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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The intercepts cut off by the plane $\overrightarrow r.(2\hat i+\hat j-\hat k)=5$ with the axes are ?

$(a)\:\:(2,1,-1)\:\qquad (b)\:\:(\large\frac{2}{5},\frac{1}{5},-\frac{1}{5})\:\:\qquad (c)\:\:(\frac{5}{2},5,-5)\:\:\qquad (d) \:\:(\frac{2}{\sqrt 6},\frac{1}{\sqrt 6},-\frac{1}{\sqrt 6})$

$\begin{array}{1 1} (2,1,-1) \\ (2/5,1/5,-1/5) \\ (5/2,5,-5) \\ (\frac{-2}{\sqrt 6} , \frac{1}{\sqrt 6} , \frac{-1}{\sqrt 6}) \end{array} $

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  • Intercept form of equation of a plane where $a,b,c$ are $x,y,z$ intercepts respectively is $\large\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.
Given equation of the plane is $\overrightarrow r.(2\hat i+\hat j-\hat k)=5$
$i.e.,$ $2x+y-z-5=0$
Converting the equation in intercept form we get
$\therefore\:$ $x\: intercept,\:y \:intercept\:and\:z\:intercept $ are given by
answered Dec 24, 2013 by rvidyagovindarajan_1

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