System of Linear Equations: Two planes with parameter, line of intersection — ACJC 2025 H2 Math Prelim Paper 1
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Question
The planes \(\pi_1\) and \(\pi_2\) have equations \(3x+c(y+z)-2=0\) and \[ \mathbf{r}=(\mathbf{i}+3\mathbf{j}-2\mathbf{k}) +s(2\mathbf{i}-\mathbf{j}+3\mathbf{k}) +t(\mathbf{i}-\mathbf{k}) \] respectively, where \(c\) is a constant, and \(s\) and \(t\) are parameters. The point \(A(1,3,-2)\) lies in both planes.
- Show that \(c=-1\).
- Show that the vector equation of the line of intersection of \(\pi_1\) and \(\pi_2\), line \(l\), is given by \(\mathbf{r}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\alpha(\mathbf{i}-\mathbf{j}+4\mathbf{k})\), where \(\alpha\) is a parameter.
- Find the position vectors of the points on the line \(l\) which are a distance of \(3\sqrt{2}\) from the point \(B(2,-3,7)\).
- Find the equation of the plane \(\pi_3\) which is parallel to \(\pi_2\) and contains the point \(B\). Hence show that the distance between the planes \(\pi_2\) and \(\pi_3\) is \(\dfrac{20}{3\sqrt{3}}\).
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(a)
Substituting \(A(1,3,-2)\) into \(3x+c(y+z)-2=0\): \[ 3(1)+c(3-2)-2 = 0 \implies c+1=0\,. \]
\(c = -1\).
(b)
With \(c=-1\), \(\pi_1: 3x-(y+z)-2=0\), i.e. \(\mathbf{r}\cdot\begin{pmatrix}3\\-1\\-1\end{pmatrix}=2\).
The direction vectors of \(\pi_2\) are \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\).
Substituting \(\mathbf{r}=\begin{pmatrix}1+2s+t\\3-s\\-2+3s-t\end{pmatrix}\) into \(\mathbf{r}\cdot\begin{pmatrix}3\\-1\\-1\end{pmatrix}=2\): \[\begin{aligned} 3(1+2s+t)-(3-s)-(-2+3s-t) &= 2\\ 4s+4t &= 0 \implies s=-t \end{aligned}\]
So the line of intersection has \(\mathbf{r}=\begin{pmatrix}1-t\\3+t\\-2-4t\end{pmatrix} = \begin{pmatrix}1\\3\\-2\end{pmatrix}+t\begin{pmatrix}-1\\1\\-4\end{pmatrix} = \begin{pmatrix}1\\3\\-2\end{pmatrix}+\alpha\begin{pmatrix}1\\-1\\4\end{pmatrix}\) (setting \(\alpha=-t\)).
\(\mathbf{r} = \mathbf{i}+3\mathbf{j}-2\mathbf{k}+\alpha(\mathbf{i}-\mathbf{j}+4\mathbf{k})\)(c)
A general point on \(l\) has position vector \(\overrightarrow{OP}=\begin{pmatrix}1+\alpha\\3-\alpha\\-2+4\alpha\end{pmatrix}\).
\[ \overrightarrow{BP} = \begin{pmatrix}1+\alpha\\3-\alpha\\-2+4\alpha\end{pmatrix}-\begin{pmatrix}2\\-3\\7\end{pmatrix} = \begin{pmatrix}-1+\alpha\\6-\alpha\\-9+4\alpha\end{pmatrix} \]Setting \(|\overrightarrow{BP}|=3\sqrt{2}\): \[\begin{aligned} (-1+\alpha)^2+(6-\alpha)^2+(-9+4\alpha)^2 &= 18\\ 18\alpha^2-86\alpha+118 &= 18\\ 9\alpha^2-43\alpha+50 &= 0\\ \alpha = 2 \quad &\text{or} \quad \alpha = \frac{25}{9} \end{aligned}\]
Points: \(\begin{pmatrix}3\\1\\6\end{pmatrix}\) and \(\dfrac{1}{9}\begin{pmatrix}34\\2\\82\end{pmatrix}\)
(d)
\(\pi_2\) has normal \(\mathbf{n}_2=\begin{pmatrix}1\\5\\1\end{pmatrix}\) (from the cross product of its direction vectors, or read from the alternative Cartesian form \(x+5y+z=14\)). \(\pi_3\parallel\pi_2\) so has the same normal.
\(\pi_3\) passes through \(B(2,-3,7)\): \[ \mathbf{r}\cdot\begin{pmatrix}1\\5\\1\end{pmatrix} = 2-15+7 = -6\,. \]
Distance between \(\pi_2\) and \(\pi_3\): \[ \frac{|14-(-6)|}{\sqrt{1^2+5^2+1^2}} = \frac{20}{\sqrt{27}} = \frac{20}{3\sqrt{3}}\,. \]