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find equation of plane containing (1,2,3) and perpendicular to the place 3x- y +2z= 3 and parallel to the line x-1/3=y/2=z/-2.write equation in dot product form

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Let the equation of the plane be $ax+by+cz+d=0$
$\Rightarrow\:$ D.R. of the normal to the plane is $ (a,b,c)$
Given: This plane is $\perp$ to the plane $3x-y+2z-3=0$
$\therefore\: (a,b,c).(3,-1.2)=0$ $ \Rightarrow\:3a-b+2c=0.....(i)$
Also given that the plane is parallel to the line $\large\frac{x-1}{3}=\frac{y}{2}=\frac{z}{-2}=\lambda\:\: (say)$
$\Rightarrow\:$ Normal to the plane is $\perp$ to the line.
It is given that the plane passes through the point $(1,2,3)$.
Solving $(i)\:\:and\:\:(ii)$ we get
$\Rightarrow\:a=-2,\:b=12, \:c=9$
Substituting the values in the equation of the plane we get
But given that the point $(1,2,3)$ lies on the plane.
$\therefore$ The required equation of the plane is
In dot product form the equation is :   $\overrightarrow r.(2\hat i-12\hat j-9\hat k)+49=0$


answered Nov 13, 2013 by rvidyagovindarajan_1
edited Nov 13, 2013 by rvidyagovindarajan_1

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