Hello I simply cant explain to myself why this equation holds. Lets say we have an orthonormal basis $\vec{i}, \vec{j}$ and 2 2-D vectors in this basis which are:
\begin{align} \vec{a} &= a_1\vec{i} + a_2 \vec{j}\\ \vec{b} &= b_1\vec{i} + b_2 \vec{j} \end{align}
I know that I can calculate the norms of vectors $\left|\vec{a}\right|$ and $\left|\vec{b}\right|$ like this:
\begin{align} \left|\vec{a}\right| = \sqrt{{a_1}^2 + {a_2}^2}\\ \left|\vec{b}\right| = \sqrt{{b_1}^2 + {b_2}^2} \end{align}
When I try to justify the equation $\vec{a} \cdot \vec{b} = \left|\vec{a}\right|\left|\vec{b}\right| \cos \alpha = a_1b_1 + a_2b_2$ for a scalar product I just can't seem to do it. I can only do it in for the easiest case $\vec{a} \cdot \vec{a}$:
\begin{align} \vec{a} \cdot \vec{a} &= \left|\vec{a}\right| \left|\vec{a}\right| \overbrace{\cos \alpha}^{=1}\\ \vec{a} \cdot \vec{a} &= \left|\vec{a}\right| \left|\vec{a}\right|\\ \vec{a} \cdot \vec{a} &= \sqrt{{a_1}^2+{{a_2}^2}}\sqrt{{a_1}^2+{{a_2}^2}}\\ \vec{a} \cdot \vec{a} &= {a_1}^2+{{a_2}^2} \end{align}
Question: How do we do it for a $\vec{a} \cdot \vec{b}$? Lets say for a start that $\alpha = 0$. What about if $\alpha \neq 0$? Please provide a similar derivation as I did for $\vec{a} \cdot \vec{a}$ using norms.
Let $$\mathbf{a}=|a|\cos\theta \mathbf{i}+|a|\sin\theta \mathbf{j}$$ and $$\mathbf{b}=|b|\cos\phi \mathbf{i}+|b|\sin\phi \mathbf{j}$$
Now, we have $$ \mathbf{a\cdot b} = |a||b|\cos\theta\cos\phi + |a||b|\sin\theta\sin\phi $$ Now use a trigonometric identity to simplify, with $\alpha=\theta-\phi$.
At 71GA's request, I'm elaborating on the relationship between my expression of $\mathbf{a}$ and the cartesian expression.
Consider the vector in the form $$ \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} $$ We now define $$|\mathbf{a}|^2 = a_1^2+a_2^2$$ Note that this forms a right triangle relationship. For vectors with positive $a_1$, we may use the formed right triangle to determine the angle of the vector from the positive $x$-axis, $\theta$. Namely, $$ \tan(\theta) = \frac{a_2}{a_1} $$ This relationship also applies if $a_1$ is negative or zero, but the angle itself must be adapted to suit. We may then solve for $a_1$ and $a_2$, by noting that \begin{align} \frac{|\mathbf{a}|^2}{a_1^2}&=1+\frac{a_2^2}{a_1^2}\\ &=1+\tan^2(\theta)\\ &=\sec^2(\theta) \end{align} and so, we have $$ a_1^2 = |\mathbf{a}|^2\cos^2(\theta) $$ and, by appropriate choice of angle, you have $$ a_1 = |\mathbf{a}|\cos(\theta) $$ Substituting back into our equation for $\tan(\theta)$, we have $$ a_2 = |\mathbf{a}|\sin(\theta) $$ As such, we have $$ \mathbf{a} = |\mathbf{a}|\cos\theta\mathbf{i}+|\mathbf{a}|\sin\theta\mathbf{j} $$ where $\theta$ is the angle of the vector from the positive $x$-axis. And we can repeat this for vector $\mathbf{b}$, with $\phi$ as the angle. Thus, the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is $\alpha=\theta-\phi$, or is different from this by $2\pi$ (in which case the relation $\mathbf{a\cdot b}=|\mathbf{a}||\mathbf{b}|\cos(\alpha)$ is unchanged by the addition or subtraction of $2\pi$).