Scalar Product of Vectors

The scalar product (or dot product) of vectors $\vec{a}$ and $\vec{b}$ is the real number:

$\boxed{\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\varphi}$

where $\varphi$ is the angle between $\vec{a}$ and $\vec{b}$, with $0° \leq \varphi \leq 180°$.

The angle $\varphi$ is measured by placing both vectors at the same starting point and taking the angle between them in the interval $[0°, 180°]$.

If $\vec{a} = (a_1, a_2)$ and $\vec{b} = (b_1, b_2)$, then:

$\boxed{\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2}$

Proof. Place both vectors at the origin: $\vec{a} = \overrightarrow{OA}$ with $A = (a_1, a_2)$, $\vec{b} = \overrightarrow{OB}$ with $B = (b_1, b_2)$. By the Law of Cosines applied to triangle $OAB$:

$|AB|^2 = |OA|^2 + |OB|^2 - 2|OA||OB|\cos\varphi$

Compute $|AB|^2$ directly:

$|AB|^2 = (b_1 - a_1)^2 + (b_2 - a_2)^2 = a_1^2 + a_2^2 + b_1^2 + b_2^2 - 2a_1 b_1 - 2a_2 b_2$

Since $|OA|^2 = a_1^2 + a_2^2 = |\vec{a}|^2$ and $|OB|^2 = b_1^2 + b_2^2 = |\vec{b}|^2$, substituting:

$|\vec{a}|^2 + |\vec{b}|^2 - 2a_1 b_1 - 2a_2 b_2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\varphi$

Simplifying: $a_1 b_1 + a_2 b_2 = |\vec{a}||\vec{b}|\cos\varphi = \vec{a} \cdot \vec{b}$. $\blacktriangleleft$

For any vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ and scalar $k$:

1. Commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
2. Square: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$
3. Distributive: $(\vec{a} + \vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}$
4. Homogeneous: $(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b})$
5. Zero vector: $\vec{0} \cdot \vec{a} = 0$

Note: the scalar product is not associative — $(\vec{a} \cdot \vec{b}) \cdot \vec{c}$ is not defined (the left side is a scalar, not a vector).

Two nonzero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if their scalar product is zero:

$\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$

Proof. Since $\vec{a} \neq \vec{0}$ and $\vec{b} \neq \vec{0}$, we have $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$. Then: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\varphi = 0 \iff \cos\varphi = 0 \iff \varphi = 90°. \quad \blacktriangleleft$

The angle $\varphi$ between nonzero vectors $\vec{a} = (a_1, a_2)$ and $\vec{b} = (b_1, b_2)$ satisfies:

$\boxed{\cos\varphi = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} = \dfrac{a_1 b_1 + a_2 b_2}{\sqrt{a_1^2 + a_2^2}\;\sqrt{b_1^2 + b_2^2}}}$

Since $0° \leq \varphi \leq 180°$, the sign of $\cos\varphi$ determines whether the angle is acute, right, or obtuse.

Find the angle between $\vec{a} = (3, 4)$ and $\vec{b} = (-4, 3)$.

Solution. Compute the dot product:

$\vec{a} \cdot \vec{b} = 3 \cdot (-4) + 4 \cdot 3 = -12 + 12 = 0$

Since $\vec{a} \cdot \vec{b} = 0$ and both vectors are nonzero, the vectors are perpendicular.

Answer: $\varphi = 90°$. $\blacktriangleleft$

Find the angle between $\vec{p} = (1, \sqrt{3})$ and $\vec{q} = (2, 0)$.

Solution.

$\vec{p} \cdot \vec{q} = 1 \cdot 2 + \sqrt{3} \cdot 0 = 2$

$|\vec{p}| = \sqrt{1 + 3} = 2, \qquad |\vec{q}| = \sqrt{4 + 0} = 2$

$\cos\varphi = \dfrac{2}{2 \cdot 2} = \dfrac{1}{2}$

Therefore $\varphi = 60°$.

Answer: $\varphi = 60°$. $\blacktriangleleft$

Given $\vec{a} = (5, -2)$ and $\vec{b} = (4, 10)$.

(a) Compute $\vec{a} \cdot \vec{b}$. (b) Are $\vec{a}$ and $\vec{b}$ perpendicular? Justify your answer. (c) Find $|\vec{a}|$ and $|\vec{b}|$.

Given $\vec{u} = (3, -4)$ and $\vec{v} = (m, 6)$ where $m$ is unknown.

(a) Find the value of $m$ for which $\vec{u} \perp \vec{v}$. (b) For the value of $m$ found in (a), find the angle $\varphi$ between $\vec{u}$ and $\vec{u} + \vec{v}$.

Hint: For (b), use the coordinate formula for the dot product with $\vec{u}$ and $\vec{u} + \vec{v}$.