Scalar product — angles, perpendiculars and the foot of a perpendicular

By the end of this lesson

By the end you can work out the scalar product of two vectors, and use it to find the angle between two vectors or two lines, to test whether they are perpendicular, and to find the foot of the perpendicular from a point to a line.

Notation first

A vector has a size and a direction. We write vectors in bold: $\mathbf{a}$ , $\mathbf{b}$ . (By hand you underline them: $\underline{a}$ .)
In 3D, $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ are one-unit steps along the $x$ -, $y$ - and $z$ -axes. So $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$ means "2 along $x$ , 3 back along $y$ , 1 up $z$ ". The same vector as a column is $\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$ . The three numbers are its components; we'll call them $a_1, a_2, a_3$ .
$\lvert \mathbf{a} \rvert$ is the magnitude (length) of $\mathbf{a}$ : $\lvert \mathbf{a} \rvert = \sqrt{a_1^2 + a_2^2 + a_3^2}$ , which is Pythagoras in 3D.
A scalar is an ordinary number, with no direction.

The scalar product

The scalar product (also called the dot product) of $\mathbf{a}$ and $\mathbf{b}$ is written $\mathbf{a} \cdot \mathbf{b}$ . You work it out by multiplying matching components and adding:

$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3.$

The answer is a single number (a scalar), not a vector. That's where the name comes from.

It also has a geometric meaning. If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ when they are drawn from the same point (with $0 \leqslant \theta \leqslant 180^\circ$ ), then

$\mathbf{a} \cdot \mathbf{b} = \lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert \cos\theta.$

Both formulas give the same number. Putting them together is what makes the scalar product so useful: the first is easy to calculate, and the second contains the angle.

Why do the two formulas agree?

Draw $\mathbf{a}$ and $\mathbf{b}$ from the same point. The third side of the triangle they make is $\mathbf{a} - \mathbf{b}$ . The cosine rule says

$\lvert \mathbf{a} - \mathbf{b} \rvert^2 = \lvert \mathbf{a} \rvert^2 + \lvert \mathbf{b} \rvert^2 - 2\lvert \mathbf{a} \rvert \lvert \mathbf{b} \rvert \cos\theta.$

Now work out the left side with components instead: $(a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2$ . The first bracket is $(a_1 - b_1)^2 = a_1^2 - 2a_1b_1 + b_1^2$ , and the other two expand the same way. Collecting the squares, the total is $\lvert \mathbf{a} \rvert^2 + \lvert \mathbf{b} \rvert^2 - 2(a_1b_1 + a_2b_2 + a_3b_3)$ .

Compare the two lines: the only difference is the last term, so $a_1b_1 + a_2b_2 + a_3b_3 = \lvert \mathbf{a} \rvert \lvert \mathbf{b} \rvert \cos\theta$ . That's the whole reason it works.

Worked example: Example 1: computing a scalar product

Find $\mathbf{a} \cdot \mathbf{b}$ for $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ .

Have a go first?

Quick check

CalculateQuestion 1 · Recall

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