Let d be the distance of the point of intersection of the lines (x+6)/3=y/2=(z+1)/1 and (x-7)/4=(y-9)/3=(z-4)/2 from the point (7,8,9). Then d²+6 is equal to :

Solution

Maths sir · 4 steps

Write both lines with a parameter, match two coordinates, and the third one checks your work.

1. Parametrise both lines. A point on the first line is (-6+3λ, 2λ, -1+λ). A point on the second line is (7+4μ, 9+3μ, 4+2μ).
2. Equate x and y. Matching x gives 3λ - 4μ = 13. Matching y gives 2λ - 3μ = 9. Solving these together gives μ = -1 and λ = 3.
3. Check z. First line: -1 + 3 = 2. Second line: 4 - 2 = 2. The z values agree, so the lines really meet, at (3, 6, 2).

   d² = (7-3)² + (8-6)² + (9-2)² = 16 + 4 + 49 = 69

4. So. d² + 6 = 69 + 6 = 75, which is option (a).

The idea to keep. To find where two lines meet, use two coordinates to get the parameters and the third to confirm. If the third does not match, the lines are skew and do not meet.

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