2016 H2 Mathematics Paper 1 Question 11

Vectors II: Lines and Planes

Answers

(ia)
λ=−89,{\lambda = - \frac{8}{9},} μ=1918,{\mu = \frac{19}{18},} t=−59{t = - \frac{5}{9}}
(ib)
−2x+y+2z=35{- 2 x + y + 2 z = 35} and −2x+y+2z=−37{- 2 x + y + 2 z = - 37}

(ii)

a=92{a = \frac{9}{2}}

Full solutions

(ia)
Let n{\mathbf{n}} denote the normal vector of p{p}
n′=d1×d2=(120)×(04−2)=(−424)=2(−212)\begin{align*} \mathbf{n'} &= \mathbf{d_1} \times \mathbf{d_2} \\ &= \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 4 \\ - 2 \end{pmatrix} \\ &= \begin{pmatrix} - 4 \\ 2 \\ 4 \end{pmatrix} \\ &= 2 \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix} \end{align*}
n=(−212)\mathbf{n} = \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}
Hence the normal vector of p{p} is parallel to the direction vector of l{l}
Hence l{l} is perpendicular to p  ■{p \; \blacksquare}
Equating the equations of p{p} and l,{l,}
(1+λ−3+2λ+4μ2−2λ)=(−1−2tt1+2t)\begin{pmatrix}1+\lambda\\-3+2\lambda+4\mu\\2-2\lambda\end{pmatrix} = \begin{pmatrix} - 1 - 2 t \\ t \\ 1 + 2 t \end{pmatrix}
λ+2t=−22λ+4μ−t=3−2μ−2t=−1\begin{align} \qquad \lambda + 2t &= -2 \\ \qquad 2\lambda + 4\mu - t &= 3 \\ \qquad -2\mu - 2t &= -1 \\ \end{align}
Solving with a GC,
λ=−89,{\lambda = - \frac{8}{9},} μ=1918,{\mu = \frac{19}{18},} t=−59  ■{t = - \frac{5}{9} \; \blacksquare}
(ib)
r⋅n=a⋅nr⋅(−212)=(1−32)⋅(−212)=−1\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \mathbf{a} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix} &= \begin{pmatrix} 1 \\ - 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix} \\ &= - 1 \end{align*}

p:r⋅(−212)=−1{p: \mathbf{r} \cdot \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix} = - 1}
r⋅(−212)∣(−212)∣=−1∣(−212)∣{\mathbf{r}\cdot \frac{\begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}}{\left|\begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}\right|} = \frac{- 1}{\left|\begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix}\right|}}
r⋅n^=−13{\mathbf{r}\cdot\mathbf{\hat{n}} = -\frac{1}{3}}
Hence the displacement from the origin to p{p} is −13{-\frac{1}{3}} units
The required planes 12 units from p{p} will then have a displacement of −13±12{-\frac{1}{3}\pm 12} units from the origin

Hence they have equations r⋅n^=353{\mathbf{r} \cdot \mathbf{\hat{n}} = \frac{35}{3}} and r⋅n^=−373{\mathbf{r} \cdot \mathbf{\hat{n}} = - \frac{37}{3}}
The cartesian equation of the required planes 12 units from p{p} are −2x+y+2z=35{- 2 x + y + 2 z = 35} and −2x+y+2z=−37{- 2 x + y + 2 z = - 37}

(ii)

n=(120)×(a4−2)=(−424−2a)\begin{align*} \mathbf{n} &= \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix}a\\4\\-2\end{pmatrix} \\ &= \begin{pmatrix}-4\\2\\4-2a\end{pmatrix} \end{align*}
If l{l} and p{p} do not meet in a unique point, then l{l} is parallel to p{p}
Hence the direction vector of l{l} is perpendicular to the normal vector of p{p}
d⋅n=0(−212)⋅(−424−2a)=08+2+8−4a=0a=92  ■\begin{gather*} \mathbf{d} \cdot \mathbf{n} = 0 \\ \begin{pmatrix} - 2 \\ 1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix}-4\\2\\4-2a\end{pmatrix} = 0 \\ 8+2+8-4a = 0 \\ a = \frac{9}{2} \; \blacksquare \end{gather*}