Shortest distance from a point to a line Problem set B Worked solutions

The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix}-2\\2\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\0\end{pmatrix}\). The point \(A\) has coordinates \((4, 0, -5)\). The point \(B\) lies on \(l\) and \(AB\) is perpendicular to \(l\).

1. (a) Find the coordinates of \(B\).

   • \(\overrightarrow{AB} = \begin{pmatrix}-6 + 2\lambda\\2 + \lambda\\5\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}2\\1\\0\end{pmatrix} = 0\) gives \(-10 + 5\lambda = 0\)
   • \(\lambda = 2\), so \(B\left(2, 4, 0\right)\)

2. (b) Hence find the shortest distance from \(A\) to \(l\).

   • \(\overrightarrow{AB} = \begin{pmatrix}-2\\4\\5\end{pmatrix}\), \(|\overrightarrow{AB}| = 3 \sqrt{5} = 6.71\)

Find the shortest distance from the point \(A(0, 1, 4)\) to the line \(\mathbf{r} = \begin{pmatrix}1\\5\\4\end{pmatrix} + \lambda\begin{pmatrix}0\\-1\\3\end{pmatrix}\).

• \(\overrightarrow{AB} = \begin{pmatrix}1\\4 - \lambda\\3\lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}0\\-1\\3\end{pmatrix} = 0\) gives \(-4 + 10\lambda = 0\)
• \(\lambda = \frac{2}{5}\), so \(B\left(1, \frac{23}{5}, \frac{26}{5}\right)\)
• \(\overrightarrow{AB} = \begin{pmatrix}1\\\frac{18}{5}\\\frac{6}{5}\end{pmatrix}\), \(|\overrightarrow{AB}| = \frac{\sqrt{385}}{5} = 3.92\)

Find the shortest distance from the point \(P(0, 4, 2)\) to the line \(\mathbf{r} = (4\mathbf{i} - 4\mathbf{j}) + \lambda(2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).

• Column form: \(\mathbf{r} = \begin{pmatrix}4\\-4\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\2\\2\end{pmatrix}\). \(\overrightarrow{PB} = \begin{pmatrix}4 + 2\lambda\\-8 + 2\lambda\\-2 + 2\lambda\end{pmatrix}\), and \(\overrightarrow{PB} \cdot \begin{pmatrix}2\\2\\2\end{pmatrix} = 0\) gives \(-12 + 12\lambda = 0\)
• \(\lambda = 1\), so \(B\left(6, -2, 2\right)\)
• \(\overrightarrow{PB} = \begin{pmatrix}6\\-6\\0\end{pmatrix}\), \(|\overrightarrow{PB}| = 6 \sqrt{2} = 8.49\)

Find the shortest distance from the point \(Q(1, 1, 3)\) to the line \(\mathbf{r} = (\mathbf{j}) + \lambda(\mathbf{j} + 2\mathbf{k})\).

• Column form: \(\mathbf{r} = \begin{pmatrix}0\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}0\\1\\2\end{pmatrix}\). \(\overrightarrow{QB} = \begin{pmatrix}-1\\\lambda\\-3 + 2\lambda\end{pmatrix}\), and \(\overrightarrow{QB} \cdot \begin{pmatrix}0\\1\\2\end{pmatrix} = 0\) gives \(-6 + 5\lambda = 0\)
• \(\lambda = \frac{6}{5}\), so \(B\left(0, \frac{11}{5}, \frac{12}{5}\right)\)
• \(\overrightarrow{QB} = \begin{pmatrix}-1\\\frac{6}{5}\\- \frac{3}{5}\end{pmatrix}\), \(|\overrightarrow{QB}| = \frac{\sqrt{70}}{5} = 1.67\)

A glider flies along a straight path. Its position vector, in kilometres from a launch field, is \(\mathbf{r} = \begin{pmatrix}3\\5\\3\end{pmatrix} + \lambda\begin{pmatrix}0\\-3\\1\end{pmatrix}\), where \(\lambda\) is the time in hours. A ground marker is at \(A(5, 1, 3)\). Find the shortest distance between the glider's path and the marker, in kilometres.

• \(\overrightarrow{AB} = \begin{pmatrix}-2\\4 - 3\lambda\\\lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}0\\-3\\1\end{pmatrix} = 0\) gives \(-12 + 10\lambda = 0\)
• \(\lambda = \frac{6}{5}\), so \(B\left(3, \frac{7}{5}, \frac{21}{5}\right)\)
• \(\overrightarrow{AB} = \begin{pmatrix}-2\\\frac{2}{5}\\\frac{6}{5}\end{pmatrix}\), \(|\overrightarrow{AB}| = \frac{2 \sqrt{35}}{5} = 2.37\)

A helicopter flies along a straight path. Its position vector, in kilometres from a base, is \(\mathbf{r} = \begin{pmatrix}6\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}0\\-1\\2\end{pmatrix}\), where \(\lambda\) is the time in hours after take-off. A beacon is at \(A(1, 4, 2)\). The helicopter is closest to the beacon at the point \(B\).

1. (a) Find the value of \(\lambda\) when the helicopter is at \(B\).

   • \(\overrightarrow{AB} = \begin{pmatrix}5\\-3 - \lambda\\-2 + 2\lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}0\\-1\\2\end{pmatrix} = 0\) gives \(-1 + 5\lambda = 0\)
   • \(\lambda = \frac{1}{5}\), so \(B\left(6, \frac{4}{5}, \frac{2}{5}\right)\)

2. (b) Find the shortest distance from \(A\) to the path, in kilometres.

   • \(\overrightarrow{AB} = \begin{pmatrix}5\\- \frac{16}{5}\\- \frac{8}{5}\end{pmatrix}\), \(|\overrightarrow{AB}| = \frac{3 \sqrt{105}}{5} = 6.15\)

The line \(l\) passes through \(P(2, 1, 0)\) and \(Q(3, -1, 2)\). Find the shortest distance from \(A(4, -2, 5)\) to \(l\).

• \(\mathbf{d} = \overrightarrow{PQ} = \begin{pmatrix}1\\-2\\2\end{pmatrix}\). A line through \(P\): \(\mathbf{r} = \begin{pmatrix}2\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}\)
• \(\overrightarrow{AB} = \begin{pmatrix}-2 + \lambda\\3 - 2\lambda\\-5 + 2\lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}1\\-2\\2\end{pmatrix} = 0\) gives \(-18 + 9\lambda = 0\)
• \(\lambda = 2\), so \(B\left(4, -3, 4\right)\)
• \(\overrightarrow{AB} = \begin{pmatrix}0\\-1\\-1\end{pmatrix}\), \(|\overrightarrow{AB}| = \sqrt{2} = 1.41\)