N2023 P1 Q9
N2023 P1 Q9
Junior College 2
13 marks
Free
The line \(l_1\) contains the point \(A\) with coordinates \((-3,1,2)\) and is parallel to the vector \(\begin{pmatrix}2\\1\\a\end{pmatrix}\), where \(a\) is a constant. The line \(l_2\) has equation \(\frac{x+2}{3}=\frac{y-1}{2}=z-5\). It is given that \(l_1\) and \(l_2\) cross at the point \(B\).
- Find the value of \(a\) and the coordinates of \(B\).[5]
- The plane \(\pi_1\) contains the point \(A\) and is perpendicular to \(l_2\).
- Find the shortest distance from \(B\) to \(\pi_1\).[3]
- Hence find the acute angle between \(l_1\) and \(\pi_1\).[2]
- The plane \(\pi_2\) is perpendicular to \(\pi_1\) and contains \(l_1\). Find a cartesian equation of \(\pi_2\).[3]
Default solution
- \(l_1:\boldsymbol r=\begin{pmatrix}-3\\ 1\\ 2\end{pmatrix}+t\begin{pmatrix}2\\ 1\\ a\end{pmatrix},\qquad l_2:\boldsymbol r=\begin{pmatrix}-2\\ 1\\ 5\end{pmatrix}+s\begin{pmatrix}3\\ 2\\ 1\end{pmatrix}\)\(-3+2t=-2+3s,\qquad1+t=1+2s\)\(t\)\({}=2s\)\(-3+4s\)\({}=-2+3s\)\(s\)\({}=1,\qquad t=2\)\(2+2a=5+1\Longrightarrow a=2\)\(B(1,3,6)\)
- \(\boldsymbol n_1=\begin{pmatrix}3\\ 2\\ 1\end{pmatrix}\)\(\pi_1:\quad3(x+3)+2(y-1)+(z-2)=0\)\(\pi_1:\quad3x+2y+z=-5\)\(d(B,\pi_1)=\frac{|3(1)+2(3)+6+5|}{\sqrt{3^2+2^2+1^2}}=\frac{20}{\sqrt{14}}\text{ units}\)
- Let \(H\) be the perpendicular foot from \(B\) to \(\pi_1\). The acute angle is \(\theta=\angle BAH\).\(AB=\sqrt{(1+3)^2+(3-1)^2+(6-2)^2}=6\)\(\sin\theta=\frac{BH}{AB}=\frac{10}{3\sqrt{14}}\)\(\theta=62.98287678\ldots{}^\circ\approx63.0^\circ\)
- \(\boldsymbol d_1=\begin{pmatrix}2\\ 1\\ 2\end{pmatrix},\qquad\boldsymbol n_2=\boldsymbol d_1\times\boldsymbol n_1=\begin{pmatrix}-3\\ 4\\ 1\end{pmatrix}\)\(\pi_2:\quad-3(x+3)+4(y-1)+(z-2)=0\)\(\pi_2:\quad-3x+4y+z=15\)
Answer:(a) \(a=2\), \(B(1,3,6)\). (b)(i) \(\frac{20}{\sqrt{14}}\) units. (b)(ii) \(63.0^\circ\). (c) \(-3x+4y+z=15\).
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