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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Determine whether the given planes are parallel or perpendicular, and in case they are neither, then find the angles between them. $(c)\; 2x -2y + 4z + 5 = 0$ and $3x -3y + 6z - 1 = 0 $

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  • The direction ratios of normal to the plane $l_1=a_1x+b_1y+c_1 x=0$ are $a_1,b_1$ and $c_1$ and $l_2=a_2x+b_2y+c_2z=0$ are $a_2,b_2,c_2$
  • If $L_1\parallel L_2$ then $\large\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Step 1:
The equation of the given planes are $2x-2y+4z+5=0$ and $3x-3y+6z-1=0$
The direction cosines of $L_1$ are $(2,-2,4)$
The direction cosines of $L_2$ are $(3,-3,6)$
Step 2:
Therefore $\large\frac{a_1}{a_2}=\frac{2}{3}$
$\large\frac{b_1}{b_2}=\frac{-2}{-3}$
$\large\frac{c_1}{c_2}=\frac{4}{6}=\frac{2}{3}$
Hence $\large\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Therefore the given planes are parallel to each other.
answered Jun 3, 2013 by sreemathi.v
 

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