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 December 15th, 2012, 08:56 AM #1 Guest   Joined: Posts: n/a Thanks: Analytic geometry. Find the equation of the $\pi$ plane which passes through the point $M(2,-1,1)$ and is normal with the planes : $\pi_1 :3x+2y-z+4=0$ and $\pi_2 : x+y+z-3=0$
December 15th, 2012, 09:36 AM   #2
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Re: Analytic geometry.

Hello, noki!

Quote:
 Find the equation of the plane which passes through the point $M(2,-1,1)$ and is normal to the planes: $\begin{Bmatrix}3x\,+\,2y\,-\,z\,+\,4=&0 \\ \\ \\ x\,+\,y\,+\,z\,-\,3=&0\end{Bmatrix}=$

The normal of the desired plane is perpendicular
[color=beige]. . [/color]to the normals of both given planes.

$\vec n \;=\;\begin{vmatrix}i&j=&k \\ \\ 3=&2=&-1 \\ \\ 1=&1=&1\end{vmatrix} \;=\;i(2\,+\,1)\,-\,j(3\,+\,1)\,+\,k(3\,-\,2) \;=\;3i\,-\,4j\,+\,k$

Hence:[color=beige] .[/color]$\vec n \:=\:\langle 3,\,-4,\,1\rangle$

The equation of the plane is:[color=beige] .[/color]$3(x\,-\,2)\,-\,4(y\,+\,1)\,+\,1(z\,-\,1) \;=\;0$

[color=beige]. . [/color]which simplifies to:[color=beige] .[/color]$3x\,-\,4y\,+\,z\;=\;11$

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