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I had an exam today and I wanted to know if I solved this question correct.

It asked me to, given a surface curve, like $z=x^2+y^2+10$, find the line tangent to it at point $(2,2,1)$ and parallel to the plane $z+x+y+10=10$.

These aren't accurate or correct numbers but just wanted to know if my approach was correct.

I first took the gradient of the surface, in this case, it would be $\langle 2x,2y,-z\rangle$. This would be the normal vector of the surface. I then plugged in the point given and got $\langle 4,4,-1\rangle $. I then dotted that vector with V such that the dot product is $0$.

My logic is that this means it would create the directional vector $V$ of the line tangent to the surface. I decided to make $V=\langle 1/4,1/4,2\rangle$. The final thing to do is cross product $V$ with the normal vector of the plane because that should produce a vector $Y$ that is is the directional derivative of a line that is both parallel to the plane and tangent to the surface curve.

With this, I deduced that the line asked of me is $r(x,y,z)=\langle 2,2,1\rangle+\langle x,y,z\rangle Y$, where $Y$ is the cross product of $V$ and the normal vector of the plane. If I am wrong, can someone tell me where I messed up on? Thanks in advance.

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    $\begingroup$ Your idea of using the cross product of normals to the two planes is sound. Unfortunately, $(2,2,1)$ doesn’t lie on the surface, so the computations that you make using it don’t actually produce a tangent to it. $\endgroup$ – amd May 15 at 8:01

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