# Finding Line tangent to surface and parallel to plane.

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.

• 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. – amd May 15 at 8:01