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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The distance of the plane $2x-3y+6z+14=0$ from origin is ?

$\begin{array}{1 1} 2 \\ 3 \\ 11 \\ 14 \end{array} $

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  • The distance of the plane $ax+by+cz+d=0$ from origin = $\bigg|\large\frac{d}{\sqrt {a^2+b^2+c^2}}\bigg|$
The distance of the plane $2x-3y+6z+14=0$ from origin is $\large \frac{14}{\sqrt {4+9+36}}$
$=\large\frac{14}{7}$$=2$
answered Dec 22, 2013 by rvidyagovindarajan_1
 

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