IIT-JEE 2008 · Paper 2 · Mathematics Q18

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MathematicsSingle correct+4 / -1Question 18 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 shortest distance between $L_1$ and $L_2$ is

\dfrac{17}{\sqrt3}

\dfrac{41}{5\sqrt3}

\dfrac{17}{5\sqrt3}

Answer and worked solution

Official answer: D

Shortest distance. Points $A(-1,-2,-1)$ on $L_1$ and $B(2,-2,3)$ on $L_2$ : $\overrightarrow{AB}=(3,0,4)$ . Projecting on the common normal $\hat n=\frac{(-1,-7,5)}{5\sqrt3}$ :
$d=\frac{|-3+0+20|}{5\sqrt3}=\frac{17}{5\sqrt3}.$

(D)

Topics:Skew lines

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IIT-JEE 2008 · Paper 2, Mathematics question 18