AI HL · AHL 4
0 of 160 skills done
Shortest distance from a point to a line
AHL 4.16
Drag P. Slide R along the line r = a + λd until PR is perpendicular to the line. Find it by hand first, then press the button to check.
Click P, then use the arrow keys to move it one unit. Click the slider and use the arrow keys to move R.
Slide R along the line.
The line \(l\) has equation \(\mathbf{r} = \begin{pmatrix}1\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\2\end{pmatrix}\). The point \(A\) has coordinates \((-1, -3, 2)\). The point \(B\) lies on \(l\) and \(AB\) is perpendicular to \(l\). Find the coordinates of \(B\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{OB} = \begin{pmatrix}1\\0\\-3\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\2\end{pmatrix} = \begin{pmatrix}1 + 2\lambda\\-\lambda\\-3 + 2\lambda\end{pmatrix}\)
B is on the line, so its position vector has this form for some value of λ.
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}2 + 2\lambda\\3 - \lambda\\-5 + 2\lambda\end{pmatrix}\)
End point minus start point.
\(\overrightarrow{AB} \cdot \begin{pmatrix}2\\-1\\2\end{pmatrix} = 0 \Rightarrow 2(2 + 2\lambda) - (3 - \lambda) + 2(-5 + 2\lambda) = 0 \Rightarrow -9 + 9\lambda = 0\)
The shortest distance is along the perpendicular, so AB · d = 0. This gives one equation in λ.
\(\lambda = 1\)
Substitute back to find B.
\(B\left(3, -1, -1\right)\)
Put λ into the general point.
Find the shortest distance from the point \(A(2, 1, 3)\) to the line \(\mathbf{r} = \begin{pmatrix}0\\2\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}\).
Have a go first. Then open the working one line at a time.
\(\overrightarrow{OB} = \begin{pmatrix}0\\2\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix} = \begin{pmatrix}\lambda\\2 - 2\lambda\\-1 + 2\lambda\end{pmatrix}\)
B is on the line, so its position vector has this form for some value of λ.
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}-2 + \lambda\\1 - 2\lambda\\-4 + 2\lambda\end{pmatrix}\)
End point minus start point.
\(\overrightarrow{AB} \cdot \begin{pmatrix}1\\-2\\2\end{pmatrix} = 0 \Rightarrow (-2 + \lambda) - 2(1 - 2\lambda) + 2(-4 + 2\lambda) = 0 \Rightarrow -12 + 9\lambda = 0\)
The shortest distance is along the perpendicular, so AB · d = 0. This gives one equation in λ.
\(\lambda = \frac{4}{3}\)
Substitute back to find B.
\(\overrightarrow{AB} = \begin{pmatrix}- \frac{2}{3}\\- \frac{5}{3}\\- \frac{4}{3}\end{pmatrix},\ |\overrightarrow{AB}| = \sqrt{5} = 2.24\)
Put λ into AB and find its length. That length is the shortest distance.
Find the shortest distance from the point \(P(1, 2, 4)\) to the line \(\mathbf{r} = (4\mathbf{i} - \mathbf{j}) + \lambda(-3\mathbf{i} - \mathbf{j} + 2\mathbf{k})\).
Have a go first. Then open the working one line at a time.
\(4\mathbf{i} - \mathbf{j} + \lambda(-3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) \Rightarrow \mathbf{r} = \begin{pmatrix}4\\-1\\0\end{pmatrix} + \lambda\begin{pmatrix}-3\\-1\\2\end{pmatrix}\)
Write the line as a column first. A missing component is 0.
\(\overrightarrow{OB} = \begin{pmatrix}4\\-1\\0\end{pmatrix} + \lambda\begin{pmatrix}-3\\-1\\2\end{pmatrix} = \begin{pmatrix}4 - 3\lambda\\-1 - \lambda\\2\lambda\end{pmatrix}\)
B is on the line, so its position vector has this form for some value of λ.
\(\overrightarrow{PB} = \overrightarrow{OB} - \overrightarrow{OP} = \begin{pmatrix}3 - 3\lambda\\-3 - \lambda\\-4 + 2\lambda\end{pmatrix}\)
End point minus start point.
\(\overrightarrow{PB} \cdot \begin{pmatrix}-3\\-1\\2\end{pmatrix} = 0 \Rightarrow -3(3 - 3\lambda) - (-3 - \lambda) + 2(-4 + 2\lambda) = 0 \Rightarrow -14 + 14\lambda = 0\)
The shortest distance is along the perpendicular, so AB · d = 0. This gives one equation in λ.
\(\lambda = 1\)
Substitute back to find B.
\(\overrightarrow{PB} = \begin{pmatrix}0\\-4\\-2\end{pmatrix},\ |\overrightarrow{PB}| = 2 \sqrt{5} = 4.47\)
Put λ into AB and find its length. That length is the shortest distance.
A drone flies along a straight line. Its position vector, in kilometres from a base, is \(\mathbf{r} = \begin{pmatrix}0\\6\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\1\end{pmatrix}\), where \(\lambda\) is the time in hours. A radio mast stands at \(A(4, 4, 0)\). Find the shortest distance between the drone's path and the mast, in kilometres.
Have a go first. Then open the working one line at a time.
\(\overrightarrow{OB} = \begin{pmatrix}0\\6\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\1\end{pmatrix} = \begin{pmatrix}2\lambda\\6 - \lambda\\1 + \lambda\end{pmatrix}\)
B is on the line, so its position vector has this form for some value of λ.
\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}-4 + 2\lambda\\2 - \lambda\\1 + \lambda\end{pmatrix}\)
End point minus start point.
\(\overrightarrow{AB} \cdot \begin{pmatrix}2\\-1\\1\end{pmatrix} = 0 \Rightarrow 2(-4 + 2\lambda) - (2 - \lambda) + (1 + \lambda) = 0 \Rightarrow -9 + 6\lambda = 0\)
The shortest distance is along the perpendicular, so AB · d = 0. This gives one equation in λ.
\(\lambda = \frac{3}{2}\)
Substitute back to find B.
\(\overrightarrow{AB} = \begin{pmatrix}-1\\\frac{1}{2}\\\frac{5}{2}\end{pmatrix},\ |\overrightarrow{AB}| = \frac{\sqrt{30}}{2} = 2.74\)
Put λ into AB and find its length. That length is the shortest distance.