Find a normal vector to a unit sphere using cartesian coordinates

Frozen Light
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Homework Statement


Consider a unit sphere centered at the origin. In terms of the Cartesian unit vectors i, j and k, find the unit normal vector on the surface

Homework Equations


A dot B = AB cos(theta)
A cross B = AB (normal vector) sin(theta)
Unit sphere radius = 1

The Attempt at a Solution


Isn't any direction a normal vector?
i x j = + k
j x i = - k
etc.
 
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Frozen Light said:

Homework Statement


Consider a unit sphere centered at the origin. In terms of the Cartesian unit vectors i, j and k, find the unit normal vector on the surface

Homework Equations


A dot B = AB cos(theta)
A cross B = AB (normal vector) sin(theta)
Unit sphere radius = 1

The Attempt at a Solution


Isn't any direction a normal vector?
i x j = + k
j x i = - k
etc.
Any nonzero vector would be a normal at some point on the surface of the sphere. My guess is that you should take an arbitrary point P(x, y, z) on the surface, and find the normal to it. If that's what is wanted in the problem, it could have been written more clearly.
 
Thank you, that would make a bit more sense.
 
The unit sphere is of the form x^2+ y^2+ z^2= 1. You can think of that as a 'Level Surface" of the function F(x, y, z)= x^2+ y^2+ z^2 and use the fact that the gradient of such a function, \nabla F, is always normal to level surfaces.
 
Or think about what direction a position vector to a point on the sphere has.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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