JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q82

← JEE Main 2017 · 2 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 82 of 90

The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular to both the lines $\dfrac{x-1}1=\dfrac{y+2}{-2}=\dfrac{z-4}3$ and $\dfrac{x-2}2=\dfrac{y+1}{-1}=\dfrac{z+7}{-1}$ , is :

\dfrac{10}{\sqrt{83}}

\dfrac5{\sqrt{83}}

\dfrac{10}{\sqrt{74}}

\dfrac{20}{\sqrt{74}}

Answer and worked solution

Official answer: A

The normal is $(1,-2,3)\times(2,-1,-1)=(5,7,3)$ , so the plane is $5x+7y+3z+5=0$ . The distance of $(1,3,-7)$ from it:
$\frac{|5+21-21+5|}{\sqrt{83}}=\frac{10}{\sqrt{83}}.$

Topics:Angles and distances Equation of a plane

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JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 82