NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice
Let P 1 = x + y + z + 1 = 0 ,   P 2 = x - y + 2 z + 1 = 0 and P 3 = 3 x + y + 4 z + 7 = 0 be three planes. The distance of the line of intersection of planes P 1 = 0 and P 2 = 0 from the plane P 3 = 0 is
Options
- A2 26 units
- B1 26 units
- C4 26 units
- D7 26 units
Correct answer
C. 4 26 units
Step-by-step solution
A vector parallel to the line of intersection is i ^ j ^ k ^ 1 1 1 1 - 1 2 = i ^ 3 - j ^ 1 + k ^ - 2 = 3 i ^ - j ^ - 2 k ^ for a point on the line of intersection Put z = 0 x + y + 1 = 0 x - y + 1 = 0 ⇒ x = - 1 , y = 0 So, equation line to the intersection is x + 1 3 = y - 0 - 1 = z - 0 - 2 ∵ line of intersection is parallel to the plane P 3 = 0 So, distance = - 3 + 0 + 0 + 7 26 = 4 26