2015 H2 Mathematics Paper 2 Question 2

Vectors II: Lines and Planes

Answers

{73.4^\circ}

Position vector of points

{\sqrt{33}}

from

{P}

{(\frac{13}{7} \mathbf{i} - \frac{5}{7} \mathbf{j} - \frac{46}{7} \mathbf{k})}

and

{(3 \mathbf{i} + \mathbf{j} - 10 \mathbf{k})}

Position vector of point on

{L}

closest to

{P}

{\frac{1}{7} \left( 17 \mathbf{i} + \mathbf{j} - 58 \mathbf{k} \right)}

{- 36 x + 2 y - 11 z = 4}

\left|\mathbf{d_L} \cdot \mathbf{d_x}\right| = \left| \mathbf{d_L} \right| \left| \mathbf{d_x} \right| \cos \theta

\begin{align*} \left|\begin{pmatrix} 2 \\ 3 \\ - 6 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right| &= \left| \begin{pmatrix} 2 \\ 3 \\ - 6 \end{pmatrix} \right| \left| \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right| \cos \theta \\ \left|2 \right| &= (7) (1) \cos \theta \end{align*}

\begin{align*} \cos \theta &= \frac{2}{(7)(1)} \\ \theta &= 73.4^\circ \; \blacksquare \end{align*}

Let

{A}

denote the points required on

{L}

\overrightarrow{OA} = \begin{pmatrix} 1 + 2 \lambda \\ - 2 + 3 \lambda \\ - 4 - 6 \lambda \end{pmatrix}

\begin{align*} \overrightarrow{AP} &= \overrightarrow{OP} - \overrightarrow{OA} \\ &= \begin{pmatrix} 2 \\ 5 \\ - 6 \end{pmatrix} - \begin{pmatrix} 1 + 2 \lambda \\ - 2 + 3 \lambda \\ - 4 - 6 \lambda \end{pmatrix} \\ &= \begin{pmatrix}1 - 2 \lambda\\7 - 3 \lambda\\- 2 + 6 \lambda\end{pmatrix} \end{align*}

\left| \overrightarrow{AP} \right| = \sqrt{33}

\begin{gather*} (1 - 2 \lambda)^2 + (7 - 3 \lambda)^2 + (- 2 + 6 \lambda)^2 = 33 \\ (1 - 4 \lambda + 4 \lambda^2) + (49 - 42 \lambda + 9 \lambda^2) + (4 - 24 \lambda + 36 \lambda^2) = 33 \end{gather*}

\begin{align*} 21 - 70 \lambda + 49 \lambda^2 &= 0 \\ 3 - 10 \lambda + 7 \lambda^2 &= 0 \\ (7 \lambda - 3)(\lambda - 1) &= 0 \end{align*}

\lambda = \frac{3}{7} \quad \textrm{or} \quad \lambda = 1

\overrightarrow{OA} = \begin{pmatrix}1+2\left(\frac{3}{7}\right)\\-2+3\left(\frac{3}{7}\right)\\-4-6\left(\frac{3}{7}\right)\end{pmatrix} \quad \textrm{or} \quad \overrightarrow{OA} = \begin{pmatrix}1+2\left(1\right)\\-2+3\left(1\right)\\-4-6\left(1\right)\end{pmatrix}

\overrightarrow{OA} = \begin{pmatrix}\frac{13}{7}\\- \frac{5}{7}\\- \frac{46}{7}\end{pmatrix} \quad \textrm{or} \quad \overrightarrow{OA} = \begin{pmatrix}3\\1\\- 10\end{pmatrix} \; \blacksquare

We observe that the point on

{L}

closest to

{P}

is the mid-point of the two points found above

Let this point be denoted

{B}

\begin{align*} \overrightarrow{OB} &= \frac{\overrightarrow{OA}_1 + \overrightarrow{OA}_2}{2} \\ &= \frac{\begin{pmatrix} \frac{13}{7} \\ - \frac{5}{7} \\ - \frac{46}{7} \end{pmatrix}+\begin{pmatrix} 3 \\ 1 \\ - 10 \end{pmatrix}}{2} \\ &= \begin{pmatrix} \frac{17}{7} \\ \frac{1}{7} \\ - \frac{58}{7} \end{pmatrix} \\ &= \frac{1}{7} \begin{pmatrix} 17 \\ 1 \\ - 58 \end{pmatrix} \; \blacksquare \end{align*}

Let

{C \left( 1, - 2, - 4 \right)}

denote the point on

{L}

\begin{align*} \overrightarrow{PC} &= \overrightarrow{OC} - \overrightarrow{OP} \\ &= \begin{pmatrix} - 1 \\ - 7 \\ 2 \end{pmatrix} \\ \end{align*}

\begin{align*} \mathbf{n} &= \mathbf{d_L} \times \overrightarrow{PC} \\ &= \begin{pmatrix} 2 \\ 3 \\ - 6 \end{pmatrix} \times \begin{pmatrix} - 1 \\ - 7 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} - 36 \\ 2 \\ - 11 \end{pmatrix} \\ \end{align*}

\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} - 36 \\ 2 \\ - 11 \end{pmatrix} &= \begin{pmatrix} 2 \\ 5 \\ - 6 \end{pmatrix} \cdot \begin{pmatrix} - 36 \\ 2 \\ - 11 \end{pmatrix} \\ &= 4 \end{align*}

Cartesian equation of plane that includes

{L}

and

{P: }

{- 36 x + 2 y - 11 z = 4 \; \blacksquare}