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If from a point $P(a,b,c)$ perpendiculars $PA,\:PB,\:$ are drawn on $yz\:\:and\:\:zx$ planes. then the equation of the plane $OAB$ is ?

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Given that $A$ is foot of $\perp$ of $P(a,b,c)$ on $yz$ plane and
$B$ is foot of $\perp$ drawn from $P$ on z$x$ plane.
$\therefore \:A\:\:and\:\:B$ are given by $A(0,b,c)$ and $B(a,0,c)$
Normal to the plane $OAB$ is given by $(0,b,c)\times(a,0,c)= (bc,ca,-ab)$
$\therefore$ The eqn. of the plane with normal $(bc,ca,-ab)$ and passing through  $O(0,0,0)$ is


answered Jan 11, 2014 by rvidyagovindarajan_1

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