Vectors - How do they get the '-k'? Watch

claret_n_blue
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#1
Report Thread starter 7 years ago
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If Q = (X,Y,Z) and P = (3,0,0). z = x² - y²

Using the partial derivatives of z = x² - y², we know the vector n = 2Xi - 2Yj - k is normal to the surface at Q.


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I understand the bit for X and Y, but I don't get how they get a '-k'. Should it not simply be 0?
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nuodai
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Report 7 years ago
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No; you can rewrite it as x^2-y^2-z=0; that is, letting f(x,y,z)=x^2-y^2-z, and so on.
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