The acute angle between the lines \(\vec{r} = (1-t)(\hat{\imath}+2\hat{\jmath}-\hat{k})+t(3\hat{\imath}+3\hat{\jmath}+\hat{k})\) and \(\vec{r} = 3\hat{k}+\mu(\hat{\imath}+\hat{\jmath})\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{4}\)
The first equation is the two-point form \(\vec{r} = \vec{a}+t(\vec{b}-\vec{a})\) of the line through \(\vec{a} = \hat{\imath}+2\hat{\jmath}-\hat{k}\) and \(\vec{b} = 3\hat{\imath}+3\hat{\jmath}+\hat{k}\), so its direction is
\(\vec{b}_1 = \vec{b}-\vec{a} = 2\hat{\imath}+\hat{\jmath}+2\hat{k}\)
The second line has direction \(\vec{b}_2 = \hat{\imath}+\hat{\jmath}\).
\( \displaystyle \cos\theta = \frac{|2+1+0|}{\sqrt{9}\,\sqrt{2}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \)
\( \displaystyle \theta = \frac{\pi}{4} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{6}\) — Incorrect. This comes from taking \(\hat{\imath}+2\hat{\jmath}-\hat{k}\) as the direction: \(\cos\theta = \frac{3}{\sqrt{6}\sqrt{2}} = \frac{\sqrt{3}}{2}\). That vector is a point of the line, not its direction.
Option (3) \(\displaystyle\frac{\pi}{3}\) — Incorrect. It needs \(\cos\theta = \frac{1}{2}\), but \(\cos\theta = \frac{1}{\sqrt{2}}\).
Option (4) \(\displaystyle\frac{\pi}{2}\) — Incorrect. \(\vec{b}_1\cdot\vec{b}_2 = 3 \ne 0\), so the lines are not perpendicular.