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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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Equation of the line through the point $(2,-1,-1)$, which is parallel to the plane $4x+y+z+2=0$ ans is $\perp$ to the line $x=\large\frac{y}{-2}=$$z-5$ is ?

$(a)\:\large\frac{x-2}{4}=\frac{y+1}{1}=\frac{z+1}{1}\:\qquad\:(b)\:\large\frac{x+2}{4}=\frac{y-1}{1}=\frac{z-1}{3}\:\qquad\:(c)\:\large\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}\:\qquad\:(d)\:\large\frac{x+2}{-1}=\frac{y-1}{1}=\frac{z-1}{3}$

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Given: Eqn. of the plane $4x+y+z+2=0$ ......(i)
Eqn. of the line is $\large\frac{x}{1}=\frac{y}{-2}=\frac{z-5}{1}$$=\lambda$.....(ii)
A line parallel to the plane (i) is $\perp$ to the normal, $ \overrightarrow n=(4,1,1)$ to (i)
A line $\perp$ to (ii) and $\overrightarrow n$ has $d.r. = \overrightarrow n\times (1,-2,1)$
(Since a vector $\perp$ to $\overrightarrow a\:\:and\:\:\overrightarrow b$ is $\overrightarrow a\times\overrightarrow b$)
$(4,1,1)\times (1,-2,1)=(3,-3,9)$
$\therefore$ $d.r.$ of the required line is $(-3,3,9)=(-1,1,3)$
$\therefore$ The eqn, of the required line through $(2,-1,-1)$ is
$\large\frac{x-2}{-1}=\frac{y+1}{1}=\frac{z+1}{3}$
answered Jan 8, 2014 by rvidyagovindarajan_1
 

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