Angle Between Vectors

Video lesson · from 45:412 worked examples

One lesson video covers all of Vectors, so it opens at 45:41 for this topic — not from the beginning.

From the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$ , we can rearrange this to find the angle between two vectors:

\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}

Remember to calculate $\vec{a} \cdot \vec{b}$ using $a_1b_1 + a_2b_2 + a_3b_3$ .

Rearranged, not memorised separately: $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$ is just the geometric form rearranged.
Both vectors point away from the vertex: to find the angle at $B$ in a triangle, use $\vec{BA}$ and $\vec{BC}$ — not $\vec{AB}$ .
A negative cosine gives an obtuse angle: that is a valid answer, not an error. Do not discard the sign.
Do not round too early: keep the scalar product and magnitudes exact until the final inverse cosine.

Example 1

Calculate angle $\theta$ between vectors $\vec{p} = 3\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ and $\vec{q} = 4\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ .

Calculate dot product $\vec{p} \cdot \vec{q}$ :

\vec{p} \cdot \vec{q} = (3)(4) + (4)(1) + (-2)(3) = 12 + 4 - 6 = 10

Calculate magnitude $|\vec{p}|$ :

|\vec{p}| = \sqrt{3^2 + 4^2 + (-2)^2} = \sqrt{9 + 16 + 4} = \sqrt{29}

Calculate magnitude $|\vec{q}|$ :

|\vec{q}| = \sqrt{4^2 + 1^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}

Substitute into formula:

\begin{aligned} \cos \theta &= \frac{10}{\sqrt{29}\sqrt{26}} \\ \cos \theta &= \frac{10}{\sqrt{754}} \\ \theta &= \cos^{-1}\left(\frac{10}{\sqrt{754}}\right) \approx 68.6^\circ \end{aligned}

Example 2

K is the point $(1, -7, 2)$ , L $(-3, 3, 4)$ and M $(2, 5, 1)$ . Find $\angle KLM$ .

For angle at L, the vectors must point away from L: $\vec{LK}$ and $\vec{LM}$ .

\vec{LK} = \vec{k} - \vec{l} = \begin{pmatrix} 1 \\ -7 \\ 2 \end{pmatrix} - \begin{pmatrix} -3 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -10 \\ -2 \end{pmatrix}

\vec{LM} = \vec{m} - \vec{l} = \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} - \begin{pmatrix} -3 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix}

\vec{LK} \cdot \vec{LM} = (4)(5) + (-10)(2) + (-2)(-3) = 20 - 20 + 6 = 6

|\vec{LK}| = \sqrt{4^2 + (-10)^2 + (-2)^2} = \sqrt{16 + 100 + 4} = \sqrt{120}

|\vec{LM}| = \sqrt{5^2 + 2^2 + (-3)^2} = \sqrt{25 + 4 + 9} = \sqrt{38}

\begin{aligned} \cos(\angle KLM) &= \frac{6}{\sqrt{120}\sqrt{38}} \\ \theta &= \cos^{-1}\left(\frac{6}{\sqrt{4560}}\right) \approx 84.9^\circ \end{aligned}