Let A and B be two distinct points on the line L: (x-6)/3=(y-7)/2=(z-7)/-2. Both A and B are at a distance 2 √17 from the foot of perpendicular drawn from the point (1,2,3) on the line L. If O is the origin, then O A⃗ · O B⃗ is equal to

Solution

Maths sir · 4 steps

Find the foot F first. A and B sit symmetrically on either side of it, so OA · OB = |OF|² − |FA|².

1. General point. Any point on L is (6+3t, 7+2t, 7-2t), with direction vector d⃗ = (3, 2, -2) and |d⃗| = √(9+4+4) = √17.
2. Foot of perpendicular. The vector from (1,2,3) to this point is (5+3t, 5+2t, 4-2t). It must be perpendicular to d⃗: 3(5+3t) + 2(5+2t) - 2(4-2t) = 17 + 17t = 0, so t = -1. The foot is F = (3, 5, 9).
3. Locate A and B. One step of t moves a distance √17, so a distance of 2√17 needs t to change by ± 2. That gives A = F + 2d⃗ = (9, 9, 5) and B = F - 2d⃗ = (-3, 1, 13).

   OA⃗·OB⃗ = (9)(-3) + (9)(1) + (5)(13) = -27 + 9 + 65 = 47

4. So. As a check, (F⃗+2d⃗)·(F⃗-2d⃗) = |F⃗|² - 4|d⃗|² = 115 - 68 = 47. The answer is 47, option (c).

The idea to keep. When two points are equidistant from a point F on a line, write them as F⃗ ± v⃗. Then their dot product is just |F⃗|² - |v⃗|², with no need to find each point.

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