JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q83

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The shortest distance between the lines $\dfrac x2=\dfrac y2=\dfrac z1$ and $\dfrac{x+2}{-1}=\dfrac{y-4}8=\dfrac{z-5}4$ lies in the interval :

[0, 1)

[1, 2)

(2, 3]

(3, 4]

Answer and worked solution

Official answer: C

$\vec d_1\times\vec d_2=(0,-9,18)$ , of length $9\sqrt5$ , and $(P_2-P_1)\cdot(0,-9,18)=54$ . The distance is $\frac{6}{\sqrt5}\approx2.68$ , in (2, 3].

Topics:Skew lines

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 83