JEE Main 2015 · 10 Apr · Shift 1 · Mathematics Q84

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If the shortest distance between the lines $\dfrac{x-1}\alpha=\dfrac{y+1}{-1}=\dfrac z1$ , $(\alpha\ne-1)$ and $x+y+z+1=0=2x-y+z+3$ is $\dfrac1{\sqrt3}$ , then a value of α is :

-\dfrac{16}{19}

-\dfrac{19}{16}

\dfrac{32}{19}

\dfrac{19}{32}

Answer and worked solution

Official answer: C

The second line has direction $(1,1,1)\times(2,-1,1)=(2,1,-3)$ and passes through $\left(-\frac43,\frac13,0\right)$ (put $z=0$ ).

The common normal is $(\alpha,-1,1)\times(2,1,-3)=(2,\,3\alpha+2,\,\alpha+2)$ . Joining $(1,-1,0)$ to that point gives $\left(-\frac73,\frac43,0\right)$ , so
$\frac{\left|-\frac{14}3+4\alpha+\frac83\right|}{\sqrt{4+(3\alpha+2)^2+(\alpha+2)^2}}=\frac1{\sqrt3}.$
$\alpha=\frac{32}{19}$ satisfies this, as keyed.

Topics:Skew lines Equation of a line in space

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JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 84