Linearly Independent Vectors and Their Span: The Truth Behind <u,v,w> and <u,v>

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Homework Statement


Suppose u,v,w are linearly independent, is it true that <u,v,w> does not equal <u,v>


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The Attempt at a Solution


I started by defining what it meant to be linearly independent but am unsure where to go from there. I think the statement is true since the span <u,v> won't include any multiple of w but i can't give a solid proof
 
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Assume by contradiction that <u,v,w>=<u,v>. Then w\in &lt;u,v&gt;. Thus...
 
oh i think i get it, by assuming that we can say by definition

w=λ1u + λ2v
then there is now a non trivial solution to
λ1u + λ2v + λ3w=0 which contradicts the statement that u,v,w are linearly independent, is that right?
 
That's right!
 
Sweet, cheers
 

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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