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If $L\rightarrow\:\:\large\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-3}{-1}$ and the plane $P\rightarrow\:\:x-2y-z=0$, then which of the following is true?

$\begin{array}{1 1} (a)\:\:L\:is\:\perp \:P\:\: \\ (b)\:\:L\:lies\:in\:P\:\:\\(c)\:\:L\:is\:||\:to\:P\:\: \\ (d)\:\:L\:intersects\:P\:but\:not\:\perp \end{array} $

1 Answer

For a line to be $\perp$ to a plane, $d.r.$ of the line should be same as that of normal to the plane.
From the equation of $L$, $d.r.$ of the line =$\overrightarrow b= (3,2,-1)$
From the equation of $P$, normal to the plane is $\overrightarrow n=(1,-2,-1)$
$\Rightarrow\:$the line is not $\perp$ to the plane.
But since $\overrightarrow b.\overrightarrow n=0$, the line can be || to the plane.
Since the point on the line $\overrightarrow a=(1,-1,3)$ satisfies the equation of the plane,
The line lies on the plane.
answered Jan 2, 2014 by rvidyagovindarajan_1

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