Question:

Let $\vec{a}$ be the direction vector of the first line. $$ \vec{a} = 2\hat{i} + 2\hat{j} + \hat{k} $$ Let $\vec{b}$ be the direction vector of the line joining $P(3, 1, 4)$ to $Q(7, 2, 12)$. $$ \vec{b} = (7-3)\hat{i} + (2-1)\hat{j} + (12-4)\hat{k} = 4\hat{i} + \hat{j} + 8\hat{k} $$ The angle $\theta$ between two lines with direction vectors $\vec{a}$ and $\vec{b}$ is given by: $$ \cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} $$ First, calculate the dot product $\vec{a} \cdot \vec{b}$: $$ \vec{a} \cdot \vec{b} = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18 $$ Next, calculate the magnitudes of $\vec{a}$ and $\vec{b}$: $$ |\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $$ $$ |\vec{b}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9 $$ Substitute these values into the formula for $\cos \theta$: $$ \cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3} $$ Thus, the angle between the lines is $\theta = \cos^{-1}\left(\frac{2}{3}\right)$.