Vectors: ASRJC 2024 Promo Q10

(i)

The lines are coplanar $\Rightarrow$ The lines are intersecting lines since they are not parallel.

Let $$\begin{pmatrix}1-\lambda\\2\\5+3\lambda\end{pmatrix} =\begin{pmatrix}a+\mu\\9+7\mu\\9+b\mu\end{pmatrix}$$ By comparing rows, $$\begin{aligned} 2&=9+7\mu &&\Rightarrow &&\mu=-1\\ 1-\lambda&=a+\mu &&\Rightarrow &&\lambda=2-a &&\dots\text{Eq(1)}\\ 5+3\lambda&=9+b\mu &&\Rightarrow &&\lambda=\frac{4-b}{3} &&\dots\text{Eq(2)} \end{aligned}$$ Eq(1) $=$ Eq(2): $$\begin{aligned} 2-a&=\frac{4-b}{3}\\ \Rightarrow\quad6-3a&=4-b\\ \Rightarrow\quad3a&=b+2\qquad(\text{Shown}) \end{aligned}$$

(ii)

Angle between the two lines is $\cos^{-1}\dfrac{11}{\sqrt{660}}$ $$\Rightarrow\theta=\cos^{-1}\left|\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}\right|$$ $$\Rightarrow\cos^{-1}\frac{11}{\sqrt{660}} =\cos^{-1}\left| \frac{ \begin{pmatrix}-1\\0\\3\end{pmatrix}\cdot \begin{pmatrix}1\\7\\b\end{pmatrix}} {\left|\begin{pmatrix}1\\0\\-3\end{pmatrix}\right| \left|\begin{pmatrix}1\\7\\b\end{pmatrix}\right|} \right|$$ $$\begin{aligned} \therefore\frac{11}{\sqrt{660}} &=\frac{|3b-1|}{\sqrt{1+9}\sqrt{1+49+b^2}}\\ \frac{11}{6}&=\frac{1-6b+9b^2}{50+b^2}\\ 43b^2-36b-544&=0\\ b&=\frac{36\pm\sqrt{1296+93568}}{86}=4\quad(\text{reject negative value of }b) \end{aligned}$$

(iii)

A normal for $\Pi_2$ is $$\begin{aligned} \begin{pmatrix}1\\0\\-3\end{pmatrix}\times \left[\begin{pmatrix}2\\1\\3\end{pmatrix}-\begin{pmatrix}1\\2\\5\end{pmatrix}\right] &=\begin{pmatrix}1\\0\\-3\end{pmatrix}\times\begin{pmatrix}1\\-1\\-2\end{pmatrix}\\ &=\begin{pmatrix}-3\\-1\\-1\end{pmatrix} =-\begin{pmatrix}3\\1\\1\end{pmatrix} \end{aligned}$$

$$\begin{aligned} \mathbf r\cdot\begin{pmatrix}3\\1\\1\end{pmatrix} &=\begin{pmatrix}1\\2\\5\end{pmatrix}\cdot\begin{pmatrix}3\\1\\1\end{pmatrix} =3+2+5=10 \end{aligned}$$ $$\therefore\Pi_2:\ \mathbf r\cdot\begin{pmatrix}3\\1\\1\end{pmatrix}=10$$

(iv)

$$\begin{aligned} \text{Angle required} &=\cos^{-1}\left| \frac{\begin{pmatrix}3\\-1\\1\end{pmatrix}\cdot\begin{pmatrix}3\\1\\1\end{pmatrix}} {\sqrt{11}\sqrt{11}}\right|\\ &=\cos^{-1}\left|\frac9{11}\right|=0.613\text{ rad (3 sf) or }35.1^\circ\text{ (nearest }0.1^\circ\text{)} \end{aligned}$$

(v)

$$\Pi_1:\ \mathbf r\cdot\frac{\begin{pmatrix}3\\-1\\1\end{pmatrix}}{\sqrt{11}} =\frac6{\sqrt{11}}<\frac7{\sqrt{11}}$$ Since $\Pi_1$ and $\Pi_3$ are parallel, then $$\Pi_3:\ \mathbf r\cdot\frac{\begin{pmatrix}3\\-1\\1\end{pmatrix}}{\sqrt{11}}=k \quad\text{where }k\text{ is a real constant.}$$

Since $\Pi_3$ is closer to the origin to $\Pi_1$, then $$k=-\left(\frac7{\sqrt{11}}-\frac6{\sqrt{11}}\right)=\frac{-1}{\sqrt{11}}$$ $$\Rightarrow\Pi_3:\ \mathbf r\cdot\frac{\begin{pmatrix}3\\-1\\1\end{pmatrix}}{\sqrt{11}} =\frac{-1}{\sqrt{11}}$$ Hence the cartesian equation for $\Pi_3$ is $3x-y+z=-1$