Three Dimensional Geometry Question 185
Question: If a plane passes through the point (1,1,1) and is perpendicular to the line $ \frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4} $ , then its perpendicular distance from the origin is
[MP PET 1998]
Options:
A) $ \frac{3}{4} $
B) $ \frac{4}{3} $
C) $ \frac{7}{5} $
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
According to $ \frac{A}{l}=\frac{B}{m}=\frac{C}{n}, $ direction ratio of plane are respectively (3, 0, 4). Equation of plane passing through point (1, 1, 1) is
$ \Rightarrow A(x-x_1)+B,(y-y_1)+C,(z-z_1)=0 $
$ \Rightarrow 3,(x-1)+0,(y-1)+4,(z-1)=0 $
$ \Rightarrow 3x+4z-7=0 $ Normal form of plane is, $ \frac{3x}{5}+\frac{4z}{5}=\frac{7}{5} $
$ \therefore $ Perpendicular distance from $ (0,0,0)=\frac{7}{5} $ .
BETA
