Three Dimensional Geometry Question 383
Question: Which one of the following is the plane containing the lien $ \frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{5} $ and parallel to z axis?
Options:
A) $ 2x-3y=0 $
B) $ 5x-2z=0 $
C) $ 5y-3z=0 $
D) $ 3x-2y=0 $
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Answer:
Correct Answer: D
Solution:
[d] The equation of the line is $ \frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{5}=r $ Where r is a constant. Any point on this line, is given by $ x=2r+2,y=3r+2 $ and $ z=5r+4 $ Since, a plane that is parallel to z-axis will have no z-co-ordinate, z=0 $ z=0\Rightarrow 5r+4=0 $ or $ r=\frac{-4}{5} $ Putting this value of r for x and y co-ordinates. $ x=2r+2=2\times (-\frac{4}{5})+2 $ or $ 5x=-8+10=2 $ $ x=\frac{2}{5},or\frac{2}{x}=5 $ ?..(1) Similarly, $ y=3r+3=3\times (-\frac{4}{5})+3 $ or, $ 5y=-12+15=3 $ $ y=\frac{3}{5}\Rightarrow \frac{3}{y}=5 $ ?..(2) From equation s(1) and (2) $ \frac{2}{x}=\frac{3}{y}\Rightarrow 3x-2y=0 $
BETA
