The angle between the lines $\vec r = 3\hat i- 2\hat j +1\hat k +μ (4\hat i + 6\hat j + 12\hat k)$ and $\vec r=7\hat i-3\hat j + 9\hat k+λ (5\hat i +8\hat j - 4\hat k)$ is:

$\cos^{-1}\frac{10}{7\sqrt{105}}$

The correct answer is Option (1) → $\cos^{-1}\frac{10}{7\sqrt{105}}$

The given equations of the lines are:

$\vec{r_1} = (3\hat i- 2\hat j +1\hat k) +μ (4\hat i + 6\hat j + 12\hat k)$

$\vec{r_2}=(7\hat i-3\hat j + 9\hat k)+λ (5\hat i +8\hat j - 4\hat k)$

∴ Direction vectors,

$d_1=4\hat i+6\hat j+12\hat k$

$d_2=5\hat i+8\hat j-4\hat k$

$\cos θ=\frac{d_1.d_2}{|d_1||d_2|}$

$=\frac{4×5+6×8+12×-4}{\sqrt{196}\sqrt{105}}$

$=\frac{10}{7\sqrt{105}}$

$⇒θ=\cos^{-1}\left(\frac{10}{7\sqrt{105}}\right)$