question_answer

Consider two lines \[{{L}_{1}}:\frac{x-7}{3}=\frac{y-7}{2}=\frac{z-3}{1}\] and\[{{L}_{2}}:\frac{x-1}{2}=\frac{y+1}{4}=\frac{z+1}{3}\]. lf a line L whose direction ratios are \[\left\langle 2,2,1 \right\rangle \] intersect the lines \[{{L}_{1}}\]and \[{{L}_{2}}\] at A and B then the distance AB is

A)  18                               

B)  24

C)  36                               

D)  50

Correct Answer: A

Solution :

\[A\equiv \,(3\lambda +7,\,\,2\lambda +7,\,\,\lambda +3)\] \[B\equiv \,(2\mu +1,\,4\mu -1,\,3\mu -1)\] \[\therefore \,\,\frac{3\lambda -2\lambda +6}{2}=\frac{2\lambda -4\mu +8}{2}\,=\frac{\lambda -3\mu +4}{1}\] \[\therefore \,\,\lambda =2\] and \[\mu =0\] \[\Rightarrow \,A=(13,\,11,5),\,\,B=\,(1,-1,\,-1)\] \[\therefore \,\,AB=18\]