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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the $\perp$ distance of the point (2,3,-5) from the plane $x+2y-2z-9=0$

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  • Distance of any point $ (x_1,y_1,z_1)$ from the plane $ ax+by+cz+d=0$ is given by $ \bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
Given point $Q(2,3,-5)$ and Eqn. of plane $x+2y-2z-9=0$
We know that the distance of any point $ (x_1,y_1,z_1)$ from the plane $ ax+by+cz+d=0$ is given by
$ \bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
$\perp$ distance =$\bigg|\large\frac{2+6+10-9}{\sqrt {1^2+2^2+(-2)^2}}\bigg|=\frac{9}{3}$
$=3$

 

answered May 22, 2013 by rvidyagovindarajan_1
edited May 23, 2013 by rvidyagovindarajan_1
 

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