JEE Advanced
Mathematics
Three Dimensional Geometry
2026
JEE Advanced 2026 (Paper 1)
JEE Advanced Mathematics Question (2026) — Solution
Question
Let P be the plane such that it contains the straight line x-1 2 = y-3 3 = z+2 1 and is perpendicular to the plane x + 2y + 3z = 4. Let P_1 be the plane which passes through the point (4, 2, 2) and is parallel to P. Then which of the following statements is (are) TRUE?
Options
- A. The equation of the plane P is 7x - 5y + z = -10
- B. The distance between the planes P and P_1 is 30
- C. The distance of the plane P from the origin is 2 3
- D. The acute angle between the plane P and the plane 2x + 2y + z = 3 is ^ -1 ( 1 3 3 )
Answer
D. The acute angle between the plane P and the plane 2x + 2y + z = 3 is ^ -1 ( 1 3 3 )
Step-by-step solution
Let the normal vector to the plane P be n . Since the plane P contains the line x-1 2 = y-3 3 = z+2 1 , its normal vector n is perpendicular to the line's direction vector b = 2 i + 3 j + k . Since P is perpendicular to the plane x + 2y + 3z = 4, n is perpendicular to its normal vector n _1 = i + 2 j + 3 k . n = (2 i + 3 j + k ) ( i + 2 j + 3 k ) = i (9-2) - j (6-1) + k (4-3) = 7 i - 5 j + k Plane P passes through the point (1, 3, -2) which lies on the given line. Equation of plane P is 7(x-1) - 5(y-3) + 1(z+2) = 0 7x - 5y + z = -10 Statement (1) is TRUE. The plane P_1 is parallel to P and passes through (4, 2, 2). Equation of plane P_1 is 7(x-4) - 5(y-2) + 1(z-2) = 0 7x - 5y + z = 20 Distance between parallel planes P and P_1 is d = |20 - (-10)| 7^2 + (-5)^2 + 1^2 = 30 75 = 2 3 . Statement (2) is FALSE. Distance of plane P from the origin is d_0 = |-10| 75 = 2 3 . Statement (3) is FALSE. Angle between the planes 7x - 5y + z + 10 = 0 and 2x + 2y + z - 3 = 0 is given by: = |7(2) + (-5)(2) + 1(1)| 7^2+(-5)^2+1^2 2^2+2^2+1^2 = 5 75 9 = 5 15 3 = 1 3 3 = ^ -1 ( 1 3 3 ). Statement (4) is TRUE.
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