Geometric proof that $a_1 b_1 + a_2 b_2 = | \vec a | |\vec b | \cos(\varphi)$

Question

I want to define the scalar product by $\vec a \cdot \vec b = | \vec a | |\vec b | \cos(\varphi)$ and derive $\vec a \cdot \vec b = a_1 b_1 + a_2 b_2$. Algebraically, this is no problem (we can just expand the vectors in the standard basis).

How do I show this by using only elementary geometry? I've drawn the areas of the rectangles $a_1 b_1$, $a_2 b_2$ and $| \vec a | \cos(\varphi) \cdot |\vec b | $but I don't see why they should be equal.

/edit: I'm really interested in the general case and in using elementary geometry, not in getting an elegant proof which involves rotations.

Answer 1

One way is to compute the (square of the) red length two ways, and equate the two.

On the one hand, Pythagoras tells us that the red length is the hypotenuse of a right triangle, whose legs have length $|a_1 - b_1|$ and $|a_2 - b_2|$, so the square of its length must be $$|a_1 - b_1|^2 + |a_2 - b_2|^2 = (a_1 - b_2)^2 + (a_2 - b_2)^2$$

On the other hand, the Law of Cosines tells us that the square of the red length is $$|a|^2 + |b|^2 - 2|a||b|\cos(\varphi) = (a_1^2 + a_2^2) + (b_1^2 + b_2^2) - 2|a||b|\cos( \varphi )$$

Since these are equal, we have $(a_1 - b_2)^2 + (a_2 - b_2)^2 = (a_1^2 + a_2^2) + (b_1^2 + b_2^2) - 2|a||b|\cos( \varphi )$

It looks like a mess, but if you multiply out the squares on the left side, you get a bunch of cancellation (with the squares on the right), and before long, $a_1b_1 + a_2b_2 = |a||b|\cos( \varphi )$ pops out.

Answer 2

Yeah, I don't think the thing to do is to look at rectangles. First, prove that the dot product is invariant under rotations (this is a little circular, since some people would say rotations are defined by being linear maps preserving the dot product and orientation).

Then, line one of those vectors up with the $x$-axis. Without a loss of generality, say $\vec{b}$ is along the $x$-axis. Then it has components $(b_1,0)$ in the standard basis. The projection of $\vec{a}$ onto the $x$-axis is $|\vec{a}|\cos(\varphi)$. That is $a_1= |\vec{a}|\cos(\varphi)$. At the same time, $|\vec{b}| = b_1$, since it is lined up on the $x$-axis. So, then we obviously have $a_1b_1= |\vec{a}||\vec{b}|\cos(\varphi)$. But, since it is rotationally invariant, this holds in general.

Answer 3

An other (more geometric) way:

Consider the following draw

We denote $\alpha :=\angle(a,b)$. Moreover, $a^\perp=(-a_2,a_1)$ is a vector that is perpendicular to $a$ with same norm than $a$ (i.e. $\alpha +\beta =\frac{\pi}{2}$ and $\|a\|=\|a^\perp\|$). You have that $$a_1b_1+a_2b_2=\left|\begin{array}{cc} b_1&-a_2\\b_2&a_1\end{array}\right|=\det(b,a^\perp)=\text{Area}\big(P(b,a^\perp)\big),$$ where $P(b,a^\perp)$ is the parallelogram generated by $b$ and $a^\perp$.

We know that $\beta =\frac{\pi}{2}-\alpha $ and that $$\text{Area}\big(P(b,a^\perp)\big)=\|b\|\|a^\perp\|\sin(\beta ).$$ Since $$\|a^\perp\|=\|a\|\quad \text{and}\quad \sin\left(\frac{\pi}{2}-\alpha \right)=\cos(\alpha ),$$ we finally get $$a_1b_1+a_2b_2=\|a\|\|b\|\cos\big(\angle(a,b)\big),$$ as wished.

---

An other (more analytically) way:

Let $a=(a_1,a_2)$ and $b=(b_1,b_2)$. Denote $\bar x$ the $x-$axis, $\hat a=\angle(\bar x,a)$ the (oriented) angle between $\bar x$ and $a$ and $\hat b=\angle (\bar x,b)$; the (oriented) angle between $\bar x$ and $b$. Finally, denote $\|\cdot \|$ the Euclidian norm.

• In one hand, $$\cos(\hat a)=\frac{a_1}{\|a\|}\quad \text{and}\quad \sin(\hat a)=\frac{a_2}{\|a\|}$$ $$\cos(\hat b)=\frac{b_1}{\|b\|}\quad \text{and}\quad \sin(\hat b)=\frac{b_2}{\|b\|}.$$

• In the other hand, $$\angle(a,b)=\hat b-\hat a,$$

and thus \begin{align*} \cos\big(\angle(a,b)\big)&=\cos(\hat b-\hat a)\\ &=\cos(\hat a)\cos(\hat b)+\sin(\hat a)\sin(\hat b)\\ &=\frac{a_1b_1}{\|a\|\|b\|}+\frac{a_2b_2}{\|a\|\|b\|}. \end{align*} Therefore, $$\|a\|\|b\|\cos\big(\angle(a,b)\big)=a_1b_1+a_2b_2,$$ as wished.