Question

The distance of the plane from the point(1, 1, 1) is 

(a) 4 units 

(b) \(\frac{1}{\sqrt3}\) units 

(c) \(\frac{4}{\sqrt3}\) units 

(d) \(\frac{1}{4\sqrt3}\) units

(ii) Find the equation of the plane passing through (1, 0. -2) and perpendicular to each of the planes 2x + y – z = 2 and x – y – z = 3

Answer 1

(i) (c) \(\frac{4}{\sqrt3}\)

(ii) Equation of the plane passing through 

(1, 0, -2) is a(x – 1) + b(y – 0) + c(z + 2) = 0……………..(1)

 Plane (1) is perpendicular to the given planes 

2a + b – c = 0 ……………..(2)

 a – b – c = 0 ……………..(3)

Solving (2) and (3) we get;

\(\frac{a}{-2}=\frac{b}{1}=\frac{c}{-3}=k\)

a = -2k; b = k ; c = -3k

(1) ⇒ -2x + y - 3z - 4 = 0

⇒2x - y + 3z + 4 = 0