Question

Find two unit vectors that are normal to the given plan and the point. sqrt((z+x)/(y-1)) = z^2 and the point is P(3, 5, 1).
I took partial derivatives and got fx = 1, fy = -z^4 and fz = 1 - 4z^3y + 4z^3 but now I'm stuck.


This was the problem taht I had and when I solved it I got i - j -15k, the book has i - j -19k ... I do not know why? Their magnatude was also something different from what I got. For magnatude I used ||a|| and divided by that number to get the unit vector. Can anyone figure out why the book has a different value for k??

Answer 1

books can be misprinted .....

Answer 2

I know... Im just assuming the probability of my error to their error would be much greater

Answer 3

They also have a magnatude taht I can not get even if I DID use their -19k value...

Answer 4

i was typing up some nice coding and got booted ....

Answer 5

for the magnatude (if the value was i-j-19k) would be sqrt(1^2 + 1^2 + 19^2) = sqrt(363) They have 463 as the squareRoot

Answer 6

Ohh nooo... I apreaciate the help!

Answer 7

http://www.wolframalpha.com/input/?i=grad%28sqrt%28%28z%2Bx%29%2F%28y-1%29%29+-+z%5E2%29

this is a good dbl chk for the gradient parts

Answer 8

Which supports my answer...

Answer 9

its always good to have support :)

Answer 10

I'm skipping tyhis one before it drives me crazy, on to the next problem. Thanks for all your help eveyone

Answer 11

I get for the gradient at 3,5,1; using the wolfs rundown of it:

\[<\frac{1}{8},-\frac{1}{8},-\frac{15}{8}>\]

Answer 12

which is the same direction as 1,-1,-15 of course

Answer 13

That is correct, just a scalar factor is 1/8 is added

Answer 14

And which in vector terms is the same ve3ctor since it is paralell..