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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  3-D Geometry

The value of $k$ for which the line $\large\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ to lie on the plane $2x-4y+3z=6$ is ?

$(a)\:2\:\:\:\qquad\:\:(b)\:\:-2\:\:\:\qquad\:\:(c)\:\:6\:\:\:\qquad\:\:(d)\:\:No\:real\:value\:of\:k$

1 Answer

For any line to lie on a plane, two conditions are to be satisfied.
1. The line should be $\perp$ to the normal $\overrightarrow n $ to the plane and
$i.e.,\:\:\overrightarrow b.\overrightarrow n=0$
2. Every point on the line should satisfy the eqn. of the plane.
Here $d.r.$ of the line $\overrightarrow b=(1,1,2)$ and a point on the line = $(4,2,k)$
Normal to the plane $\overrightarrow n=(2,-4,3)$
But $\overrightarrow b.\overrightarrow n\neq 0$
$\therefore$ the line cannot lie on the plane for any real value of $k$.
answered Jan 6, 2014 by rvidyagovindarajan_1
 

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