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The equation of the plane passing through the point $(3,2,0) $ and the line $\large\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is ?

$\begin{array}{1 1}(a)\:\:x-y+z=1\:\:\:\qquad\:\:(b)\:\:x+y+z=5\:\:\:\qquad\:\:(c)\:\:x+2y-z=1\:\:\:\qquad\:\:(d)\:\:2x-y+z=5. \end{array} $

1 Answer

Let the equation of the plane be $ax+by+cz+d=0$
$\Rightarrow\:$ The normal to the plane $\overrightarrow n$ is $(a,b,c)$
Since the plane passes through $(3,2,0)$, $ 3a+2b+d=0$.....(i)
Also the plane passes through the line $\large\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$
$\Rightarrow\:3a+6b+4c+d=0$......(ii) and
since the line lies on the plane, the line is $\perp$ to the normal $\overrightarrow n$
$\therefore (a,b,c).(1,5,4)=0$
Solving $(iii)\:\:and\:\:(iv) $ using cross product method,
Substituting the values of $a,b,c$ in (i) we get $ d=-1$
$\therefore$ The equation of the plane is $x-y+z=1$
answered Dec 28, 2013 by rvidyagovindarajan_1

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