09. Dot product, 1

Dot product

Given two vectors $v=\left(\begin{array}{c}\hfill v_1\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill v_n\hfill \end{array}\right)$ and $w=\left(\begin{array}{c}\hfill w_1\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill w_n\hfill \end{array}\right)$ in ${𝐑}^n$ , what is the angle between them?

0.43 To define angles in ${𝐑}^n$ , note that the two vectors $v$ and $w$ are contained in a unique 2-plane, and we mean the usual angle between $v$ and $w$ inside that plane.

1.45 To compute that angle, we introduce the dot product $v\cdot w$ .

We will prove this in due course, but first we'll explore it a little.

3.47 Let $v=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)$ and $w=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \end{array}\right)$ . We have $v\cdot w=1\times 0+0\times 1=0$ .

4.38 This makes sense: the angle between $v$ and $w$ is $\pi /2$ radians, and $\cos (\pi /2)=0$ . In this case, we say the vectors are orthogonal to one another (equivalent to "perpendicular" or "at right angles").

5.32 Let $v=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)$ and $w=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right)$ . We can see the angle between them should be 45 degrees ($\pi /4$ radians).

Let's confirm this: we have $v\cdot w=1\times 1+0\times 1=1$ . We also have $|v|=\sqrt{1+1}=\sqrt{2}$ by Pythagoras and $|w|=1$ , so

$$1=v\cdot w=|v||w|\cos \theta =\sqrt{2}\cos \theta ,$$ so $\cos \theta$ should be $1/\sqrt{2}$ . Indeed, $\cos (\pi /4)=1/\sqrt{2}$ .

8.12 Even if you didn't know the angle, you could figure it out as $\arccos (1/\sqrt{2})$ . You might object that $\arccos$ is multivalued, for example $\cos (\pi /4)=\cos (3\pi /4)$ . This just corresponds to the fact that there are different ways of picking "the" angle between $v$ and $w$ (e.g. clockwise or anticlockwise).