IIT-JEE 2008 · Paper 2 · Mathematics Q19

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MathematicsSingle correct+4 / -1Question 19 of 66

Passage

Paragraph for Question Nos. 17 to 19

Consider the lines
$L_1:\frac{x+1}3=\frac{y+2}1=\frac{z+1}2,\qquad L_2:\frac{x-2}1=\frac{y+2}2=\frac{z-3}3.$

The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the lines $L_1$ and $L_2$ is

\dfrac2{\sqrt{75}}

\dfrac7{\sqrt{75}}

\dfrac{13}{\sqrt{75}}

\dfrac{23}{\sqrt{75}}

Answer and worked solution

Official answer: C

Plane. Normal $(-1,-7,5)$ through $(-1,-2,-1)$ :
$-(x+1)-7(y+2)+5(z+1)=0\iff x+7y-5z+10=0.$

Distance of (1, 1, 1). $\dfrac{|1+7-5+10|}{\sqrt{75}}=\dfrac{13}{\sqrt{75}}$ .

(C)

Topics:Angles and distances Equation of a plane

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IIT-JEE 2008 · Paper 2, Mathematics question 19