The perpendicular distance, of the line (x-1)/2=(y+2)/-1=(z+3)/2 from the point P(2,-10,1), is :

Solution

Maths sir · 4 steps

Take a general point on the line and make the vector from P perpendicular to the line's direction.

1. General point. Set each fraction equal to t. Any point on the line is Q(1+2t, -2-t, -3+2t), and the direction of the line is (2, -1, 2).
2. Vector PQ. Subtract P(2,-10,1) from Q: PQ⃗ = (2t-1, 8-t, 2t-4). At the foot of the perpendicular, PQ⃗ is at right angles to (2,-1,2), so their dot product is zero.

   2(2t-1) - (8-t) + 2(2t-4) = 9t - 18 = 0 ⇒ t = 2

3. Foot and distance. With t = 2 the foot is Q(5, -4, 1), so PQ⃗ = (3, 6, 0). Its length is √(9 + 36 + 0) = √45.
4. So. The perpendicular distance is √45 = 3√5, which is option (c).

The idea to keep. For the distance from a point to a line, write a general point on the line and set (point minus P) dotted with the direction to zero. That gives you the foot, and the distance follows.

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