u} \cdot \vec{v} .

By Pythagoras and its converse, $\angle AOB = 90^\circ$ exactly when $A B^{2} = O A^{2} + O B^{2}$ , that is, exactly when $\vec{u} \cdot \vec{v} = 0$ .

Worked example 1: testing for a right angle

Is $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ orthogonal to (a) $\vec{v} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}$ ; (b) $\vec{w} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ ?

Solution (a) $\vec{u} \cdot \vec{v} = 3 \times (-8) + 4 \times 6 = -24 + 24 = 0$ : orthogonal, as the figure shows.

(b) $\vec{u} \cdot \vec{w} = 3 \times 5 + 4 \times (-2) = 15 - 8 = 7 \neq 0$ : not orthogonal.

Important

A ready-made perpendicular. For any $\vec{u} = \begin{pmatrix} a \\ b \end{pmatrix}$ , the vector $\begin{pmatrix} -b \\ a \end{pmatrix}$ is orthogonal to it, because $a (- b) + b (a) = 0$ , and so is every multiple of it, such as $\begin{pmatrix} b \\ -a \end{pmatrix}$ . It also has the same length as $\vec{u}$ .

Link with gradients. A line in the direction $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ has gradient $\frac{y_1}{x_1}$ . In the unit "Linear and Quadratic Functions", perpendicular lines had $m_{1} m_{2} = - 1$ : $\frac{y_1}{x_1} \times \frac{y_2}{x_2} = -1$ is the same as $x_{1} x_{2} + y_{1} y_{2} = 0$ . The dot product also works for vertical lines, which have no gradient.

Worked example 2: a missing component

Find $k$ so that $\begin{pmatrix} k \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ -6 \end{pmatrix}$ are orthogonal.

Solution $3k + 2 \times (-6) = 0$ , so $3 k = 12$ and $k = 4$ . Check: $\begin{pmatrix} 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -6 \end{pmatrix} = 12 - 12 = 0$ . ✓

Worked example 3: the foundation

Check the corner $A$ of the plan, and find the fourth corner $C$ so that $A B C D$ is a rectangle.

Solution $\overrightarrow{AB} = \begin{pmatrix} 12 \\ 5 \end{pmatrix}$ and $\overrightarrow{AD} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}$ : $\overrightarrow{AB} \cdot \overrightarrow{AD} = 12 \times (-5) + 5 \times 12 = -60 + 60 = 0$ . The corner is a right angle.

In a rectangle $\overrightarrow{BC} = \overrightarrow{AD}$ , so $C$ is $(12 - 5, 5 + 12) = (7, 17)$ . Also $|\overrightarrow{AB}| = \sqrt{144 + 25} = 13$ and $|\overrightarrow{AD}| = \sqrt{25 + 144} = 13$ : the plan is in fact a square of side 13 m.

Worked example 4: which angle is the right angle?

Show that the triangle with vertices $A (1, 1)$ , $B (5, 3)$ and $C (3, 7)$ is right-angled, and say where.

Solution $\overrightarrow{BA} = \begin{pmatrix} -4 \\ -2 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ : $\overrightarrow{BA} \cdot \overrightarrow{BC} = 8 - 8 = 0$ , so $\angle ABC = 90^\circ$ .

Check with lengths: $A B^{2} = 16 + 4 = 20$ , $B C^{2} = 4 + 16 = 20$ and $A C^{2} = 2^{2} + 6^{2} = 40 = 20 + 20$ . ✓ (The triangle is also isosceles.)

Worked example 5: a perpendicular of a given length

Find the vectors of length 10 that are orthogonal to $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .

Solution $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$ is orthogonal to $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ (swap and change one sign), and its length is $\sqrt{16 + 9} = 5$ . Doubling it gives length 10: the answers are $\begin{pmatrix} -8 \\ 6 \end{pmatrix}$ and $\begin{pmatrix} 8 \\ -6 \end{pmatrix}$ .

Warning

The dot product is a number, not a vector. $\begin{pmatrix} 3 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ is $2$ , never $\begin{pmatrix} 6 \\ -4 \end{pmatrix}$ .

Do not mix up the two tests. Collinear: cross-multiply and subtract, $x_{1} y_{2} - x_{2} y_{1} = 0$ . Orthogonal: multiply matching components and add, $x_{1} x_{2} + y_{1} y_{2} = 0$ .

Tip

Question	Test
Parallel (collinear)?	$x_{1} y_{2} - x_{2} y_{1} = 0$
Perpendicular (orthogonal)?	$x_{1} x_{2} + y_{1} y_{2} = 0$
How long?	$\sqrt{x^2 + y^2}$

Copy this table into your notebook: every problem in the next lesson uses one of these three lines.

Khan Academy: Vector dot product and vector length. The video works with vectors of any size; you only need two components. Watch for the result that a vector dotted with itself gives its length squared.

✦

Practice

The last two are harder.

Work out $\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ .
Are $\begin{pmatrix} 6 \\ -4 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ orthogonal?
Are $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$ orthogonal?
Find $p$ so that $\begin{pmatrix} p \\ -3 \end{pmatrix} \perp \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ .
Show that the triangle $P (0, 0)$ , $Q (4, 2)$ , $R (- 1, 2)$ has a right angle, and say where.
Find two vectors of length 15 that are orthogonal to $\begin{pmatrix} 4 \\ -3 \end{pmatrix}$ .
Find the values of $t$ for which $\begin{pmatrix} t \\ 3 \end{pmatrix}$ and $\begin{pmatrix} t \\ -12 \end{pmatrix}$ are orthogonal.
Show that $\begin{pmatrix} a \\ b \end{pmatrix}$ and $\begin{pmatrix} -b \\ a \end{pmatrix}$ are orthogonal and have the same length.

Show answersHide answers

$8 - 5 = 3$ .
$12 - 12 = 0$ : yes.
$-15 + 12 = -3 \neq 0$ : no.
$4 p - 18 = 0$ , so $p = 4.5$ .
$\overrightarrow{PQ} \cdot \overrightarrow{PR} = 4 \times (-1) + 2 \times 2 = 0$ : the right angle is at $P$ .
$\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ is orthogonal to $\begin{pmatrix} 4 \\ -3 \end{pmatrix}$ and has length 5; the answers are $\begin{pmatrix} 9 \\ 12 \end{pmatrix}$ and $\begin{pmatrix} -9 \\ -12 \end{pmatrix}$ .
$t^{2} - 36 = 0$ , so $t = 6$ or $t = - 6$ .
$a (- b) + b (a) = 0$ , and both lengths are $\sqrt{a^2 + b^2}$ .

Sources and credits

Written by Ganzaa.
Video: Vector dot product and vector length, Khan Academy, embedded from YouTube.

The diagrams, proof, worked examples and practice items were written by Ganzaa; the foundation plan is illustrative.

1Squaring a foundation

Squaring a foundation