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Mathematics 65 Online
OpenStudy (anonymous):

Find an equation of the plane trangent to the hyperboloid \[2x^2-3y^2+z^2=7\] at the point (3,2,1).

OpenStudy (anonymous):

try visiting this url it might be able to explain this concept better than i can: http://math.etsu.edu/multicalc/archives/Chap3/Chap3-6/index.htm

OpenStudy (anonymous):

did that help?

OpenStudy (anonymous):

First you need to find the unit vector. You know how to find it?

OpenStudy (anonymous):

yeah, I think I figured it out. Grad F(x,y,x)

OpenStudy (anonymous):

so would it be. \[6(x-3) -6(y - 2) +(z-1) =0\] ?

OpenStudy (anonymous):

You are right that's the equation for a plane passing through that point

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Sames right 6x-6y+z=9

OpenStudy (anonymous):

actually that doesnt look right

OpenStudy (anonymous):

The 9 should be 7

OpenStudy (anonymous):

i got the same thing as lockdown but maybe i am wrong, i do know that you should most likely condense your answer

OpenStudy (anonymous):

yeah, I condensed it. I got 12x - 12y + 2z - 14 = 0 originally, but divided by 2

OpenStudy (anonymous):

Actually yeah

OpenStudy (anonymous):

meaning combine constants on the other side of the equation

OpenStudy (anonymous):

Here is a neat trick, Lockdown, rather than writing out the equation formula. Once you get your vector <6, -6, 1> Write it as an eq 6x-6y+1= The other side can be determined in your head by the points (6*3=18) -12 + 1=7

OpenStudy (anonymous):

you know that trick is actually helpful to me too, thanks chaguanas

OpenStudy (anonymous):

Yeah, the equation is taught in schools, mathematicians don't do it the eq way

OpenStudy (anonymous):

yeah, thats a good way of thinking about it, since they always want simplified anyways.

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