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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 \(x+2y+3x-7=0 \: and \: 2x-3y+4z=0.\)

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Toolbox:
  • When two planes are $\perp$ then $a_1a_2+b_1b_2+c_1c_2=0$
  • Where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are their direction ratios.
  • The equation of the plane containing the given point is
  • $A(x-x_1)+B(y-y_1)+C(z-z_1)=0$
Step 1:
The given point is $(1,1,-1)$
The equation of the plane containing the given point is
$A(x-x_1)+B(y-y_1)+C(z-z_1)=0$
(i.e) $A(x-1)+B(y-1)+C(z-1)=0$------(1)
Step 2:
When two planes are $\perp$ then
$a_1a_2+b_1b_2+c_1c_2=0$
Where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are their direction ratios.
The plane(1) is $\perp$ to the plane
$x+2y+3z-7=0$
$\therefore A(1)+B(2)+C(3)=0$
$\Rightarrow A+2B+3C=0$-----(2)
Step 3:
The plane(1) is $\perp$ to the plane $2x-3y+4z=0$
$\Rightarrow A(2)-3(B)+4(C)=0$
$\therefore 2A-3B+4C=0$-----(3)
Step 4:
Let us Solve equ(2) and equ(3) to find the values of A,B,and C
$\large\frac{A}{\begin{vmatrix}2 & 3\\-3 &4\end{vmatrix}}=\large\frac{B}{\begin{vmatrix}1 & 3\\2 &4\end{vmatrix}}=\large\frac{C}{\begin{vmatrix}1 & 2\\2 &-3\end{vmatrix}}$
$\Rightarrow \large\frac{A}{8+9}=\frac{B}{6-4}=\frac{C}{-3-4}$
$\Rightarrow \large\frac{A}{17}=\frac{B}{2}=\frac{C}{-7}$$=\lambda$(say)
$\Rightarrow A=17\lambda,B=2\lambda$ and $C=-7\lambda$
Step 5:
Substituting these values in equ(1) we get,
$17x+2y-7z=26$
This is the required equation of the plane.

 

answered Oct 4, 2013 by sreemathi.v
edited Oct 31, 2013 by pady_1
 

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