Question

Equation of line passing through the point (2, 3, 1) and parallel to the line of intersection of the plane x – 2y – z + 5 = 0 and x + y + 3z = 6 is

• A. (x-2/5)=(y-3/-4)=(z-1/3)
• B. (x-2/-5)=(y-3/-4)=(z-1/3)
• C. (x-2/5)=(y-3/4)=(z-1/3)
• D. (x-2/4)=(y-3/3)=(z-1/2)

Answer 1

The correct option is B

(x-2/-5)=(y-3/-4)=(z-1/3)

The equation of the line can be written as $\frac{x-2}{a}=\frac{y-3}{b}=\frac{z-1}{c}$ where a, b, c are the direction cosines of the line. If the line is parallel to the given planes, then the dot product of the direction cosines of the line and the normal to the plane is 0.

a - 2b - c = 0 and a + b + 3c = 0

Solving the above equations we get, a = $\frac{-5}{3}$c and b = $\frac{-4}{3}$c

So the required line equation is $\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}$