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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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If the line drawn from the point $(-2,-1,-3)$ meets a plane at right angle at the point $(1,-3,3)$ find the equation of the plane.

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  • Any line which is perpendicular to the plane is the normal
  • Equation of a plane is $ ax+by +cz+d=0$ where $(a,b,c)$ is normal
Let the given points be $P(-2,-1,-3)$ and $(1,-3,3)$ and the line $PQ$ be perpendicular to the plane.
Therefore $ PQ=(x_2-x_1),(y_2-y_1),(z_2-z_1)=(1-(-2)),(-3-(-1)),(3-(-3))$
$\qquad=(3,-2,6)$
It is given that $Q$ is the foot of the perpendicular to $P$ on the plane.
This implies that point Q lies on the plane.
Equation of the plane is $ax+by+cz+d$,where $(a,b,c)$ is normal.
$(ie) 3x-2y+6z+d=0$
Since Q lies on the plane,let us substitute for $(x,y,z)$
$(ie) ( 3 \times 1)+(-2) \times (-3)+(6 \times 3)+ d=0$
$(ie) 3+6+18+d=0$
$=> d=-27$
Therefore equation of the plane is
$3x-2y+6z-27=0$
answered Jun 10, 2013 by meena.p
 

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