JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q15

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The square of the distance of the point of intersection of the lines $\vec r=\left(\hat i+\hat j-\hat k\right)+\lambda\left(a\hat i-\hat j\right)$ , $a\ne0$ and $\vec r=\left(4\hat i-\hat k\right)+\mu\left(2\hat i+a\hat k\right)$ from the origin is:

Answer and worked solution

Official answer: C

Equating coordinates. $(1+\lambda a,\ 1-\lambda,\ -1)=(4+2\mu,\ 0,\ -1+a\mu)$ .

z: $a\mu=0$ , and $a\ne0$ , so $\mu=0$ .
y: $\lambda=1$ .
x: $1+a=4$ , so $a=3$ .

Point $(4,0,-1)$ ; squared distance $16+0+1=17$ .

(C)

Topics:Equation of a line in space

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JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 15