question_answer

The distance of the plane passing through the point P (1, 1, 1) and perpendicular to the line \[\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}\] from the origin is

A)  \[\frac{3}{4}\]                                     

B)  \[\frac{4}{3}\]

C)  \[\frac{7}{5}\]                                     

D)  1

Correct Answer: C

Solution :

            Equation of the plane is             \[A(x-1)+B(y-1)\,+C(z-1)=0\] Since the line is perpendicular to the plane (1) \[\therefore \,3(x-1)\,+0(y-1)\,+4(z-1)=0\] \[\Rightarrow \,\,3x+0y\,+4z-7=0\]             \[\therefore \] Distance from (0, 0, 0) \[d=\,\left| \frac{-7}{5} \right|=\frac{7}{5}\]