Angles, perpendicular and parallel vectors Problem set A Worked solutions

Find the angle between each pair of vectors, to 3 significant figures.

1. (a) \(\mathbf{a} = \begin{pmatrix}2\\1\\-3\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}4\\-2\\1\end{pmatrix}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (2)(4) + (1)(-2) + (-3)(1) = 3\)
   • \(|\mathbf{a}| = \sqrt{14},\quad |\mathbf{b}| = \sqrt{21}\)
   • \(\cos\theta = \dfrac{3}{\sqrt{14}\sqrt{21}} = 0.1750\)
   • \(\theta = \cos^{-1}(0.1750) = 79.9^\circ\)

2. (b) \(\mathbf{a} = 5\mathbf{i} - \mathbf{j} + 2\mathbf{k},\ \mathbf{b} = 3\mathbf{i} + 4\mathbf{j} - \mathbf{k}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (5)(3) + (-1)(4) + (2)(-1) = 9\)
   • \(|\mathbf{a}| = \sqrt{30},\quad |\mathbf{b}| = \sqrt{26}\)
   • \(\cos\theta = \dfrac{9}{\sqrt{30}\sqrt{26}} = 0.3223\)
   • \(\theta = \cos^{-1}(0.3223) = 71.2^\circ\)

3. (c) \(\mathbf{a} = -3\mathbf{i} + 4\mathbf{j},\ \mathbf{b} = 5\mathbf{i} + 2\mathbf{j}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (-3)(5) + (4)(2) = -7\)
   • \(|\mathbf{a}| = \sqrt{25},\quad |\mathbf{b}| = \sqrt{29}\)
   • \(\cos\theta = \dfrac{-7}{\sqrt{25}\sqrt{29}} = -0.2600\)
   • \(\theta = \cos^{-1}(-0.2600) = 105^\circ\)

The points A, B and C have coordinates A(1, −2, 4), B(3, 1, −1) and C(−2, 4, 2). Find the size of angle ABC, to 3 significant figures.

• \(\overrightarrow{BA} = \begin{pmatrix}1\\-2\\4\end{pmatrix} - \begin{pmatrix}3\\1\\-1\end{pmatrix} = \begin{pmatrix}-2\\-3\\5\end{pmatrix},\quad \overrightarrow{BC} = \begin{pmatrix}-2\\4\\2\end{pmatrix} - \begin{pmatrix}3\\1\\-1\end{pmatrix} = \begin{pmatrix}-5\\3\\3\end{pmatrix}\)
• \(\overrightarrow{BA}\cdot\overrightarrow{BC} = (-2)(-5) + (-3)(3) + (5)(3) = 16\)
• \(|\overrightarrow{BA}| = \sqrt{38},\quad |\overrightarrow{BC}| = \sqrt{43}\)
• \(\cos\theta = \dfrac{16}{\sqrt{38}\sqrt{43}} = 0.3958\)
• \(\theta = \cos^{-1}(0.3958) = 66.7^\circ\)

Find the value of t for which the two vectors are perpendicular.

1. (a) \(\mathbf{a} = \begin{pmatrix}3t\\-1\\2 + t\end{pmatrix},\ \mathbf{b} = \begin{pmatrix}2\\5\\-4\end{pmatrix}\)

   • \(\mathbf{a}\cdot\mathbf{b} = 0\)
   • \((3t)(2) + (-1)(5) + (2 + t)(-4) = 0\)
   • \(-13 + 2t = 0 \Rightarrow t = 6.5\)

2. (b) \(\mathbf{a} = (1 + t)\mathbf{i} + 4\mathbf{j} + 2t\mathbf{k},\ \mathbf{b} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}\)

   • \(\mathbf{a}\cdot\mathbf{b} = 0\)
   • \((1 + t)(2) + (4)(-1) + (2t)(1) = 0\)
   • \(-2 + 4t = 0 \Rightarrow t = 0.5\)

The vectors \(\mathbf{a} = \begin{pmatrix}2t\\1\\t\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-3 + t\\-6\\2\end{pmatrix}\) are perpendicular. Find the possible values of \(t\).

• \(\mathbf{a}\cdot\mathbf{b} = 0\)
• \((2t)(-3 + t) + (1)(-6) + (t)(2) = 0\)
• \(2t^2 - 4t - 6 = 0\)
• From the calculator: Polynomial solver. t = −1 or t = 3

A radio mast has its top at A(0, 0, 12). Two cables run from A to anchor points B(8, 3, 0) and C(−5, 6, 0) on the ground. Find the angle between the two cables, to 3 significant figures. Distances are in metres.

• \(\overrightarrow{AB} = \begin{pmatrix}8\\3\\-12\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}-5\\6\\-12\end{pmatrix}\)
• \(\overrightarrow{AB}\cdot\overrightarrow{AC} = (8)(-5) + (3)(6) + (-12)(-12) = 122\)
• \(|\overrightarrow{AB}| = \sqrt{217},\quad |\overrightarrow{AC}| = \sqrt{205}\)
• \(\cos\theta = \dfrac{122}{\sqrt{217}\sqrt{205}} = 0.5784\)
• \(\theta = \cos^{-1}(0.5784) = 54.7^\circ\)

A winch pulls a block with the force \(\mathbf{F} = \begin{pmatrix}2t\\1 + t\\3\end{pmatrix}\) N, where \(t\) is a constant. The object moves through the displacement \(\mathbf{d} = \begin{pmatrix}2\\-6\\3\end{pmatrix}\) m. The work done is zero. Find the value of \(t\).

• \(\text{Work done} = \mathbf{F}\cdot\mathbf{d} = 0\)
• \((2t)(2) + (1 + t)(-6) + (3)(3) = 0\)
• \(3 - 2t = 0 \Rightarrow t = 1.5\)
• The force is perpendicular to the displacement.

The points A, B and C have coordinates A(1, 0, 2), B(4, 4, 2) and C(−2, 3, 6).

1. (a) Find the size of angle BAC.

   • \(\overrightarrow{AB} = \begin{pmatrix}3\\4\\0\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}-3\\3\\4\end{pmatrix}\)
   • \(\overrightarrow{AB}\cdot\overrightarrow{AC} = (3)(-3) + (4)(3) + (0)(4) = 3\)
   • \(|\overrightarrow{AB}| = \sqrt{25},\quad |\overrightarrow{AC}| = \sqrt{34}\)
   • \(\cos\theta = \dfrac{3}{\sqrt{25}\sqrt{34}} = 0.1029\)
   • \(\theta = \cos^{-1}(0.1029) = 84.1^\circ\)

2. (b) Hence find the area of triangle ABC.

   • \(\text{Area} = \tfrac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta\)
   • \(= \tfrac12\sqrt{25}\sqrt{34}\sin 84.09^\circ = 14.5\)

Use the scalar product to show that \(\mathbf{p} = \begin{pmatrix}2\\-6\\4\end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix}-3\\9\\-6\end{pmatrix}\) are parallel, and say whether they point in the same direction.

• \(\mathbf{p}\cdot\mathbf{q} = (2)(-3) + (-6)(9) + (4)(-6) = -84\)
• \(|\mathbf{p}||\mathbf{q}| = \sqrt{56}\sqrt{126} = 84.0\)
• \(|\mathbf{p}\cdot\mathbf{q}| = |\mathbf{p}||\mathbf{q}|\)
• So the vectors are parallel. The scalar product is negative: opposite directions (θ = 180°).