Problem Statement
Find the acute angle $\theta$ between the lines:
$$L_1: \vec{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}, \quad L_2: \vec{r} = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$$
Verified Solution & Marking Scheme
Extract direction vectors
$$\vec{d}_1 = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}, \quad \vec{d}_2 = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$$
Calculate dot product and magnitudes
$$\vec{d}_1 \cdot \vec{d}_2 = (2)(1) + (-1)(2) + (2)(-2) = 2 - 2 - 4 = -4$$
$$|\vec{d}_1| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$
$$|\vec{d}_2| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$
Apply acute angle formula
$$\cos \theta = \frac{|\vec{d}_1 \cdot \vec{d}_2|}{|\vec{d}_1| |\vec{d}_2|} = \frac{|-4|}{3 \times 3} = \frac{4}{9}$$
$$\theta = \arccos\left(\frac{4}{9}\right)$$