Problem Statement
Find a vector equation of the line of intersection of the two planes:
$$\Pi_1: 2x - y + 3z = 4$$
$$\Pi_2: x + 2y - z = 1$$
Verified Solution & Marking Scheme
Determine direction vector of the line using cross product of normal vectors
Normal to $\Pi_1$: $\vec{n}_1 = 2\hat{i} - \hat{j} + 3\hat{k}$.
Normal to $\Pi_2$: $\vec{n}_2 = \hat{i} + 2\hat{j} - \hat{k}$.
The direction vector $\vec{d}$ of the intersecting line is perpendicular to both normals:
$$\vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix}$$
$$= \hat{i}(1 - 6) - \hat{j}(-2 - 3) + \hat{k}(4 - (-1)) = -5\hat{i} + 5\hat{j} + 5\hat{k}$$
Dividing by 5, we can use the parallel direction vector $\vec{d} = -\hat{i} + \hat{j} + \hat{k}$ or $\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$.
Find a particular point on the line of intersection
Set $z = 0$ in both plane equations:
$$2x - y = 4$$
$$x + 2y = 1$$
Multiply second equation by 2: $2x + 4y = 2$.
Subtract: $-5y = 2 \implies y = -\frac{2}{5}$.
Then $x = 1 - 2(-2/5) = 1 + \frac{4}{5} = \frac{9}{5}$.
A point on the line is $\left(\frac{9}{5}, -\frac{2}{5}, 0\right)$.
Write the vector equation r = a + λd
$$\vec{r} = \begin{pmatrix} 9/5 \\ -2/5 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R}$$