Vectors: Perpendicular Lines, Line Intersection, Plane Equation and Foot of Perpendicular — HCI 2025 H2 Math Prelim P1

(a) Find the value of \(m\)

Rewrite each line in vector form, then set the scalar product of the two direction vectors to zero.

\[ l_1 : \mathbf{r} = \begin{pmatrix}15\\3\\0\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\4\end{pmatrix},\ \lambda \in \mathbb{R} \qquad l_2 : \mathbf{r} = \begin{pmatrix}5\\9\\5\end{pmatrix} + \mu\begin{pmatrix}8\\-2\\m\end{pmatrix},\ \mu \in \mathbb{R} \]

For \(l_1\) and \(l_2\) to be perpendicular,

\[ \begin{pmatrix}-1\\2\\4\end{pmatrix} \cdot \begin{pmatrix}8\\-2\\m\end{pmatrix} = 0 \Rightarrow m = 3 \]

(b) Coordinates of the intersection \(E\)

Equate the \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) components of the two lines and solve for \(\lambda\) and \(\mu\).

\[ \begin{aligned} 15 - \lambda &= 5 + 8\mu \quad &&\Rightarrow\ \lambda + 8\mu = 10\\ 3 + 2\lambda &= 9 - 2\mu \quad &&\Rightarrow\ 2\lambda + 2\mu = 6\\ 4\lambda &= 5 + 3\mu \quad &&\Rightarrow\ 4\lambda - 3\mu = 5 \end{aligned} \]

Solving, \(\lambda = 2\), \(\mu = 1\).

\[ \therefore \overrightarrow{OE} = \begin{pmatrix}15-2\\3+4\\8\end{pmatrix} = \begin{pmatrix}13\\7\\8\end{pmatrix} \]

Coordinates of \(E\) is \((13, 7, 8)\).

(c) Cartesian equation of plane \(\Pi\) through \(A\), \(B\), \(E\)

Take two direction vectors in the plane and cross them to get a normal, then form the scalar product equation.

\[ \overrightarrow{AB} = \begin{pmatrix}5\\9\\5\end{pmatrix} - \begin{pmatrix}15\\3\\0\end{pmatrix} = \begin{pmatrix}-10\\6\\5\end{pmatrix} \qquad \overrightarrow{AE} = \begin{pmatrix}13\\7\\8\end{pmatrix} - \begin{pmatrix}15\\3\\0\end{pmatrix} = \begin{pmatrix}-2\\4\\8\end{pmatrix} \]

A normal vector to \(\Pi\):

\[ \begin{pmatrix}-10\\6\\5\end{pmatrix} \times \begin{pmatrix}-2\\4\\8\end{pmatrix} = \begin{pmatrix}28\\70\\-28\end{pmatrix} = 14\begin{pmatrix}2\\5\\-2\end{pmatrix} \]

A Cartesian equation of \(\Pi\) is

\[ \Pi : \mathbf{r} \cdot \begin{pmatrix}2\\5\\-2\end{pmatrix} = \begin{pmatrix}15\\3\\0\end{pmatrix} \cdot \begin{pmatrix}2\\5\\-2\end{pmatrix} = 30 + 15 = 45 \]

The Cartesian equation of \(\Pi\) is \(2x + 5y - 2z = 45\).

(d) Position vector of the foot of perpendicular \(F\)

Travel from \(D\) along the normal direction and impose the condition that the resulting point lies on \(\Pi\).

Let \(F\) be the foot of perpendicular from \(D\) to \(\Pi\). Since \(F\) lies on a line through \(D\) perpendicular to \(\Pi\),

\[ \overrightarrow{OF} = \begin{pmatrix}-1+2t\\-3+5t\\2-2t\end{pmatrix} \quad \text{for some value of } t. \]

Since \(F\) lies on \(\Pi\),

\[ 2(-1+2t) + 5(-3+5t) - 2(2-2t) = 45 \Rightarrow t = 2 \]

Hence \(\overrightarrow{OF} = \begin{pmatrix}3\\7\\-2\end{pmatrix}\).

Let \(F\) be the foot of perpendicular from \(D\) to \(\Pi\) with a normal vector \(\mathbf{n} = \begin{pmatrix}2\\5\\-2\end{pmatrix}\).

\[ \begin{aligned} \overrightarrow{DF} &= \left(\frac{\overrightarrow{DA} \cdot \mathbf{n}}{|\mathbf{n}|}\right)\frac{\mathbf{n}}{|\mathbf{n}|}\\ &= \frac{1}{33}\left[\begin{pmatrix}16\\6\\-2\end{pmatrix} \cdot \begin{pmatrix}2\\5\\-2\end{pmatrix}\right]\begin{pmatrix}2\\5\\-2\end{pmatrix}\\ &= \frac{66}{33}\begin{pmatrix}2\\5\\-2\end{pmatrix}\\ &= \begin{pmatrix}4\\10\\-4\end{pmatrix} \end{aligned} \]

\[ \overrightarrow{OF} = \overrightarrow{OD} + \overrightarrow{DF} = \begin{pmatrix}-1\\-3\\2\end{pmatrix} + \begin{pmatrix}4\\10\\-4\end{pmatrix} = \begin{pmatrix}3\\7\\-2\end{pmatrix} \]

(e) Exact area of the circle through \(A\), \(D\), \(F\)

Since \(\angle DFA = 90^\circ\), \(AD\) is a diameter, so the radius is half of \(|AD|\).

\[ \text{Radius of the circle} = \frac{|AD|}{2} = \frac{1}{2}\left|\begin{pmatrix}-16\\-6\\2\end{pmatrix}\right| = \sqrt{74} \]

\[ \text{Area of circle} = \pi\left(\sqrt{74}\right)^2 = 74\pi \text{ units}^2 \]