Specimen 2017 H2 Mathematics Paper 1 Question 10

Vectors II: Lines and Planes

Answers

{2 x - 3 y + z = - 9.}

{P \left( - 2, 2, 1 \right).}

{\frac{x + 2}{6} = \frac{y - 2}{-11} = \frac{z - 1}{-3}.}

l: \mathbf{r} = \begin{pmatrix} 0 \\ - 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix}

Since

{\pi}

is perpendicular to

{l,}

{\begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix}}

is its normal vector.

\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} &= \begin{pmatrix} - 1 \\ 2 \\ - 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \\ &= - 9 \\ \end{align*}

Hence the cartesian equation of

{\pi}

{2 x - 3 y + z = - 9 \; \blacksquare}

The point

{P}

{l}

closest to

{A}

is the point of intersection between

{l}

and

{\pi.}

Substituting the equation of ${l}$ into the equation of ${\pi,}$

\begin{align*} \begin{pmatrix} 2 \lambda \\ - 1 - 3 \lambda \\ 2 + \lambda \end{pmatrix} \cdot \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} &= - 9 \\ 2 (2 \lambda) - 3 (- 1 - 3 \lambda) + 2 + \lambda &= - 9 \end{align*}

\begin{align*} 5 + 14 \lambda &= - 9 \\ 14 \lambda &= - 14 \\ \lambda &= - 1 \end{align*}

\begin{align*} \overrightarrow{OP} &= \begin{pmatrix} 2 (- 1) \\ - 1 - 3 (- 1) \\ 2 + (- 1) \end{pmatrix} \\ &= \begin{pmatrix} - 2 \\ 2 \\ 1 \end{pmatrix} \end{align*}

Coordinates of point

{P \left( - 2, 2, 1 \right) \; \blacksquare}

Let

{F}

denote the foot of perpendicular from

{B \left( 4, - 5, 10 \right)}

to line

{l}

and

{B'}

denote the point obtained when

{B}

is reflected about line

{l.}

\begin{align*} \overrightarrow{PB} &= \overrightarrow{OB} - \overrightarrow{OP} \\ &= \begin{pmatrix} 6 \\ - 7 \\ 9 \end{pmatrix} \end{align*}

\begin{align*} \overrightarrow{PF} &= \left(\overrightarrow{PB} \cdot \mathbf{\hat{d}}\right) \mathbf{\hat{d}} \\ \overrightarrow{OF}-\overrightarrow{OP} &= \left( \frac{\begin{pmatrix} 6 \\ - 7 \\ 9 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix}}{\left| \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \right|} \right) \frac{\begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix}}{\left| \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \right|} \\ \overrightarrow{OF} &= \begin{pmatrix} - 2 \\ 2 \\ 1 \end{pmatrix} + \frac{42}{14} \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ - 7 \\ 4 \end{pmatrix} \end{align*}

\begin{align*} \overrightarrow{OF} &= \frac{\overrightarrow{OB}+\overrightarrow{OB'}}{2} \\ \overrightarrow{OB'} &= 2\overrightarrow{OF} - \overrightarrow{OB} \\ &= 2\begin{pmatrix} 4 \\ - 7 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ - 5 \\ 10 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ - 9 \\ - 2 \end{pmatrix} \end{align*}

\begin{align*} \overrightarrow{PB'} &= \overrightarrow{OB'} - \overrightarrow{OP} \\ &= \begin{pmatrix} 6 \\ - 11 \\ - 3 \end{pmatrix} \end{align*}

The line

{n}

passes through

{P}

and has direction vector

{\overrightarrow{PB'}.}

n: \mathbf{r} = \begin{pmatrix} - 2 \\ 2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 6 \\ - 11 \\ - 3 \end{pmatrix}, \; \mu \in \mathbb{R}

Cartesian equation of

{n:}

\frac{x + 2}{6} = \frac{y - 2}{-11} = \frac{z - 1}{-3} \; \blacksquare