Find 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).

Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM

Expert Answer:

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.

Answered by  | 4th Jun, 2014, 03:23: PM