2019 H2 Mathematics Paper 1 Question 12

Vectors II: Lines and Planes

Answers

(i)

Q(811,111,211){Q \left( \frac{8}{11}, \frac{1}{11}, \frac{2}{11} \right)}, R(−3711,−3911,−2311){R \left( - \frac{37}{11}, - \frac{39}{11}, - \frac{23}{11} \right)}

(ii)

cos⁡θ=11213{\cos \theta = \frac{11}{21} \sqrt{3}}, cos⁡β=11255510{\cos \beta = \frac{11}{255} \sqrt{510}}

(iii)

1033 units{\frac{10}{3} \sqrt{3} \textrm{ units}}

(iv)

k=1.86{k=1.86}

(v)

k<1{k < 1}

Full solutions

(i)

Equation of line PQ:{PQ: }
lPQ:r=(224)+λ(−2−3−6),λ∈Rl_{PQ} : \mathbf{r} = \begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} - 2 \\ - 3 \\ - 6 \end{pmatrix}, \lambda \in \mathbb{R}
Equation of top plane: ptop:r⋅(111)=1{p_{\textrm{top}}: \mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 1}
Substituting equation of lPQ{l_{PQ}} into equation of ptop{p_{\textrm{top}}}
(2−2λ2−3λ4−6λ)⋅(111)=1\begin{pmatrix} 2 - 2 \lambda \\ 2 - 3 \lambda \\ 4 - 6 \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 1
2−2λ+2−3λ+4−6λ=12 - 2 \lambda + 2 - 3 \lambda + 4 - 6 \lambda = 1
8−11λ=1−11λ=−7λ=711\begin{align*} 8 - 11 \lambda &= 1 \\ - 11 \lambda &= - 7 \\ \lambda &= \frac{7}{11} \end{align*}
OQ→=(2−2(711)2−3(711)4−6(711))=(811111211)\begin{align*} \overrightarrow{OQ} &= \begin{pmatrix} 2 - 2 (\frac{7}{11}) \\ 2 - 3 (\frac{7}{11}) \\ 4 - 6 (\frac{7}{11}) \end{pmatrix} \\ &= \begin{pmatrix} \frac{8}{11} \\ \frac{1}{11} \\ \frac{2}{11} \end{pmatrix} \end{align*}
Coordinates of Q(811,111,211)  ■{\displaystyle Q \left( \frac{8}{11}, \frac{1}{11}, \frac{2}{11} \right) \; \blacksquare}
Equation of line RS:{RS: }
lRS:r=(−5−6−7)+μ(−2−3−6),μ∈Rl_{RS} : \mathbf{r} = \begin{pmatrix} - 5 \\ - 6 \\ - 7 \end{pmatrix} + \mu \begin{pmatrix} - 2 \\ - 3 \\ - 6 \end{pmatrix}, \mu \in \mathbb{R}
Equation of base plane: pbase:r⋅(111)=−9{p_{\textrm{base}}: \mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = - 9}
Substituting equation of lRS{l_{RS}} into equation of pbase{p_{\textrm{base}}}
(−5−2μ−6−3μ−7−6μ)⋅(111)=−9\begin{pmatrix} - 5 - 2 \mu \\ - 6 - 3 \mu \\ - 7 - 6 \mu \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = - 9
−5−2μ+−6−3μ+−7−6μ=−9- 5 - 2 \mu + - 6 - 3 \mu + - 7 - 6 \mu = - 9
−18−11μ=−9−11μ=9μ=−911\begin{align*} - 18 - 11 \mu &= - 9 \\ - 11 \mu &= 9 \\ \mu &= - \frac{9}{11} \end{align*}
OR→=(−5−2(−911)−6−3(−911)−7−6(−911))=(−3711−3911−2311)\begin{align*} \overrightarrow{OR} &= \begin{pmatrix} - 5 - 2 (- \frac{9}{11}) \\ - 6 - 3 (- \frac{9}{11}) \\ - 7 - 6 (- \frac{9}{11}) \end{pmatrix} \\ &= \begin{pmatrix} - \frac{37}{11} \\ - \frac{39}{11} \\ - \frac{23}{11} \end{pmatrix} \end{align*}
Coordinates of R(−3711,−3911,−2311)  ■{\displaystyle R \left( - \frac{37}{11}, - \frac{39}{11}, - \frac{23}{11} \right) \; \blacksquare}

(ii)

∣dPQ⋅n∣=∣dPQ∣∣n∣cos⁡θ\left|\mathbf{d_{PQ}} \cdot \mathbf{n}\right| = \left| \mathbf{d_{PQ}} \right| \left| \mathbf{n} \right| \cos \theta
∣(−2−3−6)⋅(111)∣=∣(−2−3−6)∣∣(111)∣cos⁡θ∣−11∣=(7)(3)cos⁡θ\begin{align*} \left|\begin{pmatrix} - 2 \\ - 3 \\ - 6 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right| &= \left| \begin{pmatrix} - 2 \\ - 3 \\ - 6 \end{pmatrix} \right| \left| \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right| \cos \theta \\ \left|- 11 \right| &= (7) (\sqrt{3}) \cos \theta \end{align*}
cos⁡θ=1173=11213  ■\begin{align*} \cos \theta &= \frac{11}{7\sqrt{3}} \\ &= \frac{11}{21} \sqrt{3} \; \blacksquare \end{align*}
QR→=OR→−OQ→=(−4511−4011−2511)=−511(985)\begin{align*} \overrightarrow{QR} &= \overrightarrow{OR} - \overrightarrow{OQ} \\ &= \begin{pmatrix} - \frac{45}{11} \\ - \frac{40}{11} \\ - \frac{25}{11} \end{pmatrix} \\ &= - \frac{5}{11} \begin{pmatrix} 9 \\ 8 \\ 5 \end{pmatrix} \end{align*}
∣dQR⋅n∣=∣dQR∣∣n∣cos⁡β\left|\mathbf{d_{QR}} \cdot \mathbf{n}\right| = \left| \mathbf{d_{QR}} \right| \left| \mathbf{n} \right| \cos \beta
∣(985)⋅(111)∣=∣(985)∣∣(111)∣cos⁡β∣22∣=(170)(3)cos⁡β\begin{align*} \left|\begin{pmatrix} 9 \\ 8 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right| &= \left| \begin{pmatrix} 9 \\ 8 \\ 5 \end{pmatrix} \right| \left| \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right| \cos \beta \\ \left|22 \right| &= (\sqrt{170}) (\sqrt{3}) \cos \beta \end{align*}
cos⁡β=22510=11255510  ■\begin{align*} \cos \beta &= \frac{22}{\sqrt{510}} \\ &= \frac{11}{255} \sqrt{510} \; \blacksquare \end{align*}

(iii)

Thickness of prism
=∣QR→⋅n^∣=∣−511(985)⋅(111)∣∣(111)∣=511(∣9+8+5∣1+1+1)=103=1033 units  ■\begin{align*} & = \left| \overrightarrow{QR} \cdot \mathbf{\hat{n}} \right| \\ &= \frac{\left|- \frac{5}{11} \begin{pmatrix} 9 \\ 8 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right|}{\left| \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\right|} \\ &= \frac{5}{11} \left(\frac{\left|9 + 8 + 5 \right|}{\sqrt{1 + 1 + 1}}\right) \\ &= \frac{10}{\sqrt{3}} \\ &= \frac{10}{3} \sqrt{3} \textrm{ units} \; \blacksquare \end{align*}

(iv)

θ=cos⁡−1(11213)=24.870∘\begin{align*} \theta &= \cos^{-1} \left(\frac{11}{21} \sqrt{3}\right) \\ &= 24.870^\circ \\ \end{align*}
β=cos⁡−1(11255510)=13.049∘\begin{align*} \beta &= \cos^{-1} \left(\frac{11}{255} \sqrt{510}\right) \\ &= 13.049^\circ \\ \end{align*}
k=sin⁡θsin⁡β=1.86 (3 s.f.)  ■\begin{align*} k &= \frac{\sin \theta}{\sin \beta} \\ &= 1.86 \textrm{ (3 s.f.)} \; \blacksquare \end{align*}

(v)

If β>θ,{\beta > \theta,} then sin⁡β>sin⁡θ{\sin \beta > \sin \theta} so k=sin⁡θsin⁡β<1  ■{k=\frac{\sin \theta}{\sin \beta} < 1 \; \blacksquare}