Scalar product
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 \(\theta = 111^\circ\)