Three statements regarding a normal and a tangential plane

Question

I am new to calculus, and was given the following question to answer. I have worked out an answer, but am not 100% sure about the details. Any feedback would be great! Many thanks in advance.

Given is the following surface, in parametric representation: $r(r,θ)=rcosθ\mathbf{i}+rsinθ\mathbf{j}+r\mathbf{k}; r\geq0,0\leqθ\leq2π$. We calculate the unit normal at point $Q(2,\sqrt12,4)$ to surface $\mathbf{n}$ and draw a plane tangential to the surface.

Which of the following statements are correct?

1. $\frac{\partial\mathbf{r}}{\partial{r}}\times\frac{\partial r}{\partial{θ}}$ is a normal to the surface.

2. the tangential surface is $z = \frac{x}{2}+\frac{y \sqrt{3}}{2}$

3. the unit normal is $\mathbf{n} = \frac{1}{\sqrt8}(\mathbf{i}+\mathbf{j} \sqrt3-2 \mathbf{k})$

My answers are the following:

1. Yes, as per definition

2. No: $Q(2,\sqrt{12},4)$, means that within the parametric equation, $r$ = 4 and $θ = \frac{1}{3}π$. If we take the cross product, as per the formula under number 1, we get $\frac{\partial\mathbf{r}}{\partial{r}}:(cosθ,sinθ,1)$; $\frac{\partial r}{\partial{θ}}$:$(-rsinθ, rcosθ,0)$, the cross product of which comes out as $(-rcosθ,-rsinθ,r)$. Putting in $r$ = 4 and $θ = \frac{1}{3}π$, this gives us $(-2,\sqrt{12},4)$ for the normal. Which means the tangential surface is defined by $-2(x-x_0)+\sqrt{12}(y-y_o)+4(z-z_o)=0$, i.e. $-2(x-2)+\sqrt{12}(y-\sqrt12)+4(z-4)=0$, and thus $-2x+4+\sqrt{12}y-12+4z-16=0$. This leads to $-2x+\sqrt{12}y+4z=24$, and thus $4z=2x-\sqrt{12}y+24$, in other words $z=\frac{x}{2}-\frac{\sqrt{3}y}{2}+6$.

3. the normal here is $\sqrt{4+12+16}=\sqrt{32}$, which means $\mathbf n = \frac {-2}{\sqrt{32}},\frac{2\sqrt3}{\sqrt{32}},\frac{4}{\sqrt{32}}$

Answer 1

The surface is just an upward opening cone with tip at $0$ $$ z^2 = x^2 + y^2$$

1. Yes.
2. For the point $\mathbf r = (x,y,z) = (2,\sqrt12,4)$, your parameters $r,\theta$ is correct. Your cross product is correct. But you are missing a minus sign on the second component. This changes the plane equation to the one given. So theirs is correct. If you noticed that its a cone with tip at $0$, you would notice that the tangent plane must contain the origin, so yours is wrong.
3. Their unit normal is correct, and consistent with the previous tangent plane.

Here's a 3D plot. $Black$ arrow is your normal(not unit length to exaggerate the incorrect direction), $\color{orange}{yellow}$ arrow is their unit normal. The other unit normal is in $\color{green}{green}.$ If you swap the $\sqrt{12}$ to $-\sqrt{12}$ in your normal, it points in the direction of the $\color{green}{green}$ arrow.