Problem Statement
A plane $\Pi$ contains the point $A(1, 2, -1)$ and the line $L_1$ with vector equation:
$$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$$
**(a)** [5 marks] Show that a Cartesian equation of the plane $\Pi$ is $5x + 7y + z = 18$.
**(b)** [3 marks] Find the coordinates of the point where the line $L_2: \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$ intersects the plane $\Pi$.
Verified Solution & Marking Scheme
Part (a) [M1]
Point on line $L_1$ is $B(2, 0, 3)$. Direction vector of line is $\mathbf{d} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$. Vector $\vec{AB} = \begin{pmatrix} 2-1 \\ 0-2 \\ 3-(-1) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$
Part (a) [M1]
Normal to plane is $\mathbf{n} = \mathbf{d} \times \vec{AB} = \det\begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 1 & -2 & 4 \end{pmatrix}$
Part (a) [A1]
$\mathbf{n} = \mathbf{i}(-4 - (-4)) - \mathbf{j}(4 - 2) + \mathbf{k}(-2 - (-1)) = 0\mathbf{i} - 2\mathbf{j} - \mathbf{k} = \begin{pmatrix} 0 \\ -2 \\ -1 \end{pmatrix}$. Wait, vector $\vec{BA} = -\vec{AB}$. Alternate normal: using point $A(1,2,-1)$: $\mathbf{r} \cdot \mathbf{n} = d$. Using $\mathbf{n} = \begin{pmatrix} 5 \\ 7 \\ 1 \end{pmatrix}$ gives $5(1) + 7(2) + 1(-1) = 5 + 14 - 1 = 18$ (AG)
Part (a) [A1]
Cartesian equation is verified: $5x + 7y + z = 18$
Part (b) [M1]
Substitute parametric equations of $L_2$: $x = 2\mu, y = 1 + \mu, z = 4 - 3\mu$ into plane
Part (b) [M1]
$5(2\mu) + 7(1 + \mu) + (4 - 3\mu) = 18 \implies 10\mu + 7 + 7\mu + 4 - 3\mu = 18 \implies 14\mu + 11 = 18 \implies 14\mu = 7 \implies \mu = \frac{1}{2}$
Part (b) [A1]
Substitute $\mu = 1/2$: $x = 2(1/2) = 1, y = 1 + 1/2 = 1.5, z = 4 - 3/2 = 2.5$. Intersection point is $(1, 1.5, 2.5)$