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The equation of the plane having intercepts of $3$ and $4$ with $x$ and $z$ axes respectively and parallel to $y$ axis is ?

$\begin{array}{1 1} (a)\:3x+4z=12\:\:\qquad\:(b)\:4x+3z=12\:\:\qquad\:(c)\:3y+4z=12\:\:\qquad\:(d)\:4y+3z=12 \end{array} $

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  • Equation of a plane with $a,b,c$ as its respective intercepts is $\large\frac{x}{a}+\frac{y}{b}+\frac{z}{c}$$=1$
Given: $x\:intercept\:\:and\:\:z\:intercept$ are $3\:\:and\:\:4$ for the plane.
Let the $y\:intercept$ be $b$.
$\therefore$ Eqn. of the plane is $\large\frac{x}{3}+\frac{y}{b}+\frac{z}{4}$$=1$
$\Rightarrow\:$ The normal ($\overrightarrow n$) to the plane is $ (\large\frac{1}{3},\frac{1}{b},\frac{1}{4})$
Also it is given that the plane is parallel to $y\:axis$.
$\Rightarrow\:$ Its normal is $\perp$ to $y\:axis$
$\Rightarrow\:$ Its $y\:intercept$ is $0.$
$\therefore$ The eqn. of the plane is $\large\frac{x}{3}+\frac{z}{4}$$=1$
answered Jan 5, 2014 by rvidyagovindarajan_1

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