A) \[(4,0,1)\]
B) \[(4,0,-1)\]
C) \[(1,1,1)\]
D) \[(1,1,-1)\]
Correct Answer: B
Solution :
Given liens are \[\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\] ?.. (i) and \[\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}\] ?..(ii) Let point \[\{(3r+1),\,-r+1,-1\}\] lie on the line (i) which is the intersection point of lines (i) and (ii). Therefore, this point lies on line (ii), also \[\therefore \] \[\frac{3r+1-4}{2}=\frac{-r+1}{0}=\frac{-1+1}{3}\] \[\Rightarrow \] \[3r-3=0\] \[\Rightarrow \] \[r=1\] \[\therefore \] Intersection point is \[(3+1,-1+1,-1)=(4,0,-1)\]
You need to login to perform this action.
You will be redirected in
3 sec
