# Normal unit vectors to a given surface at point P

Find two unit vectors that are normal to $\sqrt{\frac{x+z}{y-1}}=z^2$ at $P(3,5,1)$.

Attempt: I have $f(x,y,z)=\sqrt{\frac{x+z}{y-1}}-z^2$. First I found the gradient : $$\nabla f=<f_x,f_y,f_z>=<\frac{1}{2 (y-1) \sqrt{\frac{x+z}{y-1}}};-\frac{x+z}{2 (y-1)^2 \sqrt{\frac{x+z}{y-1}}};\frac{1}{2 (y-1) \sqrt{\frac{x+z}{y-1}}}-2 z>$$ $$\nabla f(3,5,1)=<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>$$

So the two unit vectors would be the positive and negative normalized gradients: $$\vec{u_1}=\frac{<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>}{||<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>||}=\frac{8}{\sqrt{227}}<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>=\frac{1}{\sqrt{227}}<1,-1,-15>$$

$$\vec{u_2}=-\frac{1}{\sqrt{227}}<1,-1,-15>$$

However, in the answer key the answer is $(\pm\frac{1}{\sqrt{365}}<1,-1,-19>$). I checked the derivatives multiple times and the gradient as well and I have no idea why my answer does not match the answer key. I do not think it is a typo, so help please.

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The answer key answers aren't even unit vectors, unless that was suppposed to be $\sqrt{363}$. Your solution appears to be correct. – Jonas Meyer May 24 '12 at 5:20
Wow. Finally, it is a typo and not me making a mistake. On Thursday I will confirm again with my professor who has the solution manual. – Koba May 24 '12 at 6:27
Yes it was a typo in the textbook. – Koba May 25 '12 at 4:59

## 1 Answer

http://ltcconline.net/greenl/courses/202/vectorFunctions/tannorm.htm

here shows the reason. T'(t) N(t) = _____ ||T'(t)||

your answer is like N(t)=f'(t)/(||f'(t)||)

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This question was already answered by the looks of the comments above. And this was more than a year ago. How is your answer helpful at all? I can't see it. – Daniel R Nov 1 '13 at 13:52