IIT-JEE 2010 · Paper 1 · Mathematics Q51

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MathematicsInteger+3Question 51 of 84

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines $\dfrac{x-1}2=\dfrac{y-2}3=\dfrac{z-3}4$ and $\dfrac{x-2}3=\dfrac{y-3}4=\dfrac{z-4}5$ is $\sqrt6$ , then $|d|$ is

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Answer and worked solution

Official answer: 6

Plane of the two lines. Normal $(2,3,4)\times(3,4,5)=(-1,2,-1)$ , through $(1,2,3)$ :
$-(x-1)+2(y-2)-(z-3)=0\iff x-2y+z=0.$

Parallel plane. For a distance to be defined, $Ax-2y+z=d$ must be parallel to it, so $A=1$ . Then
$\frac{|d|}{\sqrt6}=\sqrt6\implies|d|=6.$

Answer: 6

Topics:Angles and distances Equation of a plane

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IIT-JEE 2010 · Paper 1, Mathematics question 51