Problem Statement
Find the angle between the two planes $\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 3$ and $\vec{r} \cdot (3\hat{i} + 6\hat{j} + 2\hat{k}) = 5$.
Verified Solution & Marking Scheme
Extract Normal Vectors
$\vec{n}_1 = 2\hat{i} - \hat{j} + 2\hat{k}, \quad \vec{n}_2 = 3\hat{i} + 6\hat{j} + 2\hat{k}$
Compute Magnitudes and Dot Product
$|\vec{n}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$
$|\vec{n}_2| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$
$\vec{n}_1 \cdot \vec{n}_2 = (2)(3) + (-1)(6) + (2)(2) = 6 - 6 + 4 = 4$
Calculate Angle θ
$\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} = \frac{4}{3 \times 7} = \frac{4}{21}$
$\theta = \arccos\left(\frac{4}{21}\right)$