Angle between the lines \( \vec{r}=\left(2\hat{\imath}-\hat{\jmath}+3\hat{k}\right)+\lambda\left(\hat{\imath}+2\hat{\jmath}+\hat{k}\right) \) and \( \vec{r}=\left(\hat{\jmath}+\hat{k}\right)+\mu\left[\left(\sqrt{3}-1\right)\hat{\imath}+4\hat{\jmath}-\left(\sqrt{3}+1\right)\hat{k}\right] \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{\pi}{3} \)
The angle between two lines is the angle between their direction vectors \( \vec{b}_1=\hat{\imath}+2\hat{\jmath}+\hat{k} \) and \( \vec{b}_2=(\sqrt{3}-1)\hat{\imath}+4\hat{\jmath}-(\sqrt{3}+1)\hat{k} \); the fixed points of the lines play no part.
\( \displaystyle \cos\theta=\frac{|\vec{b}_1\cdot\vec{b}_2|}{|\vec{b}_1|\,|\vec{b}_2|} \)
\( \displaystyle \vec{b}_1\cdot\vec{b}_2=(\sqrt{3}-1)+8-(\sqrt{3}+1)=6 \)
\( \displaystyle |\vec{b}_1|=\sqrt{1+4+1}=\sqrt{6} \)
\( \displaystyle |\vec{b}_2|=\sqrt{(\sqrt{3}-1)^2+16+(\sqrt{3}+1)^2} \)
\( \displaystyle =\sqrt{(4-2\sqrt{3})+16+(4+2\sqrt{3})} \)
\( \displaystyle =\sqrt{24}=2\sqrt{6} \)
\( \displaystyle \cos\theta=\frac{6}{\sqrt{6}\times 2\sqrt{6}} \)
\( \displaystyle =\frac{6}{12}=\frac{1}{2} \)
\( \displaystyle \theta=\frac{\pi}{3} \)