2023 H2 Mathematics Paper 1 Question 9

Vectors II: Lines and Planes

Answers

a=2.a=2.
Coordinates of B=(1,3,6).B = \left( 1, 3, 6 \right).
(bi)
2014. \frac{20}{\sqrt{14}}.
(bii)
θ=63.0°. \theta = 63.0 \degree.
3x+4y+z=15.- 3 x + 4 y + z = 15.

Full solutions

(a)

l1:r=(312)+λ(21a)l2:r=(215)+μ(321)\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,B,

(3+2λ1+λ2+aλ)=(2+3μ1+2μ5+μ){ \begin{pmatrix} -3 + 2 \lambda \\ 1 + \lambda \\ 2 + a \lambda \end{pmatrix} = \begin{pmatrix} - 2 + 3 \mu \\ 1 + 2 \mu \\ 5 + \mu \end{pmatrix}}
2λ3μ=1λ2μ=0aλμ=3\begin{alignat}{6} 2 &\lambda& &\,-\,& 3 &\mu& &= 1 \\ &\lambda& &\,-\,& 2 &\mu& &= 0 \\ a &\lambda & &\,-\,& &\mu & & = 3 \end{alignat}

Solving (1)(1) and (2)(2) simultaneously,

λ=2,μ=1{ \lambda = 2, \mu = 1}

Substituting into (3),(3),

2a1=3a=2  \begin{align*} 2a - 1 &= 3 \\ a &= 2 \; \blacksquare \end{align*}
OB=(215)+1(321)=(136)\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*}
Coordinates of B=(1,3,6)  { \text{Coordinates of } B = \left( 1, 3, 6 \right) \; \blacksquare}

(bi)
AB=OBOA=(136)(312)=(424)\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 π1\pi_1 is perpendicular to l2,l_2, the direction vector of l2l_2 is a normal vector of π1\pi_1

Shortest distance from B to π1=ABd2^=(424)(321)(321)=2014 units  \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)
sinθ=2014AB=201442+22+42θ=63.0°  \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*}

(c)

n=(321)×(321)=(341)\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 π2\pi_2 contains l1,l_1, AA lies on π2\pi_2

r(341)=(312)(341)=15\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 π2:\pi_2:

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