2024 ASRJC Promo Q10

For positive real numbers \(a\) and \(b\), the lines \(l_1\) and \(l_2\) have equations given by

\({l_1:\ \mathbf r=\begin{pmatrix}1\\2\\5\end{pmatrix}+\lambda\begin{pmatrix}-1\\0\\3\end{pmatrix} \text{ for}\hspace{0.5em} \lambda\in\mathbb R, \text{ and}\hspace{0.5em} l_2:\ \mathbf r= \begin{pmatrix}a\\9\\9\end{pmatrix}+\mu\begin{pmatrix}1\\7\\b\end{pmatrix} \text{ for}\hspace{0.5em} \mu\in\mathbb R.}\)

It is given that the two lines intersect each other.

1. Show that \(3a=b+2\).[2]
2. Find the value of \(b\) if the angle between the two lines is \(\cos^{-1}\!\left(\dfrac{11}{\sqrt{660}}\right)\) radians.[3]

The equation of the plane \({{\Pi }_{1}}\) that contains both lines is now given by \(\mathbf{r}\cdot \left( \begin{matrix} \begin{matrix} 3 \\ -1 \\ 1 \\ \end{matrix} \\ \end{matrix} \right)=6\).

3. A second plane \({{\Pi }_{2}}\) meets \({{\Pi }_{1}}\) in the line \({{l}_{1}}\) and contains the point \(\left( 2,1,3 \right)\). Find an equation for \({{\Pi }_{2}}\) in scalar product form.[2]
4. Hence find the angle between the planes \({{\Pi }_{1}}\) and \({{\Pi }_{2}}\).[2]
5. The plane \({{\Pi }_{3}}\) is parallel to the plane \({{\Pi }_{1}}\). Given that the distance between both planes is \(\frac{7}{\sqrt{11}}\) units and that \({{\Pi }_{3}}\) is closer to the origin than \({{\Pi }_{1}}\), find a cartesian equation for \({{\Pi }_{3}}\).[3]