2018 H2 Mathematics Paper 2 Question 3

Vectors II: Lines and Planes

Answers

(i)

D(−5,−4,3){D \left( - 5, - 4, 3 \right)}

(ii)

4x+45y+20z=200{4 x + 45 y + 20 z = 200}

(iii)

58.6∘{58.6^\circ}

(iv)

6.88 units{6.88 \textrm{ units}}

Full solutions

(i)

BC→=OC→−OB→=(−1002)\begin{align*} \overrightarrow{BC} &= \overrightarrow{OC} - \overrightarrow{OB} \\ &= \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix} \\ \end{align*}
Since ABCD{ABCD} is a parallelogram,
AD→=BC→OD→−OA→=BC→OD→=OA→+BC→=(5−41)+(−1002)=(−5−43)\begin{align*} \overrightarrow{AD} &= \overrightarrow{BC} \\ \overrightarrow{OD} - \overrightarrow{OA} &= \overrightarrow{BC} \\ \overrightarrow{OD} &= \overrightarrow{OA} + \overrightarrow{BC} \\ &= \begin{pmatrix} 5 \\ - 4 \\ 1 \end{pmatrix} + \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} - 5 \\ - 4 \\ 3 \end{pmatrix} \end{align*}
Coordinates of D(−5,−4,3)  ■{D \left( - 5, - 4, 3 \right)\; \blacksquare}

(ii)

BE→=OE→−OB→=(−5−410)\begin{align*} \overrightarrow{BE} &= \overrightarrow{OE} - \overrightarrow{OB} \\ &= \begin{pmatrix} - 5 \\ - 4 \\ 10 \end{pmatrix} \\ \end{align*}
nBCE′=BC→×BE→=(−1002)×(−5−410)=(89040)=2(44520)\begin{align*} \mathbf{n'_{BCE}} &= \overrightarrow{BC} \times \overrightarrow{BE} \\ &= \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} - 5 \\ - 4 \\ 10 \end{pmatrix} \\ &= \begin{pmatrix} 8 \\ 90 \\ 40 \end{pmatrix} \\ &= 2 \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix} \end{align*}
r⋅n=a⋅nr⋅n=OB→⋅nr⋅(44520)=(540)⋅(44520)=200\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ \mathbf{r} \cdot \mathbf{n} &= \overrightarrow{OB} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix} &= \begin{pmatrix} 5 \\ 4 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix} \\ &= 200 \end{align*}
Cartesian equation of BCE:{BCE: } 4x+45y+20z=200  ■{4 x + 45 y + 20 z = 200 \; \blacksquare}

(iii)

AB→=OB→−OA→=(08−1)\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} 0 \\ 8 \\ - 1 \end{pmatrix} \\ \end{align*}
nbase′=AB→×BC→=(08−1)×(−1002)=(161080)=2(8540)\begin{align*} \mathbf{n'_{\textrm{base}}} &= \overrightarrow{AB} \times \overrightarrow{BC} \\ &= \begin{pmatrix} 0 \\ 8 \\ - 1 \end{pmatrix} \times \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix} \\ &= \begin{pmatrix} 16 \\ 10 \\ 80 \end{pmatrix} \\ &= 2 \begin{pmatrix} 8 \\ 5 \\ 40 \end{pmatrix} \end{align*}
∣n1⋅n2∣=∣n1∣∣n2∣cos⁡θ\left|\mathbf{n_1} \cdot \mathbf{n_2}\right| = \left| \mathbf{n_1} \right| \left| \mathbf{n_2} \right| \cos \theta
∣(44520)⋅(8540)∣=∣(44520)∣∣(8540)∣cos⁡θ∣1057∣=(2441)(1689)cos⁡θ\begin{align*} \left|\begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 5 \\ 40 \end{pmatrix} \right| &= \left| \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix} \right| \left| \begin{pmatrix} 8 \\ 5 \\ 40 \end{pmatrix} \right| \cos \theta \\ \left|1057 \right| &= (\sqrt{2441}) (\sqrt{1689}) \cos \theta \end{align*}
cos⁡θ=1057(2441)(1689)θ=58.6∘  ■\begin{align*} \cos \theta &= \frac{1057}{(\sqrt{2441})(\sqrt{1689})} \\ \theta &= 58.6^\circ \; \blacksquare \end{align*}

(iv)

Let M{M} denote the midpoint of edge AD{AD}
OM→=OA→+OD→2=(5−41)+(−5−43)2=(0−42)\begin{align*} \overrightarrow{OM} &= \frac{\overrightarrow{OA}+\overrightarrow{OD}}{2} \\ &= \frac{\begin{pmatrix} 5 \\ - 4 \\ 1 \end{pmatrix}+\begin{pmatrix} - 5 \\ - 4 \\ 3 \end{pmatrix}}{2} \\ &= \begin{pmatrix} 0 \\ - 4 \\ 2 \end{pmatrix} \end{align*}
BM→=OM→−OB→=(−5−82)\begin{align*} \overrightarrow{BM} &= \overrightarrow{OM} - \overrightarrow{OB} \\ &= \begin{pmatrix} - 5 \\ - 8 \\ 2 \end{pmatrix} \\ \end{align*}
Shortest distance from M{M} to BCE{BCE}
=∣BM→⋅n^BCE∣=∣(−5−82)⋅(44520)∣∣(44520)∣=∣−20−360+40∣16+2025+400=3402441=6.88 units  ■\begin{align*} & = \left| \overrightarrow{BM} \cdot \mathbf{\hat{n}_{BCE}} \right| \\ &= \frac{\left|\begin{pmatrix} - 5 \\ - 8 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix} \right|}{\left| \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix}\right|} \\ &= \frac{\left|- 20 - 360 + 40 \right|}{\sqrt{16 + 2025 + 400}} \\ &= \frac{340}{\sqrt{2441}} \\ &= 6.88 \textrm{ units} \; \blacksquare \end{align*}