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Home  >>  TN XII Math  >>  Vector Algebra
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Find the distane between the parallel planes $x-y+3z+5=0; 2x-2y+6z+7=0$

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  • The distance between the parallel planes $ ax+by+cz+d_1=0\: and \: ax+by+cz+d_2=0$ is $ \large\frac{|d_1-d_2|}{\sqrt{^2+b^2+c^2}}$
The distancebetween the parallel planes
$ x-y+3z+5=0\: and \: 2x-2y+6z+7=0$ i.e.,
$ 2x-2y+6z+10=0\: and \: 2x-2y+6z+7=0$
$ d_1 = 10 \: and \: d_2 = 7$
The distance between the planes is
$ d = \large\frac{|d_1.d_2|}{\sqrt{a^2+b^2+c^2}}= \large\frac{10-7}{\sqrt{4+4+36}}=\large\frac{3}{\sqrt{44}}=\large\frac{3}{2\sqrt{11}}$ units
answered Jun 17, 2013 by thanvigandhi_1
 

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