The distance of the plane from the point(1, 1, 1) is (a) 4 units (b)

1.	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» (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

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»

(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