JEE Main202225 Jul 2022Evening ShiftMathematicsThree Dimensional GeometryActual
A plane E is perpendicular to the two planes 2 x - 2 y + z = 0 and x - y + 2 z = 4 , and passes through the point P 1 , - 1 , 1 . If the distance of the plane E from the point Q a , a , 2 is 3 2 , then P Q 2 is equal to
Options
- A9
- B12
- C21
- D33
Correct answer
C. 21
Step-by-step solution
Given, plane P 1 = 2 x - 2 y + z = 0 , Whose normal vector is ≡ n ¯ 1 = 2 , - 2 , 1 Second plane, P 2 ≡ x - y + 2 z = 4 , Whose normal vector is ≡ n ¯ 2 = 1 , - 1 , 2 Now let plane perpendicular to P 1 and P 2 will have normal vector n ¯ 3 Where n ¯ 3 = n ¯ 1 × n ¯ 2 ⇒ n → 3 = i ^ j ^ k ^ 2 - 2 1 1 - 1 2 = - 3 i ^ - 3 j ^ Hence, n ¯ 3 = - 3 , - 3 , 0 Equation of plane E through P 1 , - 1 , 1 and n ¯ 3 as normal vector will be, - 3 x - 1 - 3