Let the foot of the perpendicular from the point (1, 2, 4) on the line (x + 2)/4 = (y - 1)/2 = (z + 1)/3 be P. Then the distance of P from the plane 3x + 4y + 12z + 23 = 0 is :

Solution

Maths sir · 3 steps

Write a general point on the line, make it perpendicular to the direction, and P falls out.

1. General point. Set each fraction equal to t. Any point on the line is (4t-2, 2t+1, 3t-1), and the direction of the line is (4, 2, 3).
2. Perpendicular condition. The vector from A(1, 2, 4) to this point is (4t-3, 2t-1, 3t-5). At the foot P it must be perpendicular to (4, 2, 3), so the dot product is zero: 4(4t-3) + 2(2t-1) + 3(3t-5) = 29t - 29 = 0, which gives t = 1 and P = (2, 3, 2).

   d = (|3(2) + 4(3) + 12(2) + 23|)/√(3² + 4² + 12²) = 65/√169 = 65/13

3. So. The distance of P from the plane is 65/13 = 5, which is option (a).

The idea to keep. For the foot of a perpendicular on a line, take the general point with parameter t and set its dot product with the direction ratios to zero. That one equation fixes t.

Answer from the NTA final answer key. Solution by PrepWiser.