Let L₁: (x-1)/3=(y-1)/-1=(z+1)/0 and L₂: (x-2)/2=y/0=(z+4)/α, α ∈ R, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1,1,-1) on L₂, then the value of 26 α( PB)² is _________ .

Solution

Maths sir · 4 steps

A lies on L₁, so PB is just the shadow of AB on L₂: one dot product does it.

1. Find B and alpha. Points of the lines are (1+3t, 1-t, -1) and (2+2s, 0, -4+α s). Equating y gives t = 1, so B = (4, 0, -1). Then x gives s = 1, and z gives -4+α = -1, so α = 3.
2. Spot A on L1. A(1,1,-1) is the point t = 0 on L₁. Since P is the foot of the perpendicular from A to L₂, triangle APB has a right angle at P. So PB is the projection of AB⃗ on the direction of L₂.

   PB = (|AB⃗·d⃗|)/|d⃗| = (|(3)(2) + (-1)(0) + (0)(3)|)/√(4+0+9) = 6/√13

3. Check with P. Here AB⃗ = (3, -1, 0) and d⃗ = (2, 0, 3). As a check, solving AP⃗·d⃗ = 0 gives s = 7/13. Then PB = (1 - 7/13)√13 = 6/√13, the same value.
4. So. PB² = 36/13, and 26α PB² = 26 × 3 × 36/13 = 6 × 36 = 216.

The idea to keep. If the point you drop the perpendicular from lies on the other line, the distance from the foot to the intersection point is just the projection of that segment on the line's direction.

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