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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of a plane which is at a distance $3\sqrt 3$ units from origin and the normal to which is equally inclined to coordinate axis.

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  • $\perp$ diatance of a plane from the origin is =$\bigg|\large\frac{d}{\sqrt{a^2+b^2+c^2}}\bigg|$
Let the equation of the requierd plane be
$a x+by+cz+d=0$
It is given that the plane is equally inclined coordinate axes.
Hence its direction cosines are $(1,1,1)$
=> the equation of the plane is $(x+y+z+d)=0$
It is also given that $\perp$ distance from the origin is $3 \sqrt 3$
$(ie)\;\bigg| \large\frac{d}{a^2+b^2+c^2} \bigg| $ $= 3\sqrt 3$
Substituting for $(a,b,c)$ as $(1,1,1)$
$\bigg| \large\frac{d}{1^2+1^2+1^2} \bigg| $ $= 3\sqrt 3$
$\large\frac{d}{\sqrt 3}$$=\pm 3 \sqrt 3$
$=>d=\pm 9$
Therefore the equation of the plane is $x+y+z=\pm 9$
answered Jun 12, 2013 by meena.p
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