Problem Statement
The line $L$ has equation $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$.
The plane $\Pi$ has Cartesian equation $3x - 2y + 4z = 13$.
(a) Find the coordinates of the point where $L$ intersects $\Pi$. [4 marks]
(b) Determine whether $L$ is perpendicular to $\Pi$. Justify your answer. [3 marks]
Verified Solution & Marking Scheme
Part (a): Parametric substitution
Points on $L$ have coordinates $x = 1 + 2\lambda, y = -2 + \lambda, z = 3 - \lambda$. Substitute into plane:\n$$3(1 + 2\lambda) - 2(-2 + \lambda) + 4(3 - \lambda) = 13$$\n$$3 + 6\lambda + 4 - 2\lambda + 12 - 4\lambda = 13 \implies 19 = 13$$. This is a contradiction ($$0\lambda = -6$$ has no solution).
Interpret no solution
Since $$0\lambda = -6$$ has no solution, the line $L$ does not intersect the plane $\Pi$. The line $L$ is strictly parallel to $\Pi$ and lies outside it.
Part (b): Test perpendicularity
The line is perpendicular to the plane if its direction vector $\mathbf{b} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ is a scalar multiple of the plane normal $\mathbf{n} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}$.\nSince $\frac{2}{3} \neq \frac{1}{-2} \neq \frac{-1}{4}$, $\mathbf{b}$ is not parallel to $\mathbf{n}$. Hence $L$ is not perpendicular to $\Pi$.