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Determine the equation of a plane that is parallel to the plane $z=2x+3y$ and tangent to the graph of the function $f(x,y)=x^2+xy.$

I've been doing questions involving $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$ but this parallel stuff got me.

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You need to build a tangent plane parallel to the given plane. If the two planes are parallel it's because their normal vectors are a linear combination of each other.

If you consider the plane $z=2x+3y$, then the vectors $\langle 0,1,3\rangle$ and $\langle 1,0,2\rangle$ lie on that plane. Moreover $\langle 0,1,3\rangle\times\langle 1,0,2\rangle=\langle-2,-3,1\rangle$ is a perpendicular vector to the plane. So we need to know where we could build a tangent plane to the given surface such that the normal vectors to that plane can be written as $c\langle-2,-3,1\rangle,c\in\mathbb R$.

Consider the parameterization of the surface generated by the function mentioned: $\mathbf r(x,y)=\langle x,y,z=f(x,y)=x^2+xy\rangle$. A normal vector to this surface is $\mathbb r_x\times\mathbb r_y=\langle-f_x,-f_y,1\rangle$. Computing this yields that a normal vector to the surface generated by $f$ is $\langle2x+y,x,1\rangle$. If we choose $x=-3$ and $y=4$ then, if we construct a plane tangent to the surface that contains the point $(-3,4,-3)$ then we would have what we want. I noted that you can construct this plane, good luck.

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  • $\begingroup$ What is the condition to plane parallelism? I've been looking and can't find, always get different results on the search... $\endgroup$ – João Pedro Jan 15 '15 at 0:58
  • $\begingroup$ @JoãoPedro common normal vector is sufficient $\endgroup$ – Vladimir Vargas Jan 15 '15 at 1:18

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