<p>Find the acute angle between direction vectors $\left( -6\mathbf{i}+\mathbf{j} \right)$ and $\left( \mathbf{i}-\mathbf{j}-8\mathbf{k} \right)$.</p>

Find the acute angle between direction vectors $\left( -6\mathbf{i}+\mathbf{j} \right)$ and $\left( \mathbf{i}-\mathbf{j}-8\mathbf{k} \right)$.

Vectors

International Baccalaureate Higher Level Mathematics

IB HL Math

Unit 13.1 Worked Example 3

Find the acute angle between direction vectors $(-6\mathbf{i} + \mathbf{j})$ and $(\mathbf{i} - \mathbf{j} - 8\mathbf{k})$.

$\begin{pmatrix} -6 \\ 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ -8 \end{pmatrix} = \sqrt{(-6)^2 + 1^2} \times \sqrt{1^2 + (-1)^2 + (-8)^2} \times \cos\theta$

$-7 = \sqrt{37} \times \sqrt{66} \times \cos\theta$

$\cos\theta = -\displaystyle\frac{7}{\sqrt{37}\sqrt{66}} \Rightarrow \theta = 98.14°$

Acute angle $= 180° - 98.14° = 81.9°$ (1 d.p.)

Note: obtuse $= 180° -$ acute

Unit 13.1 Worked Example 3

Unit 13.1 Worked Example 3

Video Solution: