2023 H2 Mathematics Paper 1 Question 9

Vectors II: Lines and Planes

Answers

{{a=2.}}

Coordinates of

{{B = \left( 1, 3, 6 \right).}}

{{ \frac{20}{\sqrt{14}}.}}

{{ \theta = 63.0 \degree.}}

{{- 3 x + 4 y + z = 15.}}

\begin{align*} &l_1: \mathbf{r} = \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ a \end{pmatrix} \\ &l_2: \mathbf{r} = \begin{pmatrix} - 2 \\ 1 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \end{align*}

At ${{B,}}$

{ \begin{pmatrix} -3 + 2 \lambda \\ 1 + \lambda \\ 2 + a \lambda \end{pmatrix} = \begin{pmatrix} - 2 + 3 \mu \\ 1 + 2 \mu \\ 5 + \mu \end{pmatrix}}

\begin{alignat}{6} 2 &\lambda& &\,-\,& 3 &\mu& &= 1 \\ &\lambda& &\,-\,& 2 &\mu& &= 0 \\ a &\lambda & &\,-\,& &\mu & & = 3 \end{alignat}

Solving ${{(1)}}$ and ${{(2)}}$ simultaneously,

{ \lambda = 2, \mu = 1}

Substituting into ${{(3),}}$

\begin{align*} 2a - 1 &= 3 \\ a &= 2 \; \blacksquare \end{align*}

\begin{align*} \overrightarrow{OB} &= \begin{pmatrix} - 2 \\ 1 \\ 5 \end{pmatrix} + 1 \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 \\ 3 \\ 6 \end{pmatrix} \end{align*}

{ \text{Coordinates of } B = \left( 1, 3, 6 \right) \; \blacksquare}

(bi)

\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} 1 \\ 3 \\ 6 \end{pmatrix} - \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} \end{align*}

Since ${{\pi_1}}$ is perpendicular to ${{l_2,}}$ the direction vector of ${{l_2}}$ is a normal vector of ${{\pi_1}}$

\begin{align*} & \text{Shortest distance from } B \text{ to } \pi_1 \\ &= \left| \overrightarrow{AB} \cdot \hat{\mathbf{d}_2} \right| \\ &= \left| \frac{\begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}}{\left| \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \right|} \right| \\ &= \frac{20}{\sqrt{14}} \text{ units} \; \blacksquare \end{align*}

(bii)

\begin{align*} \sin \theta &= \frac{\frac{20}{\sqrt{14}}}{\left| \overrightarrow{AB} \right|} \\ &= \frac{\frac{20}{\sqrt{14}}}{\sqrt{4^2 + 2^2 + 4^2}} \\ \theta &= 63.0 \degree \; \blacksquare \end{align*}

\begin{align*} \mathbf{n} &= \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} - 3 \\ 4 \\ 1 \end{pmatrix} \end{align*}

Since ${{\pi_2}}$ contains ${{l_1,}}$ ${{A}}$ lies on ${{\pi_2}}$

\begin{align*} \mathbf{r} \cdot \begin{pmatrix} - 3 \\ 4 \\ 1 \end{pmatrix} &= \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} - 3 \\ 4 \\ 1 \end{pmatrix} \\ &= 15 \end{align*}

Cartesian equation of ${{\pi_2:}}$

{ - 3 x + 4 y + z = 15 \; \blacksquare }