The acute angle between two lines Problem set B One question at a time

Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}3\\-2\\5\end{pmatrix} + \lambda\begin{pmatrix}2\\3\\-6\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}1\\1\\0\end{pmatrix} + \mu\begin{pmatrix}3\\-4\\12\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}2\\3\\-6\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}3\\-4\\12\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(3) + (3)(-4) + (-6)(12) = -78\), \(|\mathbf{d}_1| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{49} = 7\), \(|\mathbf{d}_2| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{169} = 13\)
• \(\cos\theta = -\frac{78}{91}\), so \(\theta = 149.0\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 149.0^\circ = 31.0^\circ\)

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

• \(\mathbf{d}_1 = \begin{pmatrix}1\\-3\\2\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}4\\2\\3\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (1)(4) + (-3)(2) + (2)(3) = 4\), \(|\mathbf{d}_1| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{14}\), \(|\mathbf{d}_2| = \sqrt{4^2 + 2^2 + 3^2} = \sqrt{29}\)
• \(\cos\theta = \frac{4}{\sqrt{406}}\), so \(\theta = 78.5\ldots^\circ\)
• cos θ is positive, so the angle is acute: \(78.5^\circ\)

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

• \(\mathbf{d}_1 = \begin{pmatrix}4\\-1\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}-3\\2\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (4)(-3) + (-1)(2) = -14\), \(|\mathbf{d}_1| = \sqrt{4^2 + (-1)^2} = \sqrt{17}\), \(|\mathbf{d}_2| = \sqrt{(-3)^2 + 2^2} = \sqrt{13}\)
• \(\cos\theta = -\frac{14}{\sqrt{221}}\), so \(\theta = 160.3\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 160.3^\circ = 19.7^\circ\)

Find the acute angle between the lines\[ l_1:\ \mathbf{r} = \begin{pmatrix}-4\\1\end{pmatrix} + \lambda\begin{pmatrix}3\\1\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}0\\-6\end{pmatrix} + \mu\begin{pmatrix}1\\5\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}3\\1\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}1\\5\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (3)(1) + (1)(5) = 8\), \(|\mathbf{d}_1| = \sqrt{3^2 + 1^2} = \sqrt{10}\), \(|\mathbf{d}_2| = \sqrt{1^2 + 5^2} = \sqrt{26}\)
• \(\cos\theta = \frac{8}{2 \sqrt{65}}\), so \(\theta = 60.3\ldots^\circ\)
• cos θ is positive, so the angle is acute: \(60.3^\circ\)

Two straight pipes in a factory are modelled by the lines below. Positions are in metres from a corner of the floor. Find the acute angle between the pipes.\[ l_1:\ \mathbf{r} = \begin{pmatrix}2\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\6\\3\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}0\\4\\5\end{pmatrix} + \mu\begin{pmatrix}-6\\-3\\2\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}2\\6\\3\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}-6\\-3\\2\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(-6) + (6)(-3) + (3)(2) = -24\), \(|\mathbf{d}_1| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7\), \(|\mathbf{d}_2| = \sqrt{(-6)^2 + (-3)^2 + 2^2} = \sqrt{49} = 7\)
• \(\cos\theta = -\frac{24}{49}\), so \(\theta = 119.3\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 119.3^\circ = 60.7^\circ\)

Two straight rail tracks on a plan are modelled by the lines below. Distances are in kilometres from the station. Find the acute angle at which the tracks meet.\[ l_1:\ \mathbf{r} = \begin{pmatrix}2\\0\end{pmatrix} + \lambda\begin{pmatrix}8\\15\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}10\\-4\end{pmatrix} + \mu\begin{pmatrix}4\\-3\end{pmatrix} \]

• \(\mathbf{d}_1 = \begin{pmatrix}8\\15\end{pmatrix},\ \mathbf{d}_2 = \begin{pmatrix}4\\-3\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (8)(4) + (15)(-3) = -13\), \(|\mathbf{d}_1| = \sqrt{8^2 + 15^2} = \sqrt{289} = 17\), \(|\mathbf{d}_2| = \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5\)
• \(\cos\theta = -\frac{13}{85}\), so \(\theta = 98.8\ldots^\circ\)
• Obtuse, so the acute angle is \(180^\circ - 98.8^\circ = 81.2^\circ\)

The line \(l_1\) passes through \(P(0, 3, 2)\) and \(Q(2, -1, 8)\). The line \(l_2\) has equation \(\mathbf{r} = \begin{pmatrix}1\\0\\5\end{pmatrix} + \mu\begin{pmatrix}-4\\-1\\2\end{pmatrix}\). Find the acute angle between \(l_1\) and \(l_2\).

• \(\mathbf{d}_1 = \overrightarrow{PQ} = \begin{pmatrix}2\\-4\\6\end{pmatrix}\), \(\mathbf{d}_2 = \begin{pmatrix}-4\\-1\\2\end{pmatrix}\)
• \(\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(-4) + (-4)(-1) + (6)(2) = 8\), \(|\mathbf{d}_1| = \sqrt{2^2 + (-4)^2 + 6^2} = \sqrt{56}\), \(|\mathbf{d}_2| = \sqrt{(-4)^2 + (-1)^2 + 2^2} = \sqrt{21}\)
• \(\cos\theta = \frac{8}{14 \sqrt{6}}\), so \(\theta = 76.5\ldots^\circ\)
• cos θ is positive, so the angle is acute: \(76.5^\circ\)

The lines below are perpendicular. Find the value of \(k\).\[ l_1:\ \mathbf{r} = \begin{pmatrix}2\\1\\4\end{pmatrix} + \lambda\begin{pmatrix}2\\k\\-1\end{pmatrix} \]\[ l_2:\ \mathbf{r} = \begin{pmatrix}4\\0\\2\end{pmatrix} + \mu\begin{pmatrix}1\\2\\5\end{pmatrix} \]

• Perpendicular lines: \(\mathbf{d}_1 \cdot \mathbf{d}_2 = 0\)
• \((2)(1) + k(2) + (-1)(5) = 0\)
• \(-3 + 2k = 0\)
• \(k = \frac{3}{2}\)