Question

The equation of the plane passing through a point (x₀, y₀, z₀) and having the vector (a, b, c) as normal is 

a(x − x₀) + b(y − y₀) + c(z − z₀) = 0 

and this plane is also parallel to the plane ax + by + cz + d = 0. Answer the following question

(i) The equation of the plane passing through the points A(2, 1, 0), B(5, 0, 1) and C(4, 1, 1) is 

(A) x + y − 2z − 3 = 0 

(B) x − y + 2z − 3 = 0 

(C) x + y + 2z − 3 = 0 

(D) x + y − 2z + 3 = 0

(ii) The equation of the plane passing through the point (2, −3, 1) and perpendicular to the 3 vector i + 4 vector j + 7 vector k is 

(A) 3x + 4y + 7z + 11 = 0 

(B) 3x + 4y + 7y − 1 = 0 

(C) 3x + 4y + 7z + 12 = 0 

(D) 3x + 4y + 7z − 12 = 0

(iii) The equation of the plane through the point (−3, −3, 1) and normal to the line joining the points (2, 6, 1) and (1, 3, 0) is 

(A) x + 3y + z − 11 = 0 

(B) 2x + y + z + 11 = 0 

(C) x + 3y + z + 11 = 0 

(D) x + 2y + 3z − 11 = 0 

Answer 1

(i) → (A), (ii) → (B), (iii) → C

Explanation :

(i) The normal to the required plane is

The plane is passing through A(2, 1, 0). Hence, its equation is 

-1(x - 2) - 1(y - 1) + 2(z - 0) = 0

- x - y + 2z + 3 = 0

x + y - 2z - 3 = 0 

(ii) The equation of the plane through (2, −3, 1) and perpendicular to the 3 vector i + 4 vector j + 7 vector k is

3(x - 2) + 4(y + 3) + 7(z - 1) = 0

3x + 4y + 7z - 1 = 0

(iii) The normal to the plane is (1 − 2, 3 − 6, −1) = (−1, −3, −1). Hence the equation of the plane is 

-1(x + 3) - 3(y + 3) - 1(z - 1) = 0

- x - 3y - z - 12 = 0 

x + 3y + z + 11 = 0