A) (3a, 3a, 3a), (a, a, a)
B) (3a, 2a, 3a), (a, a, a)
C) (3a, 2a, 3a), (a, a, 2a)
D) (2a, 3a, 3a), (2a, a, a)
Correct Answer: B
Solution :
Let the equation of line AB be \[\frac{x-0}{1}=\frac{y+a}{1}=\frac{z-0}{1}=k\] (say) \[\therefore \]Coordinates of E are\[(k,k-a,k)\]. Also, the equation of other line CD is \[\frac{x+a}{2}=\frac{y-0}{1}=\frac{z-0}{1}=\lambda \] \[\therefore \]Coordinates of F are\[(2\lambda -a,\lambda ,\lambda )\]. Direction ratios of EF are proportional to 2, 1, 2, \[(k-2\lambda +a),(k-\lambda -a),(k-\lambda )\] \[\therefore \] \[\frac{k-2\lambda +a}{2}=\frac{k-\lambda -a}{1}=\frac{k-\lambda }{2}\] On solving first and second fractions, \[\frac{k-2\lambda +a}{2}=\frac{k-\lambda -a}{1}\] \[\Rightarrow \]\[k-2\lambda +a=2k-2\lambda -2a\] \[\Rightarrow \] \[k=3a\] On solving second and third fractions, \[\frac{k-\lambda -a}{1}=\frac{k-\lambda }{2}\] \[\Rightarrow \]\[2k-2\lambda -2a=k-\lambda \] \[\Rightarrow \] \[k-\lambda =2a\] \[\Rightarrow \] \[\lambda =k-2a=3a-2a\] \[\Rightarrow \] \[\lambda =a\]and \[k=3a\] \[\therefore \]Coordinates of E = (3a, 2a, 3a) and coordinates of F = (a, a, a).You need to login to perform this action.
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