# The lines $\frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3}$

Q: The lines  $\large \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3}$

(A) are non coplanar

(B) are coplanar

(C) intersecting at (4, 0, – 1)

(D) intersecting at (1, 1, – 1)

Sol: Let $\large \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} = \lambda_1$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3} = \lambda_2$

Then 1 + 3λ1 = 4 + 2λ2 …(1)

1 – λ1 = 0 …(2)

– 1 = 3λ2 – 1 …(3)

λ1 = 1 , λ2 = 0 satisfies (1)

lines are intersecting hence coplanar and point of intersection (4, 0, – 1).

Hence (B) and (C) are the correct answers.