<s>[INST] Multi variable calculus, equation tangent to plane question So i think i may have it right but now sure...please check and help me to see if i got it right thanks! 

Question: 
  $f(x,y) = 1 + x^2 + y^2$, find vector $v$ tangent to plane of graph at $f(1,1,3)$. 

Answer: 
$f(x,y) = 1 + x^2 + y^2$ and $g(x,y,z) = 1 + x^2 + y^2 – z$ 
$g(x,y,z) = f(x,y) – z$ 
$0 = f(x,y) – z$ 
$z = f(x,y) $
$z = 1 + x^2 + y^2 $
$z = 1 + 1 + 1$ (sub in 2 points i know already $f(1,1,3)$ for $x$ and $y$). 
$z = 3 $
$g(x,y,z) = 1 + x^2 + y^2 – 3z$ 
$g(x,y,z) = 1 + x^2 + y^2 –3z$ 
$\text{grad} g(x,y,z) = 1 + x^2 + y^2 – 3z$ 
$\text{grad} g(x,y,z) = (2x, 2y, –3) $
$(x,y,z)\cdot \text{grad} g(1,1,3) = (1,1,3))\cdot \text{grad} g(1,1,3) $
$(x,y,z)\cdot (2,2,-3) = (1,1,3))\cdot (2,2,-3) $
$2x+ 2y -3z = 2 + 2 – 9 $
$2x+ 2y -3z = -5 $
$-3z = -5 - 2x -2y $
$3z = 2x + 2y +5 $
Is this right? Thanks!
 [/INST] No, your gradient is not correct (and, in your statement, $f(1,1,3)$ is misleading since $f$ is a function of TWO variables).  
Let $f(x,y) = 1 + x^2 + y^2$ then 
the gradient of 
$$g(x,y,z):=f(x,y)-z=1 + x^2 + y^2 – z$$ 
at $(1,1,f(1,1))=(1,1,3)$, is
$$\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g}{\partial z}\right)_{(1,1,3)}=
(2x,2y,-1)_{(1,1,3)}=(2,2,-1).$$
Now choose a vector ${\bf v}$ such that the scalar product ${\bf v}\cdot (2,2,-1)$ is zero.
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