The acute angle between two lines Problem set A Worked solutions

Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}1\\4\\-2\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\2\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}0\\3\\1\end{pmatrix} + \mu\begin{pmatrix}1\\2\\-2\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}2\\-1\\2\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}1\\2\\-2\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (-1)(2) + (2)(-2) = -4\), \(|\mathbf{d}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3\), \(|\mathbf{d}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\)
• \(\cos\theta = -\frac{4}{9}\), so \(\theta = 116.4\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 116.4^\circ = 63.6^\circ\)

Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}0\\-1\\4\end{pmatrix} + \lambda\begin{pmatrix}2\\3\\-1\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}3\\2\\-5\end{pmatrix} + \mu\begin{pmatrix}4\\0\\-3\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}2\\3\\-1\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}4\\0\\-3\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(4) + (3)(0) + (-1)(-3) = 11\), \(|\mathbf{d}_1| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}\), \(|\mathbf{d}_2| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{25} = 5\)
• \(\cos\theta = \frac{11}{5 \sqrt{14}}\), so \(\theta = 54.0\ldots^\circ\)
• cos θ is positive, so the angle is acute: \(54.0^\circ\)

Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}1\\-1\end{pmatrix} + \lambda\begin{pmatrix}3\\2\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}-2\\6\end{pmatrix} + \mu\begin{pmatrix}-4\\1\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}3\\2\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}-4\\1\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(-4) + (2)(1) = -10\), \(|\mathbf{d}_1| = \sqrt{3^2 + 2^2} = \sqrt{13}\), \(|\mathbf{d}_2| = \sqrt{(-4)^2 + 1^2} = \sqrt{17}\)
• \(\cos\theta = -\frac{10}{\sqrt{221}}\), so \(\theta = 132.3\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 132.3^\circ = 47.7^\circ\)

Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}0\\7\end{pmatrix} + \lambda\begin{pmatrix}2\\5\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}-3\\2\end{pmatrix} + \mu\begin{pmatrix}4\\1\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}2\\5\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}4\\1\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(4) + (5)(1) = 13\), \(|\mathbf{d}_1| = \sqrt{2^2 + 5^2} = \sqrt{29}\), \(|\mathbf{d}_2| = \sqrt{4^2 + 1^2} = \sqrt{17}\)
• \(\cos\theta = \frac{13}{\sqrt{493}}\), so \(\theta = 54.2\ldots^\circ\)
• cos θ is positive, so the angle is acute: \(54.2^\circ\)

Two straight cables on a building site are modelled by the lines below. Positions are in metres from a fixed point. Find the acute angle between the cables.\[ l_1:\ \mathbf{r} = \begin{pmatrix}0\\0\\10\end{pmatrix} + \lambda\begin{pmatrix}6\\2\\-3\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}4\\5\\0\end{pmatrix} + \mu\begin{pmatrix}3\\-2\\6\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}6\\2\\-3\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}3\\-2\\6\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (6)(3) + (2)(-2) + (-3)(6) = -4\), \(|\mathbf{d}_1| = \sqrt{6^2 + 2^2 + (-3)^2} = \sqrt{49} = 7\), \(|\mathbf{d}_2| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{49} = 7\)
• \(\cos\theta = -\frac{4}{49}\), so \(\theta = 94.7\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 94.7^\circ = 85.3^\circ\)

Two straight roads on a map are modelled by the lines below. Distances are in kilometres from the town hall. Find the acute angle at which the roads meet.\[ l_1:\ \mathbf{r} = \begin{pmatrix}1\\3\end{pmatrix} + \lambda\begin{pmatrix}3\\-4\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}8\\1\end{pmatrix} + \mu\begin{pmatrix}5\\12\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}3\\-4\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}5\\12\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(5) + (-4)(12) = -33\), \(|\mathbf{d}_1| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5\), \(|\mathbf{d}_2| = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\)
• \(\cos\theta = -\frac{33}{65}\), so \(\theta = 120.5\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 120.5^\circ = 59.5^\circ\)

The line \(l_1\) passes through \(P(1, 2, -1)\) and \(Q(4, 0, 5)\). The line \(l_2\) has equation \(\mathbf{r} = \begin{pmatrix}2\\2\\0\end{pmatrix} + \mu\begin{pmatrix}1\\-2\\2\end{pmatrix}\). Find the acute angle between \(l_1\) and \(l_2\).

• \(\mathbf{d}_1 = \overrightarrow{PQ} = \begin{pmatrix}3\\-2\\6\end{pmatrix}\), \(\mathbf{d}_2 = \begin{pmatrix}1\\-2\\2\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(1) + (-2)(-2) + (6)(2) = 19\), \(|\mathbf{d}_1| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{49} = 7\), \(|\mathbf{d}_2| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3\)
• \(\cos\theta = \frac{19}{21}\), so \(\theta = 25.2\ldots^\circ\)
• cos θ is positive, so the angle is acute: \(25.2^\circ\)

The lines below are perpendicular. Find the value of \(k\).\[ l_1:\ \mathbf{r} = \begin{pmatrix}2\\1\\4\end{pmatrix} + \lambda\begin{pmatrix}k\\2\\1\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}0\\2\\-1\end{pmatrix} + \mu\begin{pmatrix}1\\3\\-4\end{pmatrix} \]

• Perpendicular lines: \(\mathbf{d}_1 \cdot \mathbf{d}_2 = 0\)
• \(k(1) + (2)(3) + (1)(-4) = 0\)
• \(2 + k = 0\)
• \(k = -2\)