JEE Main 2021 · 24 February · morning · Single correct · +4 / −1
The distance of the point (1, 1, 9) from the point of intersection of the line (x - 3)/1 = (y - 4)/2 = (z - 5)/2 and the plane x + y + z = 17 is :
- (a) 19√2
- (b) 2√19
- (c) 38
- (d) √38
Answer and solution
Answer (d) √38
Solution
Maths sir · 3 steps
Write any point of the line with one parameter t, then let the plane equation decide t.
- General point. Set each fraction equal to t. Any point on the line is then (3 + t, 4 + 2t, 5 + 2t).
- Meet the plane. Put this point into x + y + z = 17: (3 + t) + (4 + 2t) + (5 + 2t) = 12 + 5t = 17, so t = 1. The point of intersection is (4, 6, 7).
d = √((4-1)² + (6-1)² + (7-9)²) = √(9 + 25 + 4) = √38
- So. The distance of (1, 1, 9) from the point of intersection is √38, which is option (d).
The idea to keep. To meet a line with a plane, write the line's general point in terms of one parameter and substitute it into the plane. You get one equation in t, and that fixes the point.
Answer from the NTA final answer key. Solution by PrepWiser.