System of Linear Equations: Three planes: prism & intersection analysis — DHS 2025 H2 Math Prelim Paper 1
What this question tests
Question
The point \(A\) has coordinates \((1,-2,4)\) and the plane \(\pi_1\) has equation \(2x - y + 2z = 5\).
- Find the exact shortest distance between \(A\) and \(\pi_1\).
The plane \(\pi_2\) has equation \(x + 3y - az = 3\), where \(a\) is a constant.
- Find the vector equation of line \(l\), which is the line of intersection of \(\pi_1\) and \(\pi_2\), in terms of \(a\).
The plane \(\pi_3\) has equation \(bx - 2y + 4z = 3\), where \(b\) is a constant.
- Show that \((a-6)(b-4) = 0\) if \(l\) is parallel to \(\pi_3\).
- State the conditions that \(a\) and \(b\) must follow for the three planes to form a triangular prism, where all the planes are non-parallel and they do not have any point in common. Justify your answer.
Show full worked solution▾
(a) The normal to \(\pi_1\) is \(\mathbf{n}_1 = \begin{pmatrix}2\\-1\\2\end{pmatrix}\).
By observation, \((0,-5,0)\) is a point on the plane.
\[ \text{Shortest distance} = \frac{\left[\left(\begin{pmatrix}0\\-5\\0\end{pmatrix} - \begin{pmatrix}1\\-2\\4\end{pmatrix}\right)\cdot\begin{pmatrix}2\\-1\\2\end{pmatrix}\right]}{\sqrt{2^2+(-1)^2+2^2}} = \frac{7}{3} \text{ units} \]Shortest distance \(= \dfrac{7}{3}\) units
(b)
\(2x - y + 2z = 5\) \(x + 3y - az = 3\)Direction vector of \(l\): \[ \begin{pmatrix}2\\-1\\2\end{pmatrix} \times \begin{pmatrix}1\\3\\-a\end{pmatrix} = \begin{pmatrix}a-6\\2a+2\\7\end{pmatrix} \]
Let \(z = 0\), \[\begin{aligned} 2x - y &= 5 \quad\text{--- (1)}\\ x + 3y &= 3 \quad\text{--- (2)}\\ x = \frac{18}{7},\ y &= \frac{1}{7} \quad\text{(Using GC)} \end{aligned}\]
\[ \therefore l: \mathbf{r} = \begin{pmatrix}\frac{18}{7}\\\frac{1}{7}\\0\end{pmatrix} + \lambda\begin{pmatrix}a-6\\2a+2\\7\end{pmatrix}, \quad \lambda \in \mathbb{R} \]\(l: \mathbf{r} = \begin{pmatrix}18/7\\1/7\\0\end{pmatrix} + \lambda\begin{pmatrix}a-6\\2a+2\\7\end{pmatrix}\), \(\lambda \in \mathbb{R}\)
(c) For \(l \parallel \pi_3\), the direction of \(l\) is perpendicular to \(\mathbf{n}_3 = \begin{pmatrix}b\\-2\\4\end{pmatrix}\): \[\begin{aligned} \begin{pmatrix}a-6\\2a+2\\7\end{pmatrix}\!\cdot\!\begin{pmatrix}b\\-2\\4\end{pmatrix} &= 0 \\ b(a-6) - 2(2a+2) + 28 &= 0 \\ ab - 6b - 4a - 4 + 28 &= 0 \\ ab - 6b - 4a + 24 &= 0 \\ b(a-6) - 4(a-6) &= 0 \\ (a-6)(b-4) &= 0 \qquad \text{(shown)} \end{aligned}\]
(d) For a triangular prism, the three planes must be non-parallel with no common point (a triangular cross-section). The conditions are:
- \(l \parallel \pi_3\): so \(a = 6\) or \(b = 4\) (from part (c))
- \(l\) is not on \(\pi_3\): the point \(\left(\frac{18}{7}, \frac{1}{7}, 0\right)\) must not satisfy \(\pi_3\). Substituting: \(b\cdot\frac{18}{7} - 2\cdot\frac{1}{7} \neq 3 \Rightarrow b \neq \frac{23}{18}\)
- \(\pi_1\) is not parallel to \(\pi_3\): \(\begin{pmatrix}2\\-1\\2\end{pmatrix} \neq m\begin{pmatrix}b\\-2\\4\end{pmatrix} \Rightarrow b \neq 1 \Rightarrow b \neq 4\) or \(a \neq 6\) (equivalent to \(m \neq \frac{1}{2}\))
- \(\pi_2\) is not parallel to \(\pi_3\): \(\begin{pmatrix}1\\3\\-a\end{pmatrix} \neq m\begin{pmatrix}b\\-2\\4\end{pmatrix} \Rightarrow b \neq -\frac{2}{3}\) or \(a \neq 6\)