Help on another algebra problem

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The Cosine of the angle between a and b is 4/21. Find p if a = 6i + 3j -2k and b = -2i + pj - 4k.

i started off with the general equation a . b = |a| |b| cosX where cosX is 4/21
since i, j, k all mean 1. i would get two vectors
a = (6,3,-2) and b = (-2, p, -4)

then i would take the magnitude of both a and b.

then i would take the dot product of a and b.

then i would sub them into the equation.

and then i ended up solving using the quadratic equation, but when i solved it, i ended up getting two answers and if i sub it back into the dot products and stuff it wouldn't equal.. LS = RS...

i don't kno if this is how i shouold have done it ?
thnx
 
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Yes, that is how you should do it.

I don't know what you mean by "when i solved it, i ended up getting two answers and if i sub it back into the dot products and stuff it wouldn't equal.. LS = RS...". Of course, when you square an equation you may introduce a new solution that doesn't satisfy the original equation but one of your two solutions should.
 
alright thanks
 
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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