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The equation of the plane whose foot of $\perp$ drawn from origin is $(1,2,1)$. is?

$\begin{array}{1 1} (a)\:\:x+2y+z+6=0\:\: & \:\:(b)\:\:x+2y+z-6=0\:\:\\ \:\:(c)\:\:x+2y+z+4=0\:\: &\:\: (d)\:\:x+2y+z-4=0 \end{array} $

1 Answer

Since $A(1,2,1)$ is foot of $\perp$ from origin to the required plane, $A$ lies on the plane.
Also $OA$ is $\perp$ to the plane $\therefore$ $\overrightarrow n=\overrightarrow {OA}=(1,2,3)$
Let thequation of the plane be $ax+by+cz+d=0$ where $\overrightarrow n=(1,2,1)$
$\therefore $ The eqn. becomes $x+2y+z+d=0$
Since it passes through $A(1,2,1)$ it saatisfies the eqn.
$\therefore \:1+2+1+d=0$ $\Rightarrow\:d=-4$
$\Rightarrow\: $ The eqn. of the plane is $x+2y+z-4=0$
answered Jan 2, 2014 by rvidyagovindarajan_1

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