2022TrialShort answer

Q15 (c) (i)

2 marks

Question

The line has vector equation
$\vec r=\lambda\begin{bmatrix}1\\3\\1\end{bmatrix}.$
For $P(x_1,y_1,z_1)$ , the point $Q$ on the line is such that $PQ$ is a minimum. Using vector projection methods, show that
$\overrightarrow{OQ}=\frac{x_1+3y_1+z_1}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}.$

Try the question before revealing the answer.

Solution

$\overrightarrow{OQ}=\operatorname{proj}_{\mathbf b}\overrightarrow{OP}$

$\overrightarrow{OP}=\begin{bmatrix}x_1\\y_1\\z_1\end{bmatrix},\qquad \mathbf b=\begin{bmatrix}1\\3\\1\end{bmatrix}\text{ which is a vector on }r$

$\overrightarrow{OQ}=\operatorname{proj}_{\mathbf b}\overrightarrow{OP}$
$=\frac{\overrightarrow{OP}\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\times\mathbf b$
$=\frac{(1)(x_1)+(3)(y_1)+(1)(z_1)}{1^2+3^2+1^2}\begin{bmatrix}1\\3\\1\end{bmatrix}$
$=\frac{x_1+3y_1+z_1}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}$

1 – correct projection formula and application
1 – correct answer

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Syllabus alignment

Vectors

Vectors and vector equations of linesUse coordinates and vector equations for curves, lines, line segments and spheres in two and three dimensions.
Further operations with three-dimensional vectorsUse vector magnitude, unit vectors and scalar products to solve problems and prove geometric results.

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