Is there something wrong with the problem statement? I'm doing the following question relating to parametrising surfaces and finding the tangent plane,
Question: 

I parametrised the surface in term of sphereical coordinates, such that $$x=\sin\theta \cos\phi, y = \frac{1}{\sqrt{2}}\sin\theta \sin\phi, z=\cos\theta$$
Using this information I found the normal vector, $\vec{n}$, to be:
$$\vec{n}=\left(\frac{\sqrt{2}}{2}\sin^2 \theta \cos\phi\right)\tilde{i}+\left(\sin^2 \theta \sin\phi\right)\tilde{j}+\left(\frac{\sqrt{2}}{2}\sin\theta \cos\theta\right)\tilde{k}$$
From here I used the given point $(\frac{1}{\sqrt{2}},\frac{1}{2},0)$ to find $\theta=\frac{\pi}{2}$ and $\phi=\frac{\pi}{4}$
Which then gave me,
$$\vec{n}=(\frac{1}{2},\frac{\sqrt{2}}{2}, 0) $$
Using the relation,$$(x -x_0, y-y_0,z-z_0)\dot{}\vec{n}$$
For part (a), I found the tangent plane to be:$$x + \sqrt{2}y - \sqrt{2}=0$$
But when I try to verify that $c'(\frac{\pi}{4})$ lies in the tangent plane I can never seem to verify it, always getting:$$-\sqrt{2} \neq0$$
Have I done the necessary steps to determine what the problem statement is asking? Could I have made a mistake along the way? Or could there possibly be something wrong with the problem statement?
I can provide more working if needed.
Any help would be greatly appreciated as I have no clue what else I can do. Thanks!
 A: Parametrization :
$x = sin(\theta)cos(\phi)$;
$y = (1/√2)sin(\theta)sin(\phi)$;
$z = cos(\theta)$.
Consider $\vec{r} = (x(\theta,\phi), y(\theta, \phi), z(\theta))$.
1)Tangent vector $\vec{t} $: $\frac{\partial \vec{r}}{\partial \theta}$ and 
2) Tangent vector $\vec{s}$: $\frac{\partial \vec{r}}{\partial \phi}$.
$\vec{t} =$
$(cos(\theta)cos(\phi),(1/√2)cos(\theta)sin(\phi),
-sin(\theta));$
$\vec{s} =$
$(-sin(\theta)sin(\phi),(1/√2)sin(\theta)cos(\phi),0).$
Note: vectors $t,s$ are independent (not colinear), I.e. they span a plane.
$P: (1/√2,1/2,0)$ corresponds to $\theta = π/2,$ $\phi = π/4$.
Tangent plane at $P: \vec{u} = (1/√2,1/2,0) + a \vec{t} + b \vec{s}$, $a,b \in \mathbb{R}$.
$\vec{t} = (0,0,-1)$; and $\vec{s} = (-1/√2,(1/√2)(1/√2),0)$.
Check: 
Normal vector to 
$f(x,y,z) = x^2 + 2y^2 + z^2 -1 =0$ is  $N = \nabla f(x,y,z)$.
$N = (2x,4y,2z)$ at $P(1/√2,1/2,0)$.
$N = (2/√2,2,0)$.
$N\cdot \vec{t} =0$; and $N\cdot\vec{s} =0$.
Part b) of the problem:
Surface:  $x^2 + 2y^2 + z^2 = 1$.
$c(t) = (cos(t), (1/√2)sin(t),0)$ lies on the surface,
$[cos(t)]^2 + 2 (1/2) [sin(t)]^2 = 1$, satisfies eq. of the surface.
$c'(π/4) = $
$((-1/2)√2,(1/√2)(1/2)√2,0) =$
$((-1/2)√2,1/2,0)$.
This is our $\vec{s}$ at $P$, hence $c'(π/4)$ lies in the tangent plane.
