Vectors: skew lines, reflection of a point in a plane, and plane through a line — SAJC 2025 H2 Math Prelim P1

(i) Show that \(l_1\) and \(l_2\) are skew

Skew means the lines are neither parallel nor intersecting, so check the directions, then show the system has no solution.

The direction vectors are not parallel: \[ \begin{pmatrix}-2\\1\\2\end{pmatrix} \neq k\begin{pmatrix}2\\-2\\3\end{pmatrix}\ \text{for any scalar } k, \] so \(l_1\) is not parallel to \(l_2\).

Equating the two lines gives \[ \begin{aligned} 2\lambda + 2\mu &= 1 \quad (1)\\ -\lambda - 2\mu &= -3 \quad (2)\\ -2\lambda + 3\mu &= 8 \quad (3) \end{aligned} \] Solving (1) and (2) gives \(\lambda = -2\), \(\mu = \dfrac{5}{2}\). Checking (3): \[ \text{LHS} = -2(-2) + 3\!\left(\tfrac{5}{2}\right) = \tfrac{7}{2} \neq 8 = \text{RHS}. \] So (3) is not satisfied and the lines do not intersect. Since \(l_1\) and \(l_2\) are non-parallel and non-intersecting, they are skew lines.

(ii) Reflection of \(P\) in \(\pi_1\)

Drop a perpendicular from \(P\) to the plane to find the foot \(F\), then use \(\overrightarrow{OP'} = 2\overrightarrow{OF} - \overrightarrow{OP}\).

The plane \(\pi_1:\ y - z - 2 = 0\) has normal \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\). The line through \(P\) perpendicular to \(\pi_1\) is \[ \mathbf{r} = \begin{pmatrix}1\\5\\0\end{pmatrix} + \alpha\begin{pmatrix}0\\1\\-1\end{pmatrix} = \begin{pmatrix}1\\5+\alpha\\-\alpha\end{pmatrix},\quad\alpha\in\mathbb{R}. \] Substituting into \(\mathbf{r}\cdot\begin{pmatrix}0\\1\\-1\end{pmatrix} = 2\): \[ 5 + 2\alpha = 2 \implies \alpha = -\tfrac{3}{2}, \] so \[ \overrightarrow{OF} = \begin{pmatrix}1\\\tfrac{7}{2}\\\tfrac{3}{2}\end{pmatrix}. \] Then \[ \overrightarrow{OP'} = 2\overrightarrow{OF} - \overrightarrow{OP} = 2\begin{pmatrix}1\\\frac{7}{2}\\\frac{3}{2}\end{pmatrix} - \begin{pmatrix}1\\5\\0\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}. \] The coordinates of \(P'\) are \((1, 2, 3)\).

(iii) Cartesian equation of \(\pi_2\)

A plane containing \(P\) and \(l_1\) has a normal perpendicular to both the direction of \(l_1\) and a vector joining a point on \(l_1\) to \(P\).

Take \(A(1,2,0)\) on \(l_1\). Then \[ \overrightarrow{AP} = \begin{pmatrix}1\\5\\0\end{pmatrix} - \begin{pmatrix}1\\2\\0\end{pmatrix} = \begin{pmatrix}0\\3\\0\end{pmatrix}. \] A normal to \(\pi_2\) is \[ \mathbf{n} = \begin{pmatrix}-2\\1\\2\end{pmatrix}\times\begin{pmatrix}0\\3\\0\end{pmatrix} = \begin{pmatrix}-6\\0\\-6\end{pmatrix} = -6\begin{pmatrix}1\\0\\1\end{pmatrix}. \] Using \(A(1,2,0)\): \[ \mathbf{r}\cdot\begin{pmatrix}1\\0\\1\end{pmatrix} = \begin{pmatrix}1\\2\\0\end{pmatrix}\cdot\begin{pmatrix}1\\0\\1\end{pmatrix} = 1, \] so the cartesian equation of \(\pi_2\) is \(x + z = 1\).