Let a line having direction ratios, 1, -4, 2 intersect the lines (x - 7)/3 = (y - 1)/(- 1) = (z + 2)/1 and x/2 = (y - 7)/3 = z/1 at the points A and B. Then (AB)² is equal to ___________.

Solution

Maths sir · 4 steps

Put a general point on each line, then force the vector AB to be parallel to (1, −4, 2).

1. General points. On the first line take A = (7+3s, 1-s, -2+s). On the second line take B = (2t, 7+3t, t). Then AB⃗ = (2t-3s-7, 3t+s+6, t-s+2).

   (2t-3s-7)/1 = (3t+s+6)/-4 = (t-s+2)/2

2. Two equations. The first and second ratios give 3t+s+6 = -8t+12s+28, so t - s = 2. The first and third give t-s+2 = 4t-6s-14, so 3t - 5s = 16.
3. Solve and locate. Putting t = s+2 into 3t - 5s = 16 gives s = -5 and t = -3. So A = (-8, 6, -7) and B = (-6, -2, -3), and AB⃗ = (2, -8, 4), which is 2(1,-4,2) as required.
4. So. AB² = 2² + (-8)² + 4² = 4 + 64 + 16 = 84.

The idea to keep. For a line meeting two given lines, write one parameter point on each line. Making their joining vector proportional to the given direction ratios gives two equations in two parameters.

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