Skip to main content

[Community Question] Linear-algebra: Neep help justifying a vector relation given in a question

One of our user asked:

i'm trying to do a question for which I was given the following line equations:

$\underline r = \underline a + \lambda \underline u$

$\underline r' = \underline a' + \lambda' \underline u'$

They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.

$\lvert \underline r-\underline r'\rvert^2\lvert \underline u \times \underline u'\rvert^2=\lvert (\underline a - \underline a') \cdot (\underline u \times \underline u')\lvert^2+\lvert (\underline r - \underline r') \times (\underline u \times \underline u')\lvert^2$

I know that

$\lvert (\underline r - \underline r') \times (\underline u \times \underline u')\lvert = \lvert \underline r-\underline r'\rvert\lvert \underline u \times \underline u'\rvert\sin \theta $

but cant get any further.


Labels:

Comments

Post a Comment

Popular posts from this blog

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
Post a Comment
Read more

[Community Question] Calculus: prove $\int_0^\infty \frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}{8}$ with real methods

One of our user asked: I am attempting to prove that $$J=\int_0^\infty\frac{\log^2(x)}{x^2+1}\mathrm dx=\frac{\pi^3}8$$ With real methods because I do not know complex analysis. I have started with the substitution $x=\tan u$ : $$J=\int_0^{\pi/2}\log^2(\tan x)\mathrm dx$$ $$J=\int_0^{\pi/2}\log^2(\cos x)\mathrm dx-2\int_{0}^{\pi/2}\log(\cos x)\log(\sin x)\mathrm dx+\int_0^{\pi/2}\log^2(\sin x)\mathrm dx$$ But frankly, this is basically worse. Could I have some help? Thanks.
Post a Comment
Read more