Equation of a plane which intersects the two surfaces?

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• Oct 2nd 2009, 07:13 PM
Infernorage
Equation of a plane which intersects the two surfaces?
The question asks to find the equation for a plane which intersects these two surfaces. I was sure that I solved it correctly, but my professor says it is wrong and I don't know what I am doing wrong here. Here are the two surfaces...
1) $\displaystyle x^2+2y^2-z^2+3x=1$
2) $\displaystyle 2x^2+4y^2-2z^2-5y=0$

Here's what I did...
I found three points that are on both surfaces, hence points at which they intersect. I did this by multiplying the first surface equation by two and subtracting the equations from each other to end up with it $\displaystyle 6x+5y=2$. After that, I just picked some numbers for x and y and then plugged it back into the surface equations to find z. The three points I found were $\displaystyle (1/3,0,1/3), (1/3,0,-1/3)$, and $\displaystyle (2,-2,\sqrt{17})$.

After finding these three points, I just used them to find the normal vector and then used the normal vector and the first point to find the equation. Here is the equation that I got...
$\displaystyle \frac{4}3x-\frac{10}9y+\frac{4}9=0$

Can someone please tell me whether this is correct or not. If its not correct, can someone tell me what I might be doing wrong? Thanks.
• Oct 2nd 2009, 07:44 PM
DeMath
Quote:

Originally Posted by Infernorage
The question asks to find the equation for a plane which intersects these two surfaces. I was sure that I solved it correctly, but my professor says it is wrong and I don't know what I am doing wrong here. Here are the two surfaces...
1) $\displaystyle x^2+2y^2-z^2+3x=1$
2) $\displaystyle 2x^2+4y^2-2z^2-5y=0$

Here's what I did...
I found three points that are on both surfaces, hence points at which they intersect. I did this by multiplying the first surface equation by two and subtracting the equations from each other to end up with it $\displaystyle 6x+5y=2$. After that, I just picked some numbers for x and y and then plugged it back into the surface equations to find z. The three points I found were $\displaystyle (1/3,0,1/3), (1/3,0,-1/3)$, and $\displaystyle (2,-2,\sqrt{17})$.

After finding these three points, I just used them to find the normal vector and then used the normal vector and the first point to find the equation. Here is the equation that I got...
$\displaystyle \frac{4}3x-\frac{10}9y+\frac{4}9=0$

Can someone please tell me whether this is correct or not. If its not correct, can someone tell me what I might be doing wrong? Thanks.

$\displaystyle \left\{ \begin{gathered} {x^2} + 2{y^2} - {z^2} + 3x = 1, \hfill \\ 2{x^2} + 4{y^2} - 2{z^2} - 5y = 0; \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} {z^2} = {x^2} + 2{y^2} + 3x - 1, \hfill \\ {z^2} = {x^2} + 2{y^2} - \frac{5} {2}y; \hfill \\ \end{gathered} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow {x^2} + 2{y^2} + 3x - 1 = {x^2} + 2{y^2} - \frac{5}{2}y \Leftrightarrow$

$\displaystyle \Leftrightarrow 3x + \frac{5}{2}y - 1 = 0 \Leftrightarrow 6x + 5y - 2 = 0.$