Vectors: Parallel to plane; reflection — EJC 2025 H2 Math Prelim Paper 2

(a) A normal vector to \(p\) is given by the cross product of the two direction vectors: \[ \mathbf{n}_p = \begin{pmatrix} 0\\5\\2 \end{pmatrix} \times \begin{pmatrix} 5\\0\\t \end{pmatrix} = \begin{pmatrix} 5t\\10\\-25 \end{pmatrix} = 5\begin{pmatrix} t\\2\\-5 \end{pmatrix}. \]

Since \(l_1 \parallel p\), the direction of \(l_1\) is perpendicular to \(\mathbf{n}_p\): \[\begin{aligned} \begin{pmatrix} t\\ 3-t^2\\ 1 \end{pmatrix} \cdot \begin{pmatrix} t\\2\\-5 \end{pmatrix} &= 0\\ t^2 + 2(3 - t^2) - 5 &= 0\\ t^2 + 6 - 2t^2 - 5 &= 0\\ -t^2 + 1 &= 0\\ t &= \pm 1 \end{aligned}\]

Since \(l_1\) is not on \(p\), point \(A\) does not lie on \(p\): \[\begin{aligned} \begin{pmatrix} 1\\2\\1 \end{pmatrix} \cdot \begin{pmatrix} t\\2\\-5 \end{pmatrix} &\neq 0\\ t + 4 - 5 &\neq 0\\ t &\neq 1 \end{aligned}\]

Combining, \(t = -1\). (Shown)

\(t = -1\).

(b) With \(t = -1\), \(l_1\) has direction \(\begin{pmatrix} -1\\2\\1 \end{pmatrix}\) and \(l_2\) has direction \(\mathbf{a} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\). Let \(\theta\) be the acute angle between them: \[\begin{aligned} \theta &= \cos^{-1}\!\left|\frac{1}{\sqrt{6}}\!\begin{pmatrix}-1\\2\\1\end{pmatrix} \cdot \frac{1}{\sqrt{6}}\!\begin{pmatrix}1\\2\\1\end{pmatrix}\right|\\ &= \cos^{-1}\!\left|\frac{4}{6}\right|\\ &= 0.841\text{ rad}\quad (\text{or }48.2^\circ). \end{aligned}\]

\(\theta = 0.841\) rad \(= 48.2^\circ\).

(c) \(|\mathbf{a}\cdot\mathbf{n}|\) is the perpendicular distance from \(A\) to the plane \(p\).

With \(t = -1\), \(\mathbf{n}_p = \begin{pmatrix} -1\\2\\-5 \end{pmatrix}\) and \(|\mathbf{n}_p| = \sqrt{1 + 4 + 25} = \sqrt{30}\). So \(\mathbf{n} = \dfrac{1}{\sqrt{30}}\!\begin{pmatrix} -1\\2\\-5 \end{pmatrix}\).

\[ |\mathbf{a}\cdot\mathbf{n}| = \left|\,\begin{pmatrix}1\\2\\1\end{pmatrix} \cdot \frac{1}{\sqrt{30}}\!\begin{pmatrix}-1\\2\\-5\end{pmatrix}\right|. \] \[\begin{aligned} |\mathbf{a}\cdot\mathbf{n}| &= \left|\frac{-1 + 4 - 5}{\sqrt{30}}\right|\\ &= \left|\frac{-2}{\sqrt{30}}\right|\\ &= \frac{2}{\sqrt{30}}. \end{aligned}\]

\(|\mathbf{a}\cdot\mathbf{n}|\) is the perpendicular distance from \(A\) to \(p\), and \(|\mathbf{a}\cdot\mathbf{n}| = \dfrac{2}{\sqrt{30}}\).

(d) Let \(N\) be the foot of the perpendicular from \(A\) to \(p\). Since \(p\) passes through \(O\) (because \(p: \mathbf{r}\cdot\mathbf{n}_p = 0\)), and \(\overrightarrow{ON} = \overrightarrow{OA} - s\,\mathbf{n}_p\) for some scalar \(s\): \[ \overrightarrow{ON} = \begin{pmatrix}1\\2\\1\end{pmatrix} - s\begin{pmatrix}-1\\2\\-5\end{pmatrix} = \begin{pmatrix}1+s\\2-2s\\1+5s\end{pmatrix}. \]

Since \(N\) lies on \(p\): \[\begin{aligned} \begin{pmatrix}1+s\\2-2s\\1+5s\end{pmatrix} \cdot \begin{pmatrix}-1\\2\\-5\end{pmatrix} &= 0\\ -(1+s) + 2(2 - 2s) - 5(1 + 5s) &= 0\\ -2 - 30s &= 0, \quad\text{so } s = -\tfrac{1}{15}. \end{aligned}\]

So \(\overrightarrow{ON} = \dfrac{1}{15}\!\begin{pmatrix}14\\32\\10\end{pmatrix}\).

Let \(B\) be the reflection of \(A\) in \(p\). Then \[\begin{aligned} \overrightarrow{OB} = 2\overrightarrow{ON} - \overrightarrow{OA} &= \frac{2}{15}\!\begin{pmatrix}14\\32\\10\end{pmatrix} - \begin{pmatrix}1\\2\\1\end{pmatrix}\\ &= \frac{1}{15}\!\begin{pmatrix}13\\34\\5\end{pmatrix}. \end{aligned}\]

Since \(l_2\) passes through \(O\) (fixed by reflection) and \(A\) (reflects to \(B\)), the line of reflection passes through \(O\) and \(B\): \[ \mathbf{r} = k\!\begin{pmatrix}13\\34\\5\end{pmatrix}, \quad k \in \mathbb{R}. \]