Help with physics demonstration

brad sue
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Hi ,
Please can I have some help with this demonstration:
Let a#0 ( #= different)
Show that if a x b = a x c and a.b=a.c, then b=c ( Hint: Cross both sides of the first equation with a)
When I cross by a ( as the hint suggests) , I found 0=0 ??
thank you
 
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What is "a x b" vs. "a.b"?
 
What are your definitions of The dot and cross product?
 
Integral said:
What are your definitions of The dot and cross product?
Hi , I don't know what you mean but I guess you want to know that
"a x b" means a cross product with b
and
"a.b" mean a dot b
Thank you
B
 
brad sue said:
Hi , I don't know what you mean but I guess you want to know that
"a x b" means a cross product with b
and
"a.b" mean a dot b
Thank you
B
Right. So, is "dot" vector multiplication? How is "cross product" different than vector multiplication? Some examples would be useful. What is (1,2) x (3,4)? What is (1,2).(3,4)?
Let's say a, b, and c are vectors. Let a=(1,2), b=(1,2) and c=(c1,c2).
ab = ac ---> a1b1 + a2b2 = a1c1 + a2c2 ---> 5 = c1 + 2 c2 ---> c1 = 5 - 2 c2, which means that although c = (1,2) would satisfy the equation, so would c = (5,0). Therefore what you want to demonstrate is not demonstrable on the basis of vector multiplication alone. That is why more info is needed on the exact definitions of the cross and the dot.

My guess is they are inner and outer products: a.b = a1b1 + a2b2, axb = (a1b1, a1b2, a2b1, a2b2). But can you verify this so we won't be making a mistake?
 
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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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