Let the equation of the plane containing the line be and the distance of the plane from the point be . Then is equal to _______.
Show answer and solution
Answer: 355
Solution
The line can be written as (x+10)/1 = (y-8)/(-2) = z/1, so it passes through (-10, 8, 0) with direction (1, -2, 1).
The plane ax + by + 3z = 2(a+b) must contain this line.
Condition 1: Point (-10, 8, 0) lies on the plane:
a(-10) + b(8) + 0 = 2(a+b)
-10a + 8b = 2a + 2b
-12a + 6b = 0 → b = 2a
Condition 2: Direction (1,-2,1) is parallel to the plane, so normal (a,b,3)·(1,-2,1) = 0:
a - 2b + 3 = 0
Substitute b = 2a: a - 4a + 3 = 0 → -3a = -3 → a = 1, b = 2
Plane: x + 2y + 3z = 2(1+2) = 6, i.e., x + 2y + 3z = 6.
Distance from (1, 27, 7):
c = |1 + 54 + 21 - 6| / √(1+4+9) = |70| / √14 = 70/√14 = 5√14
c² = 25×14 = 350
a² + b² + c² = 1 + 4 + 350 = 355
The plane ax + by + 3z = 2(a+b) must contain this line.
Condition 1: Point (-10, 8, 0) lies on the plane:
a(-10) + b(8) + 0 = 2(a+b)
-10a + 8b = 2a + 2b
-12a + 6b = 0 → b = 2a
Condition 2: Direction (1,-2,1) is parallel to the plane, so normal (a,b,3)·(1,-2,1) = 0:
a - 2b + 3 = 0
Substitute b = 2a: a - 4a + 3 = 0 → -3a = -3 → a = 1, b = 2
Plane: x + 2y + 3z = 2(1+2) = 6, i.e., x + 2y + 3z = 6.
Distance from (1, 27, 7):
c = |1 + 54 + 21 - 6| / √(1+4+9) = |70| / √14 = 70/√14 = 5√14
c² = 25×14 = 350
a² + b² + c² = 1 + 4 + 350 = 355
[JEE Main 2023 · 29 Jan · Shift 1]