# Scalar product

Vectors will not be assessed in the question papers in session 2021–22

The scalar product $$a.b$$ is defined as $$\textbf{a.b}=\left|\textbf{a}\right|\left|\textbf{b}\right|\cos\theta$$ where $$\theta$$ is the angle between $$\textbf{a}$$ and $$\textbf{b}$$.

From this definition it can also be shown that $$\textbf{a.b} = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}$$

The main use of the scalar product is to calculate the angle $$\theta$$.

$\left|\textbf{a}\right|\left|\textbf{b}\right|\cos \theta = \textbf{a.b}$

therefore $$\cos \theta = \frac{{\textbf{a.b}}}{{\left|\textbf{a}\right|\left|\textbf{b}\right|}}$$ where $$\textbf{a.b} = {a_x}{b_x} + {a_y}{b_y} + {a_z}{b_z}$$

## Example

Calculate the angle $$\theta$$ on the diagram below.

State the rule you are using for this question:

$\cos \theta = \frac{{p.q}}{{\left| p \right|\left| q \right|}}$

Calculate $$p.q$$

${p_x}{q_x} + {p_y}{q_y} + {p_z}{q_z} =$

$3 \times 2 + ( - 1) \times 4 + 4 \times 2$

$= 10$

Calculate $$\left| p \right|$$ and $$\left| q \right|$$

$\left| p \right| = \sqrt {9 + 1 + 16} = \sqrt {26}$

$\left| q \right| = \sqrt {4 + 16 + 4} = \sqrt {24}$

Substitute:

$\cos \theta = \frac{{10}}{{\sqrt {26} \sqrt {24} }} = 0.400$

Evaluate $$\theta$$

$\theta = 66.4^\circ$

If your answer at the substitution stage works out negative then the angle lies between $$90^\circ$$ and $$180^\circ$$. For example, if $$\cos \theta = - 0.362$$ then $$\theta = 111^\circ$$