Prove that: 1^° a·a = |a|^2 2^° cosangle(a,b) =…

Prove that: $1^\circ\quad \vec{a}\cdot\vec{a} = |\vec{a}|^2 \qquad\qquad 2^\circ\quad \cos\angle(\vec{a},\vec{b}) = \frac{\vec{a}\cdot\vec{b}}{\sqrt{\vec{a}\cdot\vec{a}}\,\sqrt{\vec{b}\cdot\vec{b}}}$.

Step 1: Proof of $1^\circ$

The angle between a vector and itself is $0^\circ$, so

\[ \vec{a}\cdot\vec{a} = |\vec{a}|\,|\vec{a}|\cos 0^\circ = |\vec{a}|^2. \]

\[\vec{a}\cdot\vec{a} = |\vec{a}|^2 \qquad\Longrightarrow\qquad |\vec a|=\sqrt{\vec a\cdot\vec a}\]

Consequently $\vec{a}\cdot\vec{a} \geq 0$, with equality exactly when $\vec{a} = \vec{0}$. (The scalar square is often written $\vec a^{\,2}$.)

Step 2: Proof of $2^\circ$

For non-zero $\vec a,\vec b$, solve the definition $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$ for $\cos\theta$ and use $1^\circ$: \[\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{\vec{a}\cdot\vec{b}}{\sqrt{\vec{a}\cdot\vec{a}}\,\sqrt{\vec{b}\cdot\vec{b}}}.\]

\[\cos\angle(\vec{a},\vec{b}) = \frac{\vec{a}\cdot\vec{b}}{\sqrt{\vec{a}\cdot\vec{a}}\,\sqrt{\vec{b}\cdot\vec{b}}}\]

The formula requires $\vec a\neq\vec 0$ and $\vec b\neq\vec 0$; then the denominator is positive.

Step 3: Special cases

In coordinates (Problem 47) the formula becomes $\cos\theta=\dfrac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\,\sqrt{b_1^2+b_2^2+b_3^2}}$.