Let Q be the foot of the perpendicular from the point P(7, -2, 13) on the plane containing the lines (x + 1)/6 = (y - 1)/7 = (z - 3)/8 and (x - 1)/3 = (y - 2)/5 = (z - 3)/7. Then (PQ)², is equal to ___________.

Solution

Maths sir · 4 steps

Get the plane from the cross product of the two directions, then just drop a perpendicular from P.

1. Normal to the plane. The plane contains both lines, so its normal is perpendicular to both directions (6, 7, 8) and (3, 5, 7). Their cross product is (9, -18, 9), which we can scale down to (1, -2, 1).
2. Equation of the plane. The plane passes through (-1, 1, 3), a point on the first line, so 1(x+1) - 2(y-1) + 1(z-3) = 0, which gives x - 2y + z = 0. Check: the point (1, 2, 3) on the second line gives 1 - 4 + 3 = 0, so both lines really lie in it.
3. Distance from P. PQ is the perpendicular distance from P(7, -2, 13) to this plane. Putting P into the left side gives 7 + 4 + 13 = 24.

   PQ² = (7 + 4 + 13)²/(1² + (-2)² + 1²) = 576/6

4. So. PQ = 24/√6 = 4√6, and PQ² = 96.

The common slip: 576. Squaring the numerator 24 but forgetting to divide by |n⃗|² = 6 gives 576, not 96.

The idea to keep. A plane containing two lines has normal equal to the cross product of their directions. Once you have the plane, the foot of a perpendicular is only a distance-to-plane formula away.

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