How do you plausibly explain that the geometric and the coordinate expressions for the scalar product are equivalent?

Question

The standard scalar product on $\mathbb{R}^3$ is defined via

$$\vec a\cdot\vec b := a_1b_1+a_2b_2+a_3b_3$$

On the other hand, it can be expressed in a more geometrical way through the lengths of the vectors and the angle $\phi$ they enclose via

$$\vec a\cdot\vec b = |\vec a| |\vec b|\cos\phi$$

However, I struggle to establish a link between the two expressions.

One possibility is to consider the second expression as a definition for the angle $\phi$, since it can be shown that

$$ \frac{\vec a\cdot\vec b}{|\vec a| |\vec b|}\in[-1;1]$$

and thus can be written as the cosine of some angle. But this does not explain why the angle in question is the one enclosed by the two vectors, we could also try to go with the sine of some angle rather than a cosine according to this argument.

Another possibility is to introduce, say, cylindrical coordinates and consider two unit vectors

$$\vec a = (\cos\phi, \sin\phi, 0),\quad \vec b = (1,0,0)$$

which clearly enclose the angle $\phi$ and compute the scalar product according to its definition which gives us $\cos\phi$ as a result, then argue that the scalar product is also proportional to the lengths of both vectors involved and finally show that it is invariant under rotations.

Both approaches require rather heavy machinery considering that the scalar product is being introduced in school already.

So my question is: What is a good way to explain to school students (11th grade, that is, around 17 years old) why the two expressions for the scalar product are equivalent?

Answer 1

I'll hazard a somewhat physicsy answer. Consider two vectors $\vec{A}$ and $\vec{B}$. Without loss of generality, choose coordinates where the positive $x$-axis aligns with $\vec{A}$. Hence, $\vec{A} = \langle A, 0 \rangle$. Suppose $\vec{B}$ makes counter-clockwise angle $\theta$ with respect to $\vec{A}$ and denote $\vec{B} = \langle B_1, B_2 \rangle$. See the picture below:

Apparently $B_2 = B \sin \theta$ and $B_1 = B \cos \theta$. Notice that $B = \sqrt{B_1^2+B_2^2}$ and $A = \sqrt{A_1^2+A_2^2}$ where we have assumed $A_1 >0$ and $A_2=0$ hence $A = A_1$. In total, $$ A_1B_1+A_2B_2 = AB\cos \theta. $$ Naturally, if you don't wish to begin with the rather greedy step of choosing coordinates to make the problem trivial then we'd have to fight through the change of coordinates in essence. Perhaps I will return to this and add that later.

Edit (2-22-2020) next, let us suppose $\vec{A}$ is at angle $\alpha$ measured CCW from the positive $x$-axis. See below:

We can use the usual adding-angles formulas for sine and cosine to expand the geometric expressions $A_1 = A\cos( \alpha + \theta )$ and $A_2 = A \sin( \alpha + \theta)$: $$ A_1 = A\cos( \alpha + \theta ) = A (\cos \alpha \cos \theta- \sin \alpha \sin \theta) $$ $$ A_2 = A\sin( \alpha + \theta ) = A (\cos \alpha \sin \theta+ \sin \alpha \cos \theta) $$ Now we have all we need to investigate the geometric content of the algebraic expression $A_1B_1+A_2B_2$, let's see what happens: \begin{align} A_1B_1+A_2B_2 &= (A\cos \alpha ) B (\cos \alpha \cos \theta- \sin \alpha \sin \theta) \\ \notag & \qquad + (A\sin \alpha) B (\cos \alpha \sin \theta+ \sin \alpha \cos \theta) \\ \notag &= AB( \cos^2 \alpha \cos \theta -\cos \alpha \sin \theta + \cos \alpha \sin \theta + \sin^2 \alpha \cos \theta) \\ \notag &= AB(\cos^2 \alpha+\sin^2 \alpha) \cos \theta \\ \notag &= AB \cos \theta. \end{align} Naturally this more complicated (and unphysicsy) argument collapses to my initial physicsy argument when we set $\alpha = 0$. I think students would find the way $\alpha$ disappears quite rewarding if they could discover it for themselves.

Answer 2

Cosine rule!

Think of vectors $\vec a$ and $\vec b$ as two sides of a triangle, with tails at a common vertex. The remaining side is given by $\vec a - \vec b$. Then cosine rule gives us $|\vec a|^2 + |\vec b|^2 - 2|\vec a||\vec b|\text{cos}\,\theta=|\vec a-\vec b|^2$, where $\theta$ is the angle between $\vec a$ and $\vec b$. Now write out $|\vec a|^2$, $|\vec b|^2$ and $|\vec a - \vec b|^2$ in terms of their orthogonal components, and the result follows after a little bit of algebra (which is accessible to 11th graders).

Answer 3

Here's a way to do it by computing the length of $\vec{a} + \vec{b}$ in two different ways: one of which is purely symbolic and the other uses some geometric knowledge.

This argument is somewhat similar to Stephan Kubicki's argument, just with a $+$ instead of a $-$ .

$\vec{a} + \vec{b}$ can be defined geometrically and component-wise in a straightforward way.

First a word on notation, $a$ means the vector $\vec{a}$ to reduce visual clutter. Additionally, I'll be using $\langle x, y \rangle$ consistently to represent the dot product $x \cdot y$ .

The very first thing I think is to define the inner product (101). In this definition, $n$ is the number of dimensions.

$$ \langle a, b \rangle \stackrel{\text{def}}{=\!=} \sum_ {k =1} ^ n a_kb_k \tag{101} $$

In order to motivate this definition, one can show that the squared length of a vector is the dot product of $a$ with itself (102).

$$ |a|^2 = \langle a , a \rangle \tag{102} $$

As another bit of motivation, you can also show that two vectors are perpendicular if and only if their dot product is zero and work through a couple of examples (111).

$$ \text{$x$ and $y$ are perpendicular} \iff \langle x, y \rangle = 0 \tag{111} $$

With that out of the way, we can ask about the squared length of $a+b$ (103).

$$ |a+b|^2 \tag{103} $$

This is equivalent to the dot product of $a+b$ with itself (103a).

$$ \langle a + b, a + b \rangle \tag{103a} $$

First, let's look at this problem symbolically. You can distribute over the left and right arguments (104)

$$ \langle a +b, a+b \rangle = \langle a, a \rangle + \langle a, b \rangle + \langle b, a \rangle + \langle b, b \rangle \tag{104} $$

I think it's straightforward to show that $\langle a, b \rangle = \langle b, a \rangle$, giving (104a).

$$ \langle a +b, a+b \rangle = \langle a, a \rangle + 2\langle a, b \rangle + \langle b, b \rangle \tag{104a} $$

Next, ask students to picture computing $| a + b | $ by splitting $b$ into two vectors, one of which is parallel to $a$ (let's call it $b_\text{sam}$) and one of which is perpendicular to $a$ (let's call it $b_\text{dif}$) (105). For this sort of argument I think a picture would help.

$$ | a + b | = \sqrt{ |a + b_\text{sam}|^2 + |0 + b_\text{dif}|^2 } \tag{105} $$

If $\theta$ is the angle between $a$ and $b$, then we can rewrite this expression.

$$ | a + b | = \sqrt{ (|a| + |b|\cos{\theta})^2 + (|b|\sin{\theta})^2 } \tag{105a} $$

Next we can expand it out.

$$ | a + b | = \sqrt{ (|a||a| + 2|a||b|\cos{\theta} + |b||b|\cos{\theta}\cos{\theta}) + (|b||b|\sin{\theta}\sin{\theta}) } \tag{105b} $$

We can exploit the fact that $\cos{\theta}\cos{\theta} + \sin{\theta}\sin{\theta} = 1 $ (105c).

$$ |a+b| = \sqrt{|a||a| + 2|a||b|\cos{\theta} + |b||b|} \tag{105c} $$

Square both sides (105d)

$$ |a+b||a+b| = |a||a| + 2|a||b|\cos{\theta} + |b||b| \tag{105d} $$

Use the dot product instead of the squared length (105e).

$$ \langle a+b, a+b \rangle = \langle a, a \rangle + 2|a||b|\cos{\theta} + \langle b, b \rangle \tag{105e} $$

Next we compare the (105e) and (104a), reproduced below for convenience.

$$ \langle a +b, a+b \rangle = \langle a, a \rangle + 2\langle a, b \rangle + \langle b, b \rangle \tag{104a} $$

$$ \langle a+b, a+b \rangle = \langle a, a \rangle + 2|a||b|\cos{\theta} + \langle b, b\rangle \tag{105e} $$

Therefore, as desired, $ \langle a, b \rangle = |a||b|\cos{\theta} $ .

Answer 4

Suppose we take as the definition of the dot product that

$$\textbf{a}\cdot\textbf{b} = ab\cos\phi. \qquad (1)$$

It's then fairly straightforward to show that the dot product is bilinear, i.e., that

$$(p\textbf{a}+q\textbf{b})\cdot\textbf{c}=p\textbf{a}\cdot\textbf{c}+q\textbf{b}\cdot\textbf{c}\qquad (2)$$

(and likewise for the right-hand factor). Properties (1) and (2) are both clearly independent of how we rotate or translate our coordinate system.

From property (1), $\hat{\textbf{x}}\cdot\hat{\textbf{x}}=1$, $\hat{\textbf{x}}\cdot\hat{\textbf{y}}=0$, and so on. And then from property (2), the coordinate expression for the dot product follows.

Answer 5

If you can get across the intuition that the scalar product is rotation invariant, and then that dotting with xhat or yhat pulls out the coordinate, after rotating a so that it is xhat this is the definition of cosine. (Xhat and yhat are the unit length vectors in the coordinate directions.)

As for the rotation invariance, one intuitive approach to it breaks down into: calculating the scalar product between orthogonal* and parallel unit vectors, and noting that by distributivity this is sufficient to characterize the scalar product, that the concepts of orthogonal and parallel unit vectors are rotation invariant, and that rotation is a linear transformation. Of course, some of this would have to be phrased in a lot more detail for a high school audience.

*For the scalar product between two orthogonal unit vectors, write one of them as (cos(a), sin(a)) and the other as (cos(a + pi), sin(a + pi)).

Another important simplification I made is that one can work in the plane. You don't have to explain why this is the case, I think.