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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the equation of the plane passing through the point (1,1,-1) and perpendicular to the planes : $ x+2y+3z-7=0 \: and \: 2x-3y+4z=0 $

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Toolbox:
  • Vector equation of a line passing through a point and parallel to a vector is $\overrightarrow r=\overrightarrow a+ \lambda \overrightarrow b$ where $\lambda \in R$
Step 1:
The equation of the plane containing the given point is
$A(x-1)+B(y-1)+C(z+1)=0$--------(1)
Applying the condition of perpendicularity to the plane (1) with the given planes $x+2y+3z-7=0$ and $2x-3y+4z=0$
$A+2B+3C=7$ and $2A-3B+4C=0$
Step 2:
Solving the equation we get,
$\large\frac{A}{\begin{vmatrix}2 & 3\\-3 & 4\end{vmatrix}}=\large\frac{B}{\begin{vmatrix}3 & 1\\4 & 2\end{vmatrix}}=\large\frac{C}{\begin{vmatrix}1 & 2\\2 & -3\end{vmatrix}}$
$\Rightarrow \large\frac{A}{8+9}=\frac{B}{6-4}=\frac{C}{-3-4}$
$\Rightarrow \large\frac{A}{17}=\frac{B}{2}=\frac{C}{-7}$$=\lambda$(say)
$\therefore A=17\lambda,b=2\lambda$ and $C=-7\lambda$
Step 3:
Substituting these values in equ(1) we get,
$17\lambda(x-1)+2\lambda(y-1)-7\lambda(z+1)=0$
$\Rightarrow 17x+2y-7z=26$
This is the required solution of the plane.
answered Nov 11, 2013 by sreemathi.v
 

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