2012 H2 Mathematics Paper 1 Question 9

Vectors II: Lines and Planes

Answers

(i)

l:r=(789)+λ(121),λ∈R{l: \mathbf{r} = \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}}

(ii)

ON→=(547){\overrightarrow{ON}=\begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix}}
AN:NB=1:3{AN:NB = 1:3}

(iii)

x−7−1=y−84=z−9−1{\displaystyle \frac{x - 7}{-1} = \frac{y - 8}{4} = \frac{z - 9}{-1}}

Full solutions

(i)

Let A,B{A,B} denote the points (7,8,9){\left( 7, 8, 9 \right)} and (−1,−8,1){\left( - 1, - 8, 1 \right)}.
AB→=OB→−OA→=(−1−81)−(789)=(−8−16−8)=−8(121)\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 1 \\ - 8 \\ 1 \end{pmatrix} - \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} \\ &= \begin{pmatrix} - 8 \\ - 16 \\ - 8 \end{pmatrix} \\ &= - 8 \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}
Equation of l:{l:}
l:r=(789)+λ(121),λ∈R  ■l: \mathbf{r} = \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R} \; \blacksquare

(ii)

AC→=OC→−OA→=(−60−6)\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} - 6 \\ 0 \\ - 6 \end{pmatrix}
AN→=(AC→⋅d^)d^\overrightarrow{AN} = \left(\mathbf{\overrightarrow{AC}} \cdot \mathbf{\hat{d}}\right) \mathbf{\hat{d}}
ON→−OA→=((−60−6)⋅(121)∣(121)∣)(121)∣(121)∣\overrightarrow{ON}-\overrightarrow{OA} = \left( \frac{\begin{pmatrix} - 6 \\ 0 \\ - 6 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}}{\left| \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \right|} \right) \frac{\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}}{\left| \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \right|}
ON→=(789)+−126(121)\overrightarrow{ON} = \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} + \frac{- 12}{6} \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}
ON→=(547)  ■\overrightarrow{ON} = \begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix}\; \blacksquare
AN→=ON→−OA→=(−2−4−2)=−2(121)\begin{align*} \overrightarrow{AN} &= \overrightarrow{ON} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 2 \\ - 4 \\ - 2 \end{pmatrix} \\ &= - 2 \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}
BN→=ON→−OB→=(6126)=6(121)\begin{align*} \overrightarrow{BN} &= \overrightarrow{ON} - \overrightarrow{OB} \\ &= \begin{pmatrix} 6 \\ 12 \\ 6 \end{pmatrix} \\ &= 6 \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \end{align*}
Hence AN:BN=1:3  ■{AN:BN = 1:3 \; \blacksquare}

(iii)

Let C′{C'} denote the point obtained when C{C} is reflected in the line AB{AB}
By ratio theorem,
ON→=OC→+OC′→2{\overrightarrow{ON} = \frac{\overrightarrow{OC}+\overrightarrow{OC'}}{2}}
OC′→=2ON→−OC→{\overrightarrow{OC'} = 2\overrightarrow{ON} - \overrightarrow{OC}}
OC′→=2(547)−(183){\overrightarrow{OC'} = 2\begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 1 \\ 8 \\ 3 \end{pmatrix}}
OC′→=(9011){\overrightarrow{OC'} = \begin{pmatrix} 9 \\ 0 \\ 11 \end{pmatrix}}
We observe that A{A} is the point of intersection between the two lines
AC′→=OC′→−OA→=(2−82)=2(1−41)\begin{align*} \overrightarrow{AC'} &= \overrightarrow{OC'} - \overrightarrow{OA} \\ &= \begin{pmatrix} 2 \\ - 8 \\ 2 \end{pmatrix} \\ &= 2 \begin{pmatrix} 1 \\ - 4 \\ 1 \end{pmatrix} \end{align*}
lAC′:r=(789)+ν(−14−1),ν∈Rl_{AC'}: \mathbf{r} = \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} + \nu \begin{pmatrix} - 1 \\ 4 \\ - 1 \end{pmatrix}, \nu \in \mathbb{R}
Cartesian equation of line of reflection: x−7−1=y−84=z−9−1  ■{\displaystyle \frac{x - 7}{-1} = \frac{y - 8}{4} = \frac{z - 9}{-1} \; \blacksquare}