System of Linear Equations: Line–plane: parallel conditions & distance — ASRJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
The line \(\ell\) has the equation \(\dfrac{x-a}{2} = y - 2 = \dfrac{1-2z}{b}\), where \(a\) and \(b\) are real constants. The plane \(\pi\) has the equation \(3x - y + 4z = 10\).
- Given that the line \(\ell\) and the plane \(\pi\) do not intersect, show that \(a \neq \dfrac{10}{3}\) and \(b = \dfrac{5}{2}\).
- It is given that \(a=1\) and \(b=3\). Find the point of intersection between the line \(\ell\) and the plane \(\pi\).
- It is given now that \(a=1\), \(b=\dfrac{5}{2}\).
- Find the distance between the line \(\ell\) and the plane \(\pi\).
- Determine whether the line \(\ell\) and the origin \(O\) lie on the same side of the plane \(\pi\). State an equation of the other plane that is equidistant from line \(\ell\) and parallel to plane \(\pi\).
Show full worked solution▾
(b) Intersection point with \(a=1\), \(b=3\).
Line \(\ell\): \(\mathbf{r} = \begin{pmatrix}1\\2\\1/2\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-3/2\end{pmatrix}\), so \(\overrightarrow{OP} = \begin{pmatrix}1+2\lambda\\2+\lambda\\\frac{1}{2}-\frac{3}{2}\lambda\end{pmatrix}\).
Substituting into plane \(3x - y + 4z = 10\): \[\begin{aligned} 3(1+2\lambda) - (2+\lambda) + 4\!\left(\tfrac{1}{2} - \tfrac{3}{2}\lambda\right) &= 3 + 6\lambda - 2 - \lambda + 2 - 6\lambda = 3 - \lambda \end{aligned}\]
Setting equal to 10: \(3 - \lambda = 10 \implies \lambda = -7\).
\[ \overrightarrow{OP} = \begin{pmatrix}1\\2\\\tfrac{1}{2}\end{pmatrix} - 7\begin{pmatrix}2\\1\\-\tfrac{3}{2}\end{pmatrix} = \begin{pmatrix}-13\\-5\\11\end{pmatrix} \](c)(i)
Taking the point \(A\left(1, 2, \dfrac{1}{2}\right)\) from the line \(\ell\), and the point \(B(2, 0, 1)\) from the plane \(\pi\):
\[ \overrightarrow{AB} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ \frac{1}{2} \end{pmatrix} \] \[\begin{aligned} \text{Required distance} &= \frac{\left| \begin{pmatrix} 1 \\ -2 \\ \frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \right|}{\left| \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \right|} \\[6pt] &= \frac{|3 + 2 + 2|}{\sqrt{3^2 + (-1)^2 + 4^2}} \\[6pt] &= \frac{7}{\sqrt{26}} \text{ units} \end{aligned}\](c)(ii)
Equation of plane \(\pi\): \(\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} = 10 \Rightarrow \mathbf{r} \cdot \dfrac{\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}}{\sqrt{26}} = \dfrac{10}{\sqrt{26}}\)
Equation of plane parallel to \(\pi\) containing the line \(\ell\): \(\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ \frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} = 3\)
\[ \Rightarrow \mathbf{r} \cdot \frac{\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}}{\sqrt{26}} = \frac{3}{\sqrt{26}} \]Since they are of the same sign, plane \(\pi\) and line \(\ell\) are on the same side of the origin, therefore both the origin and the line \(\ell\) lie on the same side of the plane.
Another plane \(\pi'\): \(\quad 3x - y + 4z = -4\)