Find the angle between the lines $\frac{x-2}{3} = \frac{y+1}{-2}, z=2$ and $\frac{x-1}{1} = \frac{2y+3}{3} = \frac{z+5}{2}$.

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{-1} = \frac{z - 3}{1}$ and $\frac{x + 3}{-3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

Which of the following lines in ${R^3}$ is coincident with the line $\vec{r} = (1, 1, 1) + k(2, 3, 4), k \in R$?

The shortest distance between the lines ${r_1} = 4i - 3j - k + \lambda (i - 4j + 7k)$ and ${r_2} = i - j - 10k + \mu (2i - 3j + 8k)$ is

Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.