Problem Statement
Two planes $\Pi_1$ and $\Pi_2$ have Cartesian equations:
$$\Pi_1: 2x - y + 2z = 8$$
$$\Pi_2: x + 2y - 2z = 5$$
(a) Find the cosine of the acute angle between $\Pi_1$ and $\Pi_2$. [3 marks]
(b) Find a vector equation of the line of intersection $L$ of the two planes. [5 marks]
Verified Solution & Marking Scheme
Part (a): Find angle between normal vectors
Normal vectors: $\vec{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\vec{n}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$.
Dot product:
$$\vec{n}_1 \cdot \vec{n}_2 = (2)(1) + (-1)(2) + (2)(-2) = 2 - 2 - 4 = -4$$
Magnitudes: $|\vec{n}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$, and $|\vec{n}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$.
Acute angle cosine:
$$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} = \frac{|-4|}{3 \times 3} = \frac{4}{9}$$
Part (b): Find direction vector of line L using cross product
The direction vector $\vec{d}$ is perpendicular to both normals:
$$\vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2 - 4) - \hat{j}(-4 - 2) + \hat{k}(4 - (-1)) = -2\hat{i} + 6\hat{j} + 5\hat{k}$$
Part (b): Find a particular point on the intersection
Set $y = 0$ in the plane equations:
$$2x + 2z = 8 \implies x + z = 4$$
$$x - 2z = 5$$
Subtracting the second equation from the first: $3z = -1 \implies z = -\frac{1}{3}$.
Then $x = 4 - z = 4 - (-1/3) = \frac{13}{3}$.
Particular point: $\vec{a} = \begin{pmatrix} 13/3 \\ 0 \\ -1/3 \end{pmatrix}$.
Write vector equation of line L
$$\vec{r} = \begin{pmatrix} 13/3 \\ 0 \\ -1/3 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 6 \\ 5 \end{pmatrix}, \quad \lambda \in \mathbb{R}$$