Find the point on the graph of$ z=x^2−y^2$ at which vector $\vec{n}=⟨−8,12,−2⟩$ is normal to the tangent plane $P$.

Question

The normal vector to a tangent plane is the $∇F(x,y,z)$, hence $\langle 2x,-2y,-1 \rangle $ from $F(x,y,z)=x^2-y^2-z$. So, if the normal vector is equal to $\langle -8,12,2 \rangle$, then $-8=2x, 12=-2y$ and $-1=-2$ (?) It is wrong, but I don't see how. Any help is appreciated. Thank you

Answer 1

You need to have: $\vec{n} \parallel \nabla{F}$ at $(x,y,z) \in C, C: z= x^2-y^2$. This means: $\langle -8,12,-2 \rangle = k\langle 2x,-2y,-1\rangle \implies 2kx=-8, -2ky=12, -k = -2\implies k = 2, x = -2, y = -3, z = (-2)^2 - (-3)^2 = 4 - 9 = -5.$ Thus the point you are solving for is: $(-2,-3,-5)$.