question_answer

Perpendicular are drawn from points on the line \[\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}\] to the plane \[x+y+z=13\]. The feet of 2-13 perpendiculars lie on the line

A) \[\frac{x}{2}=\frac{y-1}{8}=\frac{z-2}{-13}\]

B)   \[\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}\]

C) \[\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}\]        

D) \[\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}\]

Correct Answer: D

Solution :

Any point on line \[\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda \] is \[\operatorname{P}\left( 2\lambda -2,\,\,-\lambda -1,\,\,3\lambda  \right).\] This point lies on the plane \[\operatorname{x}+y+ z = 3.\] So, \[\lambda  =6 \,\Rightarrow \,\,\lambda  = 3/2\] Thus, point P is \[\left( 1,\,-5/2,\,9/2 \right).\] Equation of line RQ is \[\frac{x+2}{1}=\frac{y+2}{1}=\frac{z}{1}=t\] So, any point on this line is \[\operatorname{Q}(t -2,t -1,t).\] This point lies on the plane \[\operatorname{x}+y+ z = 3.\] \[\therefore \,\,\,t=2\] point Q is (0, 1, 2). Therefore, direction ratios of PQ are (1, -7/2, 5/2) or (2, -7, 5). i.e. feet of perpendiculars lies on \[\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}\]