Question

The equation of the plane passing through the point (3, - 3, 1) and perpendicular to the line joining the points (3, 4, - 1) and (2, - 1, 5) is:

• A. - x - 5y + 6z + 18 = 0
• B. x - 5y + 6z + 18 = 0
• C. x + 5y - 6z + 18 = 0
• D. - x - 5y - 6z + 18 = 0

Answer 1

The correct option is C

x + 5y - 6z + 18 = 0

The equation of the plane passing through the point (3,- 3, 1) is:

a(x - 3) + b(y + 3) + c(z - 1) = 0 and the direction ratios of the line joining the points

(3, 4, - 1) and (2, - 1, 5) is 2 - 3, -1 -4, 5 + 1, i.e., -1, - 5, 6.

Since the plane is perpendicular to the line whose direction ratios are 1, - 5, 6, therefore, direction ratios of the normal to the plane is -1, - 5, 6.

So, required equation of plane is: -1(x - 3) - 5(y + 3) + 6(z - 1) = 0

i.e. x + 5y - 6z + 18 = 0.