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