JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

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?

A) \[2x-3y=0\]

B) \[5x-2z=0\]

C) \[5y-3z=0\]

D) \[3x-2y=0\]

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)

\[\frac{2}{x}=\frac{3}{y}\Rightarrow 3x-2y=0\]