2014 H2 Mathematics Paper 1 Question 9

Vectors II: Lines and Planes

Answers

(i)

−x+2y+z=0{- x + 2 y + z = 0}

(ii)

m:r=(630)+μ(412),μ∈R{m: \mathbf{r} = \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}, \mu \in \mathbb{R}}

(iii)

(187,157,−127){\left( \frac{18}{7}, \frac{15}{7}, - \frac{12}{7} \right)}

Full solutions

(i)

l:r=(1−13)+λ(2−14),λ∈Rl: \mathbf{r} = \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ - 1 \\ 4 \end{pmatrix}, \lambda \in \mathbb{R}
Since p{p} and q{q} are perpendicular, the normal vector of p{p} is a direction vector of q{q}
nq′=dl×np=(2−14)×(12−3)=(−5105)=5(−121)\begin{align*} \mathbf{n_q'} &= \mathbf{d_l} \times \mathbf{n_p} \\ &= \begin{pmatrix} 2 \\ - 1 \\ 4 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ - 3 \end{pmatrix} \\ &= \begin{pmatrix} - 5 \\ 10 \\ 5 \end{pmatrix} \\ &= 5 \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}
r⋅n=a⋅nr⋅(−121)=(1−13)⋅(−121)=0\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} &= \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} - 1 \\ 2 \\ 1 \end{pmatrix} \\ &= 0 \end{align*}
Cartesian equation of q:{q:} −x+2y+z=0  ■{- x + 2 y + z = 0 \; \blacksquare}

(ii)

p:x+2y−3z=12q:−x+2y+z=0\begin{align} \qquad &p: x + 2 y - 3 z = 12 \\ \qquad &q: - x + 2 y + z = 0 \end{align}
Using a GC, equation for m:{m:}
r=(630)+μ(412),μ∈R  ■\mathbf{r} = \begin{pmatrix} 6 \\ 3 \\ 0 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}, \mu \in \mathbb{R} \; \blacksquare

(iii)

OB→=(6+4μ3+μ2μ){\overrightarrow{OB} = \begin{pmatrix}6+4\mu\\3+\mu\\2\mu\end{pmatrix}}
AB→=OB→−OA→=(6+4μ3+μ2μ)−(1−13)=(5+4μ4+μ−3+2μ)\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix}6+4\mu\\3+\mu\\2\mu\end{pmatrix} - \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix}5+4\mu\\4+\mu\\-3+2\mu\end{pmatrix} \end{align*}
∣AB→∣2=(5+4μ)2+(4+μ)2+(−3+2μ)2=(25+40μ+16μ2)+(16+8μ+μ2)+(9−12μ+4μ2)=50+36μ+21μ2  ■\begin{align*} & \left|\overrightarrow{AB}\right|^2 \\ = & (5+4\mu)^2+(4+\mu)^2+(-3+2\mu)^2 \\ =& (25 + 40 \mu + 16 \mu^2) + (16 + 8 \mu + \mu^2) + (9 - 12 \mu + 4 \mu^2) \\ =& 50 + 36 \mu + 21 \mu^2 \; \blacksquare \end{align*}
Let y=50+36μ+21μ2{y = 50 + 36 \mu + 21 \mu^2}
When point on m{m} is nearest tto A{A}, y{y} is a minimum
Hence dydμ=0{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}\mu} = 0}
dydμ=42μ+36{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}\mu} = 42\mu + 36}
When dydμ=0,{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}\mu} = 0,} μ=−67{\displaystyle \mu = -\frac{6}{7}}
d2ydμ2=42>0{\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}\mu^2} = 42 > 0} ⇒y{\Rightarrow y} is a minimum
When μ=−67,{\mu = -\frac{6}{7},}
OB→=(6+4(−67)3−672(−67)){\overrightarrow{OB} = \begin{pmatrix}6+4(-\frac{6}{7})\\3-\frac{6}{7}\\2(-\frac{6}{7})\end{pmatrix}}
OB→=(187157−127){\overrightarrow{OB} = \begin{pmatrix}\frac{18}{7}\\\frac{15}{7}\\- \frac{12}{7}\end{pmatrix}}
Coordinates of point on m{m} nearest to A{A}: (187,157,−127)  ■{\displaystyle \left( \frac{18}{7}, \frac{15}{7}, - \frac{12}{7} \right)\; \blacksquare}