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

The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix}-2\\5\\-4\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\1\end{pmatrix}\). The point \(A\) has coordinates \((2, -3, 4)\). The point \(B\) lies on \(l\) and \(AB\) is perpendicular to \(l\).

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

   • \(\overrightarrow{AB} = \begin{pmatrix}-4 + 2\lambda\\8 - \lambda\\-8 + \lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}2\\-1\\1\end{pmatrix} = 0\) gives \(-24 + 6\lambda = 0\)
   • \(\lambda = 4\), so \(B\left(6, 1, 0\right)\)

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

   • \(\overrightarrow{AB} = \begin{pmatrix}4\\4\\-4\end{pmatrix}\), \(|\overrightarrow{AB}| = 4 \sqrt{3} = 6.93\)

Find the shortest distance from the point \(A(3, 1, 5)\) to the line \(\mathbf{r} = \begin{pmatrix}6\\2\\0\end{pmatrix} + \lambda\begin{pmatrix}0\\2\\1\end{pmatrix}\).

• \(\overrightarrow{AB} = \begin{pmatrix}3\\1 + 2\lambda\\-5 + \lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}0\\2\\1\end{pmatrix} = 0\) gives \(-3 + 5\lambda = 0\)
• \(\lambda = \frac{3}{5}\), so \(B\left(6, \frac{16}{5}, \frac{3}{5}\right)\)
• \(\overrightarrow{AB} = \begin{pmatrix}3\\\frac{11}{5}\\- \frac{22}{5}\end{pmatrix}\), \(|\overrightarrow{AB}| = \frac{\sqrt{830}}{5} = 5.76\)

Find the shortest distance from the point \(P(3, 0, 6)\) to the line \(\mathbf{r} = (5\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}) + \lambda(-\mathbf{i} + 2\mathbf{j} + \mathbf{k})\).

• Column form: \(\mathbf{r} = \begin{pmatrix}5\\-4\\-2\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\1\end{pmatrix}\). \(\overrightarrow{PB} = \begin{pmatrix}2 - \lambda\\-4 + 2\lambda\\-8 + \lambda\end{pmatrix}\), and \(\overrightarrow{PB} \cdot \begin{pmatrix}-1\\2\\1\end{pmatrix} = 0\) gives \(-18 + 6\lambda = 0\)
• \(\lambda = 3\), so \(B\left(2, 2, 1\right)\)
• \(\overrightarrow{PB} = \begin{pmatrix}-1\\2\\-5\end{pmatrix}\), \(|\overrightarrow{PB}| = \sqrt{30} = 5.48\)

Find the shortest distance from the point \(Q(5, 3, 9)\) to the line \(\mathbf{r} = (4\mathbf{i} + 2\mathbf{k}) + \lambda(2\mathbf{i} + \mathbf{k})\).

• Column form: \(\mathbf{r} = \begin{pmatrix}4\\0\\2\end{pmatrix} + \lambda\begin{pmatrix}2\\0\\1\end{pmatrix}\). \(\overrightarrow{QB} = \begin{pmatrix}-1 + 2\lambda\\-3\\-7 + \lambda\end{pmatrix}\), and \(\overrightarrow{QB} \cdot \begin{pmatrix}2\\0\\1\end{pmatrix} = 0\) gives \(-9 + 5\lambda = 0\)
• \(\lambda = \frac{9}{5}\), so \(B\left(\frac{38}{5}, 0, \frac{19}{5}\right)\)
• \(\overrightarrow{QB} = \begin{pmatrix}\frac{13}{5}\\-3\\- \frac{26}{5}\end{pmatrix}\), \(|\overrightarrow{QB}| = \frac{\sqrt{1070}}{5} = 6.54\)

A hot-air balloon travels along a straight path. Its position vector, in kilometres from the launch site, is \(\mathbf{r} = \begin{pmatrix}2\\2\\2\end{pmatrix} + \lambda\begin{pmatrix}-2\\2\\0\end{pmatrix}\), where \(\lambda\) is the time in hours. A weather station stands at \(A(0, 1, 4)\). Find the shortest distance between the balloon's path and the station, in kilometres.

• \(\overrightarrow{AB} = \begin{pmatrix}2 - 2\lambda\\1 + 2\lambda\\-2\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}-2\\2\\0\end{pmatrix} = 0\) gives \(-2 + 8\lambda = 0\)
• \(\lambda = \frac{1}{4}\), so \(B\left(\frac{3}{2}, \frac{5}{2}, 2\right)\)
• \(\overrightarrow{AB} = \begin{pmatrix}\frac{3}{2}\\\frac{3}{2}\\-2\end{pmatrix}\), \(|\overrightarrow{AB}| = \frac{\sqrt{34}}{2} = 2.92\)

An aircraft flies along a straight path. Its position vector, in kilometres from an airport, is \(\mathbf{r} = \begin{pmatrix}5\\0\\1\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\0\end{pmatrix}\), where \(\lambda\) is the time in hours after noon. A radar station is at \(A(4, 8, 2)\). The aircraft is closest to the station at the point \(B\).

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

   • \(\overrightarrow{AB} = \begin{pmatrix}1 + 3\lambda\\-8 + \lambda\\-1\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}3\\1\\0\end{pmatrix} = 0\) gives \(-5 + 10\lambda = 0\)
   • \(\lambda = \frac{1}{2}\), so \(B\left(\frac{13}{2}, \frac{1}{2}, 1\right)\)

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

   • \(\overrightarrow{AB} = \begin{pmatrix}\frac{5}{2}\\- \frac{15}{2}\\-1\end{pmatrix}\), \(|\overrightarrow{AB}| = \frac{\sqrt{254}}{2} = 7.97\)

The line \(l\) passes through \(P(1, 0, 2)\) and \(Q(3, 2, 3)\). Find the shortest distance from \(A(5, 3, 6)\) to \(l\).

• \(\mathbf{d} = \overrightarrow{PQ} = \begin{pmatrix}2\\2\\1\end{pmatrix}\). A line through \(P\): \(\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + \lambda\begin{pmatrix}2\\2\\1\end{pmatrix}\)
• \(\overrightarrow{AB} = \begin{pmatrix}-4 + 2\lambda\\-3 + 2\lambda\\-4 + \lambda\end{pmatrix}\), and \(\overrightarrow{AB} \cdot \begin{pmatrix}2\\2\\1\end{pmatrix} = 0\) gives \(-18 + 9\lambda = 0\)
• \(\lambda = 2\), so \(B\left(5, 4, 4\right)\)
• \(\overrightarrow{AB} = \begin{pmatrix}0\\1\\-2\end{pmatrix}\), \(|\overrightarrow{AB}| = \sqrt{5} = 2.24\)