question_answer

A line with direction cosines proportional to 2,1,2 meets each of the lines\[x=y+a=z\]and \[x+a=2y=2z\]. The coordinates of each of the points of intersection are given by

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).