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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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A plane passes through the points $(2,0,0) (0,3,0)$ and $(0,0,4)$.The equation of plane is__________.

$\begin{array}{1 1} (A)\;\large\frac{x}{2}+\frac{y}{3}-\frac{z}{4}=1 \\(B)\;\large\frac{x}{8}+\frac{y}{7}+\frac{z}{4}=1 \\ (C)\;\large\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1 \\ (D)\;\large\frac{x}{2}-\frac{y}{3}-\frac{z}{4}=1 \end{array} $

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  • The cartesian equation of a plane intercepting length $a,b $ and $c$ with $x,y$ and $z$ axes $\large\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$=1$
The given points are $A(2,0,0),B(0,3,0)$ and $C(0,0,4)$
Here it is clear that the intercepting length $a,b,c$ with the $x,y,z$ axes is $2,3,4$
Therefore The equation of the plane is
$\large\frac{x}{2}+\frac{y}{3}+\frac{z}{4}$$=1$
answered Jun 12, 2013 by meena.p
 

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