Vectors & 3D GeometryeasyPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Distance Along a Line in 3D: Square = 6 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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{z}{2} is equal to
A33
B66correct
C88
D1212
Solution
Step 1: The measuring line through A(−2,−8,6)A(-2,-8,6) has direction (1,−1,2)(1,-1,2):
x+21=y+8−1=z−62=λ,\frac{x+2}{1}=\frac{y+8}{-1}=\frac{z-6}{2}=\lambda,
 so a general point is B=(λ−2, −λ−8, 2λ+6)B=(\lambda-2,\ -\lambda-8,\ 2\lambda+6). Step 2: BB must also lie on the first line: B=(μ+1, 2μ+1, −μ)B=(\mu+1,\ 2\mu+1,\ -\mu). Solving the two parameterisations gives
B=(−3,−7,4).B=(-3,-7,4).
 Step 3:
AB=(−3+2)2+(−7+8)2+(4−6)2=1+1+4=6,AB=\sqrt{(-3+2)^2+(-7+8)^2+(4-6)^2}=\sqrt{1+1+4}=\sqrt6,
 so AB2=6AB^2=6. Correct answer: (2)
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