Find point of tangency on a Surface Find the point on the surface $x^3-2y^2+z^2=27$ where the tangent plane is perpendicular to the line given parametrically as $x=3t-5,y=2t+7,z=1-\sqrt{2}t$.
I have no idea where to start. I am aware that two vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if $\mathbf{u}\cdot\mathbf{v}=0$. Can this be used here?
Edit
$$\nabla f(x,y,z)=\begin{pmatrix}3x^2 & -4y & 2z\end{pmatrix},$$ so $$\left\{\begin{array}{l}3x^2 &= 3\\ -4y &= 2\\ 2z &= -\sqrt{2}\end{array}\right.$$ However, the solution $(\pm 1, -1/2, -\sqrt{2}/2)$ Is not a point on the surface of the original function.
 A: A good place to start here is to reformat your tangent line into a more vector like form.
In this case it would be.
$$
(x(t), y(t), z(t)) = (3, 2, -{\sqrt 2})t + (-5, 7, 1)
$$
We can see that this line is a tangent vector and an initial point. Now because the tangent plane is perpendicular to our given line, we can say the normal vector of our tangent plane will be the same as the line equation's vector. Because the line will be perpendicular to the plane and a normal vector is a line perpendicular to the plane, an equal vector for both will be a plane perpendicular to the line.
An important piece of information to note is that this is not the only vector possible for the normal. This is just a vector that goes the same direction. Any multiple of this vector is also a valid solution.
This means all you have to do is setup a system of equations where the normal vector is equal to the vector of the line equation.
I've setup the equations needed for the multiple based formula and solved using mathematica. I'm assuming you're only wanting real and not complex solutions. Here's the equations I used for reference
$${\nabla}f(x,y,z) = (3x^2,-4y,2z)$$
$$
\left\{
\begin{array}{1}
3x^2 &= 3k \\
-4y &= 2k \\
-z{\sqrt 2} &= -k{\sqrt 2}
\end{array}
\right\}
$$
Solving for $(x,y,z)$ gives us $(-{\sqrt k}, -\frac{k}{2}, -\frac{k}{\sqrt 2})$ and $({\sqrt k}, -\frac{k}{2}, -\frac{k}{\sqrt 2})$ as the two sets of solutions. We can now progress to solve these for all values of k which will result in a point on the surface. So solve both of these to make $f(x,y,z)=27$ true. This gives 3 solutions, only one is real which is $k=9$. Substitute that in for $k$ to get the point on the surface.
