JEE Main 2017 · 9 Apr · Shift 1 · Mathematics Q83

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MathematicsSingle correct+4 / -1Question 83 of 90

If the line, $\dfrac{x-3}1=\dfrac{y+2}{-1}=\dfrac{z+\lambda}{-2}$ lies in the plane, $2x-4y+3z=2$ , then the shortest distance between this line and the line, $\dfrac{x-1}{12}=\dfrac y9=\dfrac z4$ is :

Answer and worked solution

Official answer: C

The direction (1, −1, −2) is perpendicular to the normal (2, −4, 3), and $(3,-2,-\lambda)$ on the plane gives $\lambda=4$ . Then
$(P_2-P_1)\cdot(\vec d_1\times\vec d_2)=(-2,2,4)\cdot(14,-28,21)=0,$
so the lines intersect and the shortest distance is 0.

Topics:Skew lines Equation of a plane

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JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 83