Calculating the Cross Product Vector: Is It Possible to Find VxW?

Christie
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Okay so, I am wondering if it is possible to find the actual cross product (not the magnitude of the cross product) from this information
1. magnitude of both vectors
2.angle between vectors
3.plane the vectors lie in
Is there any way to calculate that cross product vector?
Thank you very much.
I know that
Mag (VxW)= Mag(v)Mag(w)sinthea
but am wondering if there is anyway to actually isolate VxW from this information.
 
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It is the normal vector of the plane that has the appropriate magnitude.

Depending on what you mean by saying that the angle is known, there may or may not be a possible sign ambiguity.
 
What would b that ambiguity?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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