Angles, perpendicular and parallel vectors Problem set B One question at a time

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

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

   • \(\mathbf{a}\cdot\mathbf{b} = (3)(-1) + (-2)(5) + (4)(2) = -5\)
   • \(|\mathbf{a}| = \sqrt{29},\quad |\mathbf{b}| = \sqrt{30}\)
   • \(\cos\theta = \dfrac{-5}{\sqrt{29}\sqrt{30}} = -0.1695\)
   • \(\theta = \cos^{-1}(-0.1695) = 99.8^\circ\)

2. (b) \(\mathbf{a} = 2\mathbf{i} + 6\mathbf{j} - 3\mathbf{k},\ \mathbf{b} = -4\mathbf{i} + \mathbf{j} + 5\mathbf{k}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (2)(-4) + (6)(1) + (-3)(5) = -17\)
   • \(|\mathbf{a}| = \sqrt{49},\quad |\mathbf{b}| = \sqrt{42}\)
   • \(\cos\theta = \dfrac{-17}{\sqrt{49}\sqrt{42}} = -0.3747\)
   • \(\theta = \cos^{-1}(-0.3747) = 112^\circ\)

3. (c) \(\mathbf{a} = 6\mathbf{i} - \mathbf{j},\ \mathbf{b} = 2\mathbf{i} + 7\mathbf{j}\)

   • \(\mathbf{a}\cdot\mathbf{b} = (6)(2) + (-1)(7) = 5\)
   • \(|\mathbf{a}| = \sqrt{37},\quad |\mathbf{b}| = \sqrt{53}\)
   • \(\cos\theta = \dfrac{5}{\sqrt{37}\sqrt{53}} = 0.1129\)
   • \(\theta = \cos^{-1}(0.1129) = 83.5^\circ\)

The points P, Q and R have coordinates P(−3, 0, 2), Q(1, 4, −3) and R(2, −2, 5). Find the size of angle PQR, to 3 significant figures.

• \(\overrightarrow{QP} = \begin{pmatrix}-3\\0\\2\end{pmatrix} - \begin{pmatrix}1\\4\\-3\end{pmatrix} = \begin{pmatrix}-4\\-4\\5\end{pmatrix},\quad \overrightarrow{QR} = \begin{pmatrix}2\\-2\\5\end{pmatrix} - \begin{pmatrix}1\\4\\-3\end{pmatrix} = \begin{pmatrix}1\\-6\\8\end{pmatrix}\)
• \(\overrightarrow{QP}\cdot\overrightarrow{QR} = (-4)(1) + (-4)(-6) + (5)(8) = 60\)
• \(|\overrightarrow{QP}| = \sqrt{57},\quad |\overrightarrow{QR}| = \sqrt{101}\)
• \(\cos\theta = \dfrac{60}{\sqrt{57}\sqrt{101}} = 0.7908\)
• \(\theta = \cos^{-1}(0.7908) = 37.7^\circ\)

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

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

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

2. (b) \(\mathbf{a} = (-1 + 2t)\mathbf{i} + 3\mathbf{j} + t\mathbf{k},\ \mathbf{b} = 3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\)

   • \(\mathbf{a}\cdot\mathbf{b} = 0\)
   • \((-1 + 2t)(3) + (3)(2) + (t)(-4) = 0\)
   • \(3 + 2t = 0 \Rightarrow t = -1.5\)

The vectors \(\mathbf{a} = \begin{pmatrix}t\\4\\2 + t\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}-4 + t\\2\\-1\end{pmatrix}\) are perpendicular. Find the possible values of \(t\).

• \(\mathbf{a}\cdot\mathbf{b} = 0\)
• \((t)(-4 + t) + (4)(2) + (2 + t)(-1) = 0\)
• \(t^2 - 5t + 6 = 0\)
• From the calculator: Polynomial solver. t = 2 or t = 3

A crane has its hook point at A(0, 0, 15). Two ropes run from A to the ground at B(6, −4, 0) and C(−3, −7, 0). Find the angle between the two ropes, to 3 significant figures. Distances are in metres.

• \(\overrightarrow{AB} = \begin{pmatrix}6\\-4\\-15\end{pmatrix},\quad \overrightarrow{AC} = \begin{pmatrix}-3\\-7\\-15\end{pmatrix}\)
• \(\overrightarrow{AB}\cdot\overrightarrow{AC} = (6)(-3) + (-4)(-7) + (-15)(-15) = 235\)
• \(|\overrightarrow{AB}| = \sqrt{277},\quad |\overrightarrow{AC}| = \sqrt{283}\)
• \(\cos\theta = \dfrac{235}{\sqrt{277}\sqrt{283}} = 0.8393\)
• \(\theta = \cos^{-1}(0.8393) = 32.9^\circ\)

A tow rope pulls a barge with the force \(\mathbf{F} = \begin{pmatrix}5\\3t\\-2 + t\end{pmatrix}\) N, where \(t\) is a constant. The object moves through the displacement \(\mathbf{d} = \begin{pmatrix}-4\\2\\2\end{pmatrix}\) m. The work done is zero. Find the value of \(t\).

• \(\text{Work done} = \mathbf{F}\cdot\mathbf{d} = 0\)
• \((5)(-4) + (3t)(2) + (-2 + t)(2) = 0\)
• \(-24 + 8t = 0 \Rightarrow t = 3\)
• The force is perpendicular to the displacement.

The points P, Q and R have coordinates P(0, 2, −1), Q(3, 2, 3) and R(−3, 5, 2).

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

   • \(\overrightarrow{PQ} = \begin{pmatrix}3\\0\\4\end{pmatrix},\quad \overrightarrow{PR} = \begin{pmatrix}-3\\3\\3\end{pmatrix}\)
   • \(\overrightarrow{PQ}\cdot\overrightarrow{PR} = (3)(-3) + (0)(3) + (4)(3) = 3\)
   • \(|\overrightarrow{PQ}| = \sqrt{25},\quad |\overrightarrow{PR}| = \sqrt{27}\)
   • \(\cos\theta = \dfrac{3}{\sqrt{25}\sqrt{27}} = 0.1155\)
   • \(\theta = \cos^{-1}(0.1155) = 83.4^\circ\)

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

   • \(\text{Area} = \tfrac12|\overrightarrow{PQ}||\overrightarrow{PR}|\sin\theta\)
   • \(= \tfrac12\sqrt{25}\sqrt{27}\sin 83.37^\circ = 12.9\)

Use the scalar product to show that \(\mathbf{p} = \begin{pmatrix}4\\-2\\6\end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix}10\\-5\\15\end{pmatrix}\) are parallel, and say whether they point in the same direction.

• \(\mathbf{p}\cdot\mathbf{q} = (4)(10) + (-2)(-5) + (6)(15) = 140\)
• \(|\mathbf{p}||\mathbf{q}| = \sqrt{56}\sqrt{350} = 140\)
• \(|\mathbf{p}\cdot\mathbf{q}| = |\mathbf{p}||\mathbf{q}|\)
• So the vectors are parallel. The scalar product is positive: the same direction (θ = 0°).