Problem Statement
The line $L$ has vector equation $\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$.
The plane $\Pi$ has Cartesian equation $2x - y + z = 9$.
**(a)** [3 marks] Find the coordinates of the point of intersection $P$ of the line $L$ and the plane $\Pi$.
**(b)** [5 marks] Find the acute angle between the line $L$ and the plane $\Pi$, giving your answer in radians.
Verified Solution & Marking Scheme
Part (a) [M1]
Parametric form of line $L$: $x = 2 + \lambda, y = 2\lambda, z = -1 + 3\lambda$
Part (a) [M1]
Substitute into plane: $2(2 + \lambda) - (2\lambda) + (-1 + 3\lambda) = 9 \implies 4 + 2\lambda - 2\lambda - 1 + 3\lambda = 9 \implies 3 + 3\lambda = 9 \implies \lambda = 2$
Part (a) [A1]
Intersection point $P = (2+2, 2(2), -1+3(2)) = (4, 4, 5)$
Part (b) [M1]
Direction of line $\mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$, normal of plane $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$
Part (b) [M1]
Dot product $\mathbf{b} \cdot \mathbf{n} = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3$
Part (b) [A1]
Magnitudes: $|\mathbf{b}| = \sqrt{1+4+9} = \sqrt{14}$, $|\mathbf{n}| = \sqrt{4+1+1} = \sqrt{6}$
Part (b) [M1]
Angle $\theta$ between line and plane satisfies $\sin\theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|} = \frac{3}{\sqrt{14}\sqrt{6}} = \frac{3}{\sqrt{84}} = \frac{3}{2\sqrt{21}}$
Part (b) [A1]
Acute angle $\theta = \arcsin\left(\frac{3}{2\sqrt{21}}\right) = \arcsin\left(\frac{\sqrt{21}}{14}\right) \approx 0.334\text{ rad}$