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Q.
A plane which passes through the point $ (3,\,2,\,0) $ and the line $ \frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4} $ is:
Jharkhand CECEJharkhand CECE 2006
Solution:
If a plane contains a line, then the normal to the plane is perpendicular to the line.
Any plane passing through $(3,2,0)$ is
$a(x-3)+b(y-2)+c(z-0)=0 \ldots$ (i)
It passes through $(3,6,4)$
$\therefore 0 . a+4 b+4 c=0 \ldots$ (ii)
Normal to plane (i) is perpendicular to given line
$\therefore a+5 b+4 c=0 \ldots$ (ii)
On solving Eqs. (i) and (ii), we get
$\frac{a}{1}=\frac{b}{-1}=\frac{c}{1}=k$
So, $a=k, b=-k, c=k$
Putting the value of $a, b, c$ in Eq. (i),
we get $x-y+z=1$
In any line of a plane it has only one direction cosines and direction ratios may be more than one.