Three Dimensional Geometry Question 220
Question: The equation of straight line passing through the point (a, b, c) and parallel to z- axis, is
[MP PET 1995; Pb. CET 2000]
Options:
A) $ \frac{x-a}{1}=\frac{y-b}{1}=\frac{z-c}{0} $
B) $ \frac{x-a}{0}=\frac{y-b}{1}=\frac{z-c}{1} $
C) $ \frac{x-a}{1}=\frac{y-b}{0}=\frac{z-c}{0} $
D) $ \frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1} $
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Answer:
Correct Answer: D
Solution:
The line through $ (a,b,c) $ is $ \frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n} $ ?..(i) Since the line is parallel to z-axis, therefore any point on this line will be of the form $ (a,b,z_1). $ Also any point on line (i) is $ (lr+a,mr+b,nr+c). $ Hence $ \begin{matrix} lr+a=a \\ mr+b=b \\ \end{matrix}\Rightarrow l=m=0 $ Hence the line will be $ \frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1} $ .
BETA
