A) \[\frac{x}{0}=\frac{y}{0}=\frac{z+4}{1}\]
B) \[\frac{x}{1}=\frac{y}{2}=\frac{z}{-3}\]
C) \[\frac{x}{1}=\frac{y}{1}=\frac{z-4}{-7}\]
D) None of these
Correct Answer: D
Solution :
[d] The plane is \[x+3y+z-4+\lambda (2x-y)=0\] This always passes through the intersection of the planes \[x+3y+z-4=0\]and \[2x-y=0,\] which is a line. Now, \[2x-y=0\] \[\Rightarrow \,\,\,\frac{x}{1}=\frac{y}{2}\] \[\therefore \,\,\,x+3y+z-4=0\] \[\Rightarrow \,\,\,x+3\times 2x+z-4=0\] \[\Rightarrow \,\,\,7x+z-4=0\] \[\therefore \,\,\,\,7x=-(z-4)\] \[\Rightarrow \,\,\,\frac{x}{1}=\frac{z-4}{-7}\] Hence, line is \[\frac{x}{1}=\frac{y}{2}=\frac{z-4}{-7}.\]
You need to login to perform this action.
You will be redirected in
3 sec
