A line passes through A(4,-6,-2) and B(16,-2,4). The point P(a, b, c), where a, b, c are non-negative integers, on the line A B lies at a distance of 21 units, from the point A. The distance between the points P(a, b, c) and Q(4,-12,3) is equal to __________.

Solution

Maths sir · 4 steps

Turn AB into a unit vector, walk 21 units from A, and keep the point whose coordinates are non-negative.

1. Direction of AB. AB⃗ = (16-4, -2+6, 4+2) = (12, 4, 6). Its length is √(144 + 16 + 36) = √196 = 14, so the unit vector along the line is 1/7(6, 2, 3).
2. Walk 21 units. A point 21 units from A is A ± 21 × 1/7(6,2,3) = A ± 3(6,2,3) = A ± (18, 6, 9).
3. Pick the right point. The plus sign gives (22, 0, 7) and the minus sign gives (-14, -12, -11). Only (22, 0, 7) has non-negative coordinates, so P = (22, 0, 7).

   PQ = √((22-4)² + (0+12)² + (7-3)²) = √(324 + 144 + 16) = √484

4. So. PQ = √484 = 22.

The idea to keep. To find a point at a given distance along a line, divide the direction vector by its length first. Then distance times unit vector gives the step, and both signs give the two candidate points.

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