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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the equation of the plane through the intersection of the planes $3x -y + 2z -4 = 0$ and $x + y + z -2 = 0$ and the point $(2, 2, 1).$

$\begin{array}{1 1} 7x-5y+4z-8=0 \\ 11x-y+8z-16=0 \\ 7x-5y+4z+8=0 \\ 7x-5y-4z-8=0\end{array} $

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Toolbox:
  • The equation of a plane passing through the intersection of two planes is $(a_1x_1+b_1y_1+c_1z_1+d_1)+\lambda(a_2x_2+b_2y_2+c_2z_2+d_2)=0$
  • Where $\lambda$ is any real number.
Step 1:
Given planes are $3x-y+2z=4$ and $x+y+z=2$
We know equation of a plane passing through the intersection of two planes is $(a_1x_1+b_1y_1+c_1z_1+d_1)+\lambda(a_2x_2+b_2y_2+c_2z_2+d_2)=0$
Where $\lambda$ is any real number.
Therefore $(3x-y+2x-4)+\lambda(x+y+z-2)=0$-------(1)
The given plane passes through the point $(2,2,1)$
Now substituting for $x,y$ and $z$ we get,
$(3\times 2-2+2\times 1-4)+\lambda(2+2+1-2)=0$
$\Rightarrow (6-2+2-4)+\lambda(3)=0$
Step 2:
On simplifying we get
$2+3\lambda=0$
Therefore $\lambda=\large\frac{-2}{3}$
Now substituting the value of $\lambda$ in equ(1)
We get
$(3x-y+2z-4)-\large\frac{2}{3}$$(x+y+z-2)=0$
$\Rightarrow (9x-3y+6z-12)-(2x+2y+2z-4)=0$
On simplifying we get
$7x-5y+4z-8=0$
This is the required equation of the plane.
answered May 31, 2013 by sreemathi.v
 

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