Why is this the equation of the tangent plane?

Question

I want to find the equation of the tangent plane of the surface patch $\sigma (r, \theta)=(r\cosh \theta , r\sinh \theta , r^2)$ at the point $(1,0,1)$.

I have done the following:

The point $(1,0,1)$ corresponds to $\sigma (1,0)$.

We have that $$\sigma_r=(\cosh \theta , \sinh \theta , 2r) \rightarrow \sigma_r(1,0)=(1,0,2) \\ \sigma_{\theta}=(r \sinh \theta , r \cosh \theta , 0) \rightarrow \sigma_{\theta}(1,0)=(0,1,0)$$

$$\sigma_r (1,0) \times \sigma_{\theta} (1,0)=(-2,0,1)$$

The equation of the tangent plane is given by the formula $$(-2, 0, 1) \cdot (x-1, y-0, z-1)=0 \\ \Rightarrow -2x+z+1=0$$

$$$$

In the solution of the book, the answer is $-2x - 2y + z =0$.

Where is the mistake at my calculations?

Answer 1

$(1,0,1)$ does not lie on $-2x-2y+z=0.$

Answer 2

Your answer is correct. I guess it's a typo in the book