How do you show that 2 lines $L_{1}$ and $L_{2}$ are perpendicular(orthogonals) but using this result $ ||u+v||=||u-v||$ My question has 2 parts, I m struggling with the second one.
Let's see the first:Prove that the vectors $u$ and $v$ are orthogonal, if and only 
if $\| u + v \| = \| u - v \|$ 
<=
If $u,v$ are orthogonal vectors, then:(by Phythagoras'theorem)
$$\| u + v \|^2 = \|u\|^2 + \|v\|^2$$
$$\| u - v \|^2 = \|u\|^2 + \|-v\|^2 = \|u\|^2 + \|v\|^2$$
now $\| u + v \|^2 = \| u - v \|^2$, but the norm is ever positive therefore:
$\| u + v \| = \| u - v \|$.
=> Now, if $\| u + v \| = \| u - v \|$ we have:
$\| u + v \|^2 = \|u\|^2 + 2u\cdot v + \|v\|^2$
$\| u - v \|^2 = \|u\|^2 - 2u\cdot v + \|v\|^2$ 
By the equality
$\|u\|^2 + 2u\cdot v + \|v\|^2 = \|u\|^2 - 2u\cdot v + \|v\|^2$ if and only if
$2u\cdot v = -2u\cdot v \Leftrightarrow 4u\cdot v = 0 \Leftrightarrow u\cdot v = 0$
this is $u$ and $v$ are orthogonal
I understand this part. OK
Let's see the second part of the question.
Considering the previous result of the proof(orthogonal vectors)
Show that $L_{1}=(1,2,3)+t(1,-1,1)$
And $L_{2}=(0,3,2)+t(5,2-3)$ are perpendiculars.
I think that the main point here is to consider that $uv=0$ means orthogonal but I have no idea how to use this result to show the Lines $L_{1}$ and $L_{2}$ are perpendiculars(orthogonals). Suggestions or hints will be welcome. Thanks

$$\|(1,-1,1)+(5,2,-3)\|=\sqrt{41}=\|(1,-1,1)-(5,2,-3)\|.$$
