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?

  • $\begingroup$ What do you mean by "Plausibilize" here? $\endgroup$
    – jonathanjo
    Commented Feb 18, 2020 at 0:24
  • $\begingroup$ jonathanjo: Well, make it intuitively clear, possibly without a rigorous proof. $\endgroup$
    – Photon
    Commented Feb 18, 2020 at 19:36
  • 1
    $\begingroup$ I think it is not unreasonable to claim that the formula for the dot-product is essentially a slick notation for the law of cosines (I agree with Stephan K's answer). It's more than that I suppose, but it is at least that... page 25 of supermath.info/CalculusIIIf2014.pdf might be helpful. $\endgroup$ Commented Feb 21, 2020 at 3:08

5 Answers 5


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:enter image description here

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: enter image description here

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.

  • $\begingroup$ That one is similar to my second idea but already integrates the lengths of the vectors, so definitely an improvement. I like physicsy explanations since I have a physics background myself, so obviously I am biased towards thinking that physicists can explain maths more intuitively, though less rigorously. ;) Thanks! $\endgroup$
    – Photon
    Commented Feb 21, 2020 at 10:34

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).

  • 1
    $\begingroup$ Thanks, good idea! Unfortunately, the cosine rule is not part of the curriculum but it still seems like a better approach than the two I had in mind! $\endgroup$
    – Photon
    Commented Feb 17, 2020 at 12:55
  • 8
    $\begingroup$ Are you teaching vectors to students who have not had trig? $\endgroup$ Commented Feb 17, 2020 at 19:57
  • $\begingroup$ mrstoptraffic: Valid remark. ;) Well, the curriculum is such that they know the definitions of the trigonometric functions in a right-angled triangle, also they know them as functions (their graphs, symmetry, periodicity etc.). Also they know the Pythagoras' theorem in right-angled triangles, but they know neither the cosine rule nor the sine rule in general triangles, unfortunately. $\endgroup$
    – Photon
    Commented Feb 18, 2020 at 6:26
  • $\begingroup$ Then teach them the cosine rule first. It cannot hurt to know more than the curriculum, especially if it is an essential lemma for what you want to show them. $\endgroup$ Commented Feb 18, 2020 at 8:27
  • $\begingroup$ Well, as of now I am only tutoring some students, so I am not their actual teacher. In school both expressions were introduced without any comments regarding why they both define the same thing. Therefore in the current situation I have to either use some quick intuitive argument or skip the justification altogether since I have only 2h of teaching time per week with this student. However, at the same time, I am thinking about how I would introduce the topic once I have become an actual teacher. I'm not clear about how much room I'll have to cover topics which are not part of the curriculum. $\endgroup$
    – Photon
    Commented Feb 18, 2020 at 19:22

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} $ .


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.

  • $\begingroup$ That's an interesting idea, thanks! I fear, it won't work for my students because they don't know the concepts of basis vectors and basis expansion (they are not part of the curriculum) but still interesting to tackle the problem from the other side! $\endgroup$
    – Photon
    Commented Feb 21, 2020 at 10:46
  • $\begingroup$ @Photon: It doesn't seem to me that the notion of a basis is needed here. If you want to avoid the notation $\hat{\textbf{x}}$, you can just write $\langle1,0,0\rangle$. $\endgroup$
    – user507
    Commented Feb 21, 2020 at 13:29

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.