The distance of the point (7,10,11) from the line (x-4)/1=(y-4)/0=(z-2)/3 along the line (x-9)/2=(y-13)/3=(z-17)/6 is

Solution

Maths sir · 4 steps

Walk from the point in the given direction, find where you hit the line, then measure that walk.

1. Line through the point. The direction (2, 3, 6) has length √(4 + 9 + 36) = 7. Any point reached from P(7, 10, 11) in this direction is Q = (7 + 2t, 10 + 3t, 11 + 6t).
2. Points on the line. On the first line, y is always 4, since its direction has 0 in the y place. Its points are (4 + s, 4, 2 + 3s).
3. Find the meeting point. Matching y: 10 + 3t = 4, so t = -2. This gives Q = (3, 4, -1). Check it on the first line: x = 3 gives s = -1, and then z = 2 - 3 = -1, which matches.

   PQ = √((7-3)² + (10-4)² + (11+1)²) = √(16 + 36 + 144) = √196

4. So. PQ = 14. The shortcut gives the same: |t| × 7 = 2 × 7 = 14. The answer is (d).

The idea to keep. "Distance along a line" is not the perpendicular distance. Draw the line through the point in the given direction, find where it meets the other line, then measure.

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