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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 15

The square of the distance of the point (−2,−8,6)(-2,-8,6) from the line x−11=y−12=z−1\dfrac{x-1}1=\dfrac{y-1}2=\dfrac z{-1} along the line x+51=y+5−1=z2\dfrac{x+5}1=\dfrac{y+5}{-1}=\dfrac z2 is equal to:

Choose one option
Answer and worked solution

Official answer: B

Move along the given direction. (−2+t, −8−t, 6+2t)(-2+t,\,-8-t,\,6+2t) meets the line when t−3=−9−t2=−6−2tt-3=\frac{-9-t}2=-6-2t, i.e. t=−1t=-1 (all three agree).

Distance squared. t2∣(1,−1,2)∣2=6t^2|(1,-1,2)|^2=6.

(B)

Topics:Angles and distancesEquation of a line in space
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