Vectors: Skew lines; reflection; plane — SAJC 2025 H2 Math Prelim Paper 1

(i) \[ l_1:\;\mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + \lambda\begin{pmatrix}-2\\1\\2\end{pmatrix},\qquad l_2:\;\mathbf{r} = \begin{pmatrix}0\\5\\-8\end{pmatrix} + \mu\begin{pmatrix}2\\-2\\3\end{pmatrix},\quad \lambda,\mu\in\mathbb{R} \]

\(\begin{pmatrix}-2\\1\\2\end{pmatrix} \neq k\begin{pmatrix}2\\-2\\3\end{pmatrix}\) for any scalar \(k\), so \(l_1\) is not parallel to \(l_2\).

Equating the two lines: \[\begin{aligned} 2\lambda + 2\mu &= 1 \quad (1)\\ -\lambda - 2\mu &= -3 \quad (2)\\ -2\lambda + 3\mu &= 8 \quad (3) \end{aligned}\] No solution from GC.

OR: Solution for equations (1) and (2) gives \(\lambda = -2\), \(\mu = \dfrac{5}{2}\). Checking (3): \[ \text{LHS} = -2(-2) + 3\!\left(\tfrac{5}{2}\right) = \tfrac{7}{2} \neq 8 = \text{RHS} \] Hence (3) is not satisfied. Since \(l_1\) and \(l_2\) are non-parallel and non-intersecting, they are skew lines.

(ii) Let \(F\) be the foot of perpendicular from \(P\) to \(\pi_1\).

Equation of line through \(P\) and \(F\): \[ \mathbf{r} = \begin{pmatrix}1\\5\\0\end{pmatrix} + \alpha\begin{pmatrix}0\\1\\-1\end{pmatrix} = \begin{pmatrix}1\\5+\alpha\\-\alpha\end{pmatrix},\quad\alpha\in\mathbb{R} \]

Substituting into equation of plane \(\pi_1\): \[ \begin{pmatrix}1\\5+\alpha\\-\alpha\end{pmatrix} \cdot\begin{pmatrix}0\\1\\-1\end{pmatrix} = 2 \implies 5 + 2\alpha = 2 \implies \alpha = -\tfrac{3}{2} \] \[ \therefore\;\overrightarrow{OF} = \begin{pmatrix}1\\\tfrac{7}{2}\\\tfrac{3}{2}\end{pmatrix} \] \[ \overrightarrow{OP'} = 2\overrightarrow{OF} - \overrightarrow{OP} = 2\begin{pmatrix}1\\\frac{7}{2}\\\frac{3}{2}\end{pmatrix} - \begin{pmatrix}1\\5\\0\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix} \]

Alternative method to find foot of perpendicular:

\(Q(0,2,0)\) is a point on plane \(\pi_1\). \[ \overrightarrow{PF} = \left[\overrightarrow{PQ}\cdot\frac{1}{\sqrt{2}} \begin{pmatrix}0\\1\\-1\end{pmatrix}\right] \frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\-1\end{pmatrix} = \frac{1}{2}\left[ \begin{pmatrix}-1\\-3\\0\end{pmatrix} \cdot\begin{pmatrix}0\\1\\-1\end{pmatrix} \right]\begin{pmatrix}0\\1\\-1\end{pmatrix} = \frac{-3}{2}\begin{pmatrix}0\\1\\-1\end{pmatrix} \] \[ \overrightarrow{OF} = \frac{-3}{2}\begin{pmatrix}0\\1\\-1\end{pmatrix} + \begin{pmatrix}1\\5\\0\end{pmatrix} = \begin{pmatrix}1\\\frac{7}{2}\\\frac{3}{2}\end{pmatrix} \] \[ \overrightarrow{OP'} = 2\overrightarrow{OF} - \overrightarrow{OP} = \begin{pmatrix}1\\2\\3\end{pmatrix} \] Coordinates of \(P'\) are \((1,2,3)\).

(iii) Let \(A(1,2,0)\) lie on \(l_1\). \[ \overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \begin{pmatrix}1\\5\\0\end{pmatrix} - \begin{pmatrix}1\\2\\0\end{pmatrix} = \begin{pmatrix}0\\3\\0\end{pmatrix} \]

Normal of plane \(\pi_2\): \[ \mathbf{n} = \begin{pmatrix}-2\\1\\2\end{pmatrix}\times\begin{pmatrix}0\\3\\0\end{pmatrix} = \begin{pmatrix}(1)(0)-(2)(3)\\(2)(0)-(-2)(0)\\(-2)(3)-(1)(0)\end{pmatrix} = \begin{pmatrix}-6\\0\\-6\end{pmatrix} = -6\begin{pmatrix}1\\0\\1\end{pmatrix} \]</p> <p>Equation of plane containing \(P\) and \(l_1\): \[ \mathbf{r}\cdot\begin{pmatrix}1\\0\\1\end{pmatrix} = \begin{pmatrix}1\\2\\0\end{pmatrix}\cdot\begin{pmatrix}1\\0\\1\end{pmatrix} = 1 \]