(i) Distance of the point(0, 0, 1) from the plane x + y + z = 3 (a) 1/√3 units (b) 2/√3 units (c) √3 units

Question

(i) Distance of the point(0, 0, 1) from the plane x + y + z = 3

(a) \(\frac{1}{\sqrt{3}}\) units

(b) \(\frac{2}{\sqrt{3}}\) units

(c) \(\sqrt{3}\) units

(d) \(\frac{\sqrt{3}}{2}\) units

(ii) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0 

Answer 1

(i). (b). \(\frac{2}{\sqrt{3}}\) units

(ii). Equation of the plane passing through the intersection is of the form

x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0 _____(1)

(1 + 2λ)x + (1 + 3λ)j + (1 + 4λ)z – 1 – 5λ = 0

The Dr’s of the required plane is

(1 + 2λ), (1 + 3λ), (1 + 4λ)

The Dr’s of the Perpendicular plane is 1, -1, 1

⇒ (1 + 2λ)(1) + (1 + 3λ)(-1) + (1 + 4λ)(1) = 0

⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0

⇒ 3λ + 1 = 0 ⇒ λ = −\(\frac{1}{3}\)

(1) ⇒ x + y + z – \(\frac{1}{3}\)(2x + 3y + 4z – 5) = 0

⇒ 3x + 3y + 3z – 2x – 3y – 4z + 5 = 0

⇒ x – z + 2 = 0.