Find the points on the graph of $z = xy^{3}+8y^{-1}$ where the tangent plane is parallel to $ 2x+7y+2z=0$
I have no idea how to approach this question and have not yet taken any form of linear algebra. Is it possible to solve this question only using calculus?
We have $$f(x,y,z)=xy^3+8y^{-1}-z$$
First we want to find the gradient of this function:
\begin{align}\vec\nabla f&=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)\\ &=\left(y^3, 3xy^2-8y^{-2},-1\right) \end{align}
Now we want to find values where this is parallel to the vector $(2,7,2)$. Two vectors are parallel if their cross product is zero, so we want to find:
$$\left(y^3, 3xy^2-8y^{-2},-1\right)\times (2,7,2)=0 $$
We can find this through the following formula:
$$\underline u\times \underline v = \underline{i}(u_yv_z-u_zv_y)+ \underline j(u_xv_z-u_zv_x)+\underline k(u_xv_y-u_yv_x)$$
where $\underline i, \underline j, \underline k$ are the unit vectors
So, we have \begin{align}\underline{i}\left(2\left(3xy^2-8y^{-2}\right)+7\right)+ \underline j\left(2y^3+2\right)+\underline k\left(7y^3-2\left(3xy^2-8y^{-2}\right)\right)&=0\underline i+0\underline j+0\underline k\end{align}
We can equate coefficients to get the following simultaneous equations
\begin{align}2\left(3xy^2-8y^{-2}\right)+7&=0\tag{$\underline i$}\\ 2y^3+2&=0\tag{$\underline j$}\\ 7y^3-2\left(3xy^2-8y^{-2}\right)&=0\tag{$\underline k$}\end{align}
Can you continue from here?