Vectors: line parallel to a plane, angle between lines, and reflection of a line in a plane — EJC 2025 H2 Math Prelim P2

(a) Show that \(t = -1\)

Build a normal to \(p\) from its two direction vectors, then impose that \(l_1\)'s direction is perpendicular to it and that \(A\) is not on \(p\).

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)

(b) Acute angle between \(l_1\) and \(l_2\)

Take the dot product of the two unit direction vectors and apply \(\cos^{-1}\) with an absolute value to keep the angle acute.

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:

\[ \begin{aligned} \theta &= \cos^{-1}\!\left|\frac{1}{\sqrt{6}}\!\begin{pmatrix}-1\\2\\1\end{pmatrix} \cdot \frac{1}{\sqrt{6}}\!\begin{pmatrix}1\\2\\1\end{pmatrix}\right|\\ &= \cos^{-1}\!\left|\frac{4}{6}\right|\\ &= 0.841\text{ rad}\quad (\text{or }48.2^\circ). \end{aligned} \]

(c) Geometrical meaning and value of \(|\mathbf{a}\cdot\mathbf{n}|\)

Since \(p\) passes through the origin, \(|\mathbf{a}\cdot\mathbf{n}|\) is the perpendicular distance from \(A\) to \(p\); evaluate it with the unit normal.

\(|\mathbf{a}\cdot\mathbf{n}|\) 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 \(\mathbf{n} = \dfrac{1}{\sqrt{30}}\!\begin{pmatrix} -1\\2\\-5 \end{pmatrix}\).

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

(d) Line of reflection of \(l_2\) in \(p\)

Reflect \(A\) in \(p\) via the foot of perpendicular \(N\); since \(O\) lies on \(p\) it is fixed, so the reflected line passes through \(O\) and the image \(B\).

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 \(B\):

\[ \mathbf{r} = k\!\begin{pmatrix}13\\34\\5\end{pmatrix}, \quad k \in \mathbb{R}. \]