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Maths Question 15 – JEE-MAIN 2026

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The square of the distance of the point of intersection of the lines r→=(i^+j^−k^)+λ(ai^−j^),a≠0 and r→=(4i^−k^)+μ(2i^+ak^) from the origin is:

To find the point of intersection of two lines in vector form, equate their corresponding coordinates.

Step 1: Equate components and solve for parameters✦ Active

The given lines are r→1=(i^+j^−k^)+λ(ai^−j^) and r→2=(4i^−k^)+μ(2i^+ak^). Equating the components of the position vectors at the point of intersection:

1+aλ=4+2μ(x-component) 1−λ=0(y-component) −1=−1+aμ(z-component)

From the y-component, 1−λ=0⟹λ=1. From the z-component, −1=−1+aμ⟹aμ=0. Since a≠0, we must have μ=0. Substitute λ=1 and μ=0 into the x-component equation: 1+a(1)=4+2(0)⟹1+a=4⟹a=3. Thus, the value of a is 3.

Step 2: Determine the point of intersection○ Expand

Substitute λ=1 (or μ=0 and a=3) into either line equation. Using r→1 with λ=1 and a=3:

r→=(i^+j^−k^)+1(3i^−j^) r→=i^+j^−k^+3i^−j^ r→=4i^−k^

The point of intersection is P(4,0,−1).

Step 3: Calculate the square of the distance from the origin○ Expand

The point of intersection is (4,0,−1). The origin is (0,0,0). The distance d from the origin is:

d=(4−0)2+(0−0)2+(−1−0)2 d=42+02+(−1)2 d=16+0+1 d=17

The square of the distance is d2=(17)2=17.

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