Perpendicular Vectors

Video lesson · from 42:312 worked examples

One lesson video covers all of Vectors, so it opens at 42:31 for this topic — not from the beginning.

Theory

If $\vec{a}$ and $\vec{b}$ are perpendicular then the angle between them is $90^\circ$ .

Since $\cos 90^\circ = 0$ , $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 90^\circ = 0$ .

If $\vec{a}$ and $\vec{b}$ are perpendicular then $\vec{a} \cdot \vec{b} = 0$ .

Conversely, if $\vec{a} \cdot \vec{b} = 0$ then $\vec{a}$ and $\vec{b}$ are perpendicular.

⚠️ Common Examiner Traps

Perpendicular means the scalar product is zero: this is both the test and the equation you solve for an unknown.
Set it equal to zero, then solve: for a question with an unknown component, form $\vec{a} \cdot \vec{b} = 0$ and solve for it.
State the conclusion: after showing the product is zero, say that the vectors are therefore perpendicular.
Zero product does not mean zero vector: two non-zero vectors can have a scalar product of zero — that is exactly what perpendicular means.

Worked examples

Example 1

Two vectors are defined as $\vec{a} = 4\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}$ and $\vec{b} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ . Show that $\vec{a}$ and $\vec{b}$ are perpendicular.

\vec{a} \cdot \vec{b} = (4)(2) + (2)(1) + (-5)(2) = 8 + 2 - 10 = 0

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

Example 2

$\vec{PQ} = \begin{pmatrix} 4 \\ a \\ 7 \end{pmatrix}$ and $\vec{RS} = \begin{pmatrix} 2 \\ -3 \\ a \end{pmatrix}$ where $a$ is a constant. Given that $\vec{PQ}$ and $\vec{RS}$ are perpendicular, find the value of $a$ .

Since they are perpendicular, $\vec{PQ} \cdot \vec{RS} = 0$ .

\begin{aligned} (4)(2) + (a)(-3) + (7)(a) &= 0 \\ 8 - 3a + 7a &= 0 \\ 8 + 4a &= 0 \\ 4a &= -8 \\ a &= -2 \end{aligned}