2011 H2 Mathematics Paper 1 Question 11

Vectors II: Lines and Planes

Answers

(i)

x+y+2z=−3{x + y + 2 z = - 3}

(ii)

k=−7{k=- 7}

(iii)

(−1,6,−4){\left( - 1, 6, - 4 \right)}

(iv)

22.2∘{22.2^\circ}

Full solutions

(i)

Let A,B,C{A,B,C} denote the points (4,−1,−3){\left( 4, - 1, - 3 \right)}, (−2,−5,2){\left( - 2, - 5, 2 \right)}, and (4,−3,−2){\left( 4, - 3, - 2 \right)}
AB→=OB→−OA→=(−2−52)−(4−1−3)=(−6−45)\begin{align*} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} - 2 \\ - 5 \\ 2 \end{pmatrix} - \begin{pmatrix} 4 \\ - 1 \\ - 3 \end{pmatrix} \\ &= \begin{pmatrix} - 6 \\ - 4 \\ 5 \end{pmatrix} \end{align*}
AC→=OC→−OA→=(4−3−2)−(4−1−3)=(0−21)\begin{align*} \overrightarrow{AC} &= \overrightarrow{OC} - \overrightarrow{OA} \\ &= \begin{pmatrix} 4 \\ - 3 \\ - 2 \end{pmatrix} - \begin{pmatrix} 4 \\ - 1 \\ - 3 \end{pmatrix} \\ &= \begin{pmatrix} 0 \\ - 2 \\ 1 \end{pmatrix} \end{align*}
n′=AB→×AC→=(−6−45)×(0−21)=(6612)=6(112)\begin{align*} \mathbf{n'} &= \overrightarrow{AB} \times \overrightarrow{AC} \\ &= \begin{pmatrix} - 6 \\ - 4 \\ 5 \end{pmatrix} \times \begin{pmatrix} 0 \\ - 2 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 6 \\ 6 \\ 12 \end{pmatrix} \\ &= 6 \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \end{align*}
r⋅n=OA→⋅nr⋅(112)=(4−1−3)⋅(112)r⋅(112)=−3\begin{align*} \mathbf{r} \cdot \mathbf{n} &= \overrightarrow{OA} \cdot \mathbf{n} \\ \mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} &= \begin{pmatrix} 4 \\ - 1 \\ - 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ \mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} &= - 3 \end{align*}
Cartesian equation of p:x+y+2z=−3  ■{p: x + y + 2 z = - 3 \; \blacksquare}

(ii)

Equating the equations of l1{l_1} and l2,{l_2,}
(1+2λ2−4λ−3+λ)=(−2+μ1+5μ3+kλ)\begin{pmatrix} 1 + 2 \lambda \\ 2 - 4 \lambda \\ - 3 + \lambda \end{pmatrix} = \begin{pmatrix}- 2 + \mu\\1 + 5 \mu\\3+k\lambda\end{pmatrix}
1+2λ=−2+μ2−4λ=1+5μ−3+λ=3+kλ\begin{align} \qquad 1 + 2 \lambda &= - 2 + \mu \\ \qquad 2 - 4 \lambda &= 1 + 5 \mu \\ \qquad - 3 + \lambda &= 3 + k \lambda \end{align}
Solving equations (1) and (2) simultaneously,
λ=−1,μ=1\lambda = - 1, \mu = 1
Substituting into (3),
−3−1=3+kk=−7  ■\begin{align*} -3-1 &= 3 + k \\ k &= -7 \; \blacksquare \end{align*}

(iii)

Substituting equation of l1{l_1} into equation of p{p}
LHS=(1+2λ2−4λ−3+λ)⋅(112)=1+2λ+2−4λ+2(−3+λ)=−3=RHS\begin{align*} & \textrm{LHS} \\ = & \begin{pmatrix} 1 + 2 \lambda \\ 2 - 4 \lambda \\ - 3 + \lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ = & 1 + 2 \lambda + 2 - 4 \lambda + 2 (- 3 + \lambda) \\ = & - 3 \\ = & \textrm{RHS} \end{align*}
Hence l{l} lies in p  ■{p \; \blacksquare}
Let X{X} denote the point of intersection of l2{l_2} and p{p}
Substituting equation of l2{l_2} into equation of p{p}
(−2+μ1+5μ3−7μ)⋅(112)=−3\begin{pmatrix} - 2 + \mu \\ 1 + 5 \mu \\ 3 - 7 \mu \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = - 3
−2+μ+1+5μ+2(3−7μ)=−3- 2 + \mu + 1 + 5 \mu + 2 (3 - 7 \mu) = - 3
5−8λ=−3−8λ=−8λ=1\begin{align*} 5 - 8 \lambda &= - 3 \\ - 8 \lambda &= - 8 \\ \lambda &= 1 \\ \end{align*}
OX→=(−2+(1)1+5(1)3−7(1))=(−16−4)\begin{align*} \overrightarrow{OX} &= \begin{pmatrix} - 2 + (1) \\ 1 + 5 (1) \\ 3 - 7 (1) \end{pmatrix} \\ &= \begin{pmatrix} - 1 \\ 6 \\ - 4 \end{pmatrix} \end{align*}
Coordinates of point of intersection: X(−1,6,−4)  ■{X \left( - 1, 6, - 4 \right) \; \blacksquare}

(iv)

∣d2⋅n∣=∣d2∣∣n∣sin⁡θ\left|\mathbf{d_2} \cdot \mathbf{n}\right| = \left| \mathbf{d_2} \right| \left| \mathbf{n} \right| \sin \theta
∣(15−7)⋅(112)∣=∣(15−7)∣∣(112)∣sin⁡θ∣−8∣=(53)(6)sin⁡θ\begin{align*} \left|\begin{pmatrix} 1 \\ 5 \\ - 7 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right| &= \left| \begin{pmatrix} 1 \\ 5 \\ - 7 \end{pmatrix} \right| \left| \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right| \sin \theta \\ \left|- 8 \right| &= (5 \sqrt{3}) (\sqrt{6}) \sin \theta \end{align*}
sin⁡θ=8(53)(6)θ=22.2∘  ■\begin{align*} \sin \theta &= \frac{8}{(5 \sqrt{3})(\sqrt{6})} \\ \theta &= 22.2^\circ \; \blacksquare \end{align*}