Question

Let P₁: 2x + y – z = 3 and P₂: x + 2y + z = 2 be two planes. Then, which of the following statement(s) is (are) TRUE? 

(A) The line of intersection of P₁ and P₂ has direction ratios 1, 2, – 1.

(B) The line (3x - 4)/9 = (1 - 3y)/9 = z/3 is perpendicular to the line of intersection of P₁ and P₂. 

(C) The acute angle between P₁ and P₂ is 60°. 

(D) If P₃ is the plane passing through the point (4, 2, –2) and perpendicular to the line of intersection of P₁ and P₂, then the distance of the point (2, 1, 1) from the plane P₃ is 2/√3.

Answer 1

Answer is (C), (D)

Let us check all four options as follows:

• Option (A): It is given that 

 P₁ ⇒ 2x + y – z = 3 

 P₂ ⇒ x + 2y + z = 2

Let the direction ratios of the line of intersection of plane be a, b and c. Then, the equations of the normal of the plan are

2a + b – c = 0

and a + 2b + c = 0

That is,

and 3b = −3c ⇒ c = −b

Therefore, a = 1, b = −1 and c = 1.

The direction ratios of the line of intersection are 1, −1, 1.

Hence, option (A) is false.

• Option (B): It is given that

Therefore, the given line is parallel to the line of intersection of the two planes P₁ and P₂. 

Hence, option (B) is false. 

Now, acute angle between the two planes P₁ and P₂ is

Therefore, the acute angle between the two planes P₁ and P₂ is 60°. Hence, option (C) is correct. 

• Option (D): It is given that plane P₃ passes through the point (4, 2, −2). Therefore, the equation of plane P₃ is 

(x – 4) – (y – 2) + (z + 2) = 0

Thus, P₃ ⇒ x – y + z = 0

Now, the distance of plane P₃ from the point (2, 1, 1) is obtained as

 (2 - 1 + 1)/√(1 + 1 + 1) = 2/√3

Thus, option (D) is true.