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.
You should be familiar with some basic properties of the cross product. Especially the identity
$$ || \mathbf{a} \times \mathbf{b} ||^2 = ||\mathbf{a}||^2 \, || \mathbf{b}||^2 - |\mathbf{a} \cdot \mathbf{b}|^2 $$
As can be seen, for instance from:
$$ \begin{aligned} || \mathbf{a} \times \mathbf{b} ||^2 & = (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) \\ &= \mathbf{b} \cdot ((\mathbf{a} \times \mathbf{b}) \times \mathbf{a}) \\ &= -\mathbf{b} \cdot (\mathbf{a} \times (\mathbf{a} \times \mathbf{b})) \\ &= -\mathbf{b} \cdot (\mathbf{a}\,(\mathbf{a}\cdot\mathbf{b}) - \mathbf{b}\,(\mathbf{a}\cdot\mathbf{a})) \\ &= ||\mathbf{a}||^2\,||\mathbf{b}||^2 - |\mathbf{a}\cdot\mathbf{b}|^2 \end{aligned} $$
Now, with $\mathbf{a} = \mathbf{r} - \mathbf{r'}$ and $\mathbf{b} = \mathbf{u} \times \mathbf{u'}$, use the bilinearity of the inner product and
$$ \mathbf{r} = \mathbf{a} + \lambda \,\mathbf{u}; \,\, \mathbf{r'} = \mathbf{a'} + \lambda' \,\mathbf{u'} \\ \mathbf{u} \cdot (\mathbf{u} \times \mathbf{u'}) = \mathbf{u'} \cdot (\mathbf{u} \times \mathbf{u'}) = 0 $$