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KEAM 2010 Math Paper

Question : 14 of 120

Marks: +1, -0

2x-3y+5z-7=0

Solution:

Comparing given plane

2x-3y+5z-7=0

with the plane

lx+my+nz+d=0

\Rightarrow \;\; l=2, \;\; m=-3, \;\; z=5

since, the required line is perpendicular to the given plane. Therefore, these are the direction cosine of the line.

∴ Equation of line is

\frac{x-3}{2}=\frac{y-0}{-3}=\frac{z+4}{5}

\Rightarrow \;\; \frac{x-3}{2}=-\frac{y}{3}=\frac{z+4}{5}

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