I've been sitting here quite a while trying to figure out how to prove
$$ |s\vec{\mathbf{v}}| = |s||\vec{\mathbf{v}}| $$
I was not told in what dimension to prove this is, so I am assuming a 3D plane. I think I have a solution but I'm not sure if my method is write. Could someone confirm I am doing it right?
So I started by setting $$\vec{\mathbf{v}} = (x, y, z)$$ $$ |\vec{\mathbf{v}}| = \sqrt{x^2 + y^2 + z^2} $$ (x, y, z) being points in a 3D graph.
Then $$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2x^2 + |s|^2y^2 + |s|^2z^2} $$ $$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2(x^2 + y^2 + z^2)} $$ $$ |s\vec{\mathbf{v}}| = \sqrt{|s|^2} \sqrt{x^2 + y^2 + z^2} $$ $$ |s\vec{\mathbf{v}}| = |s| \sqrt{x^2 + y^2 + z^2} $$ Since $$ |\vec{\mathbf{v}}| = \sqrt{x^2 + y^2 + z^2} $$ Therefore $$ |s||\vec{\mathbf{v}}| = |s|\sqrt{x^2 + y^2 + z^2} $$ Hence $$ |s\vec{\mathbf{v}}| = |s||\vec{\mathbf{v}}| $$
My result does get me what I was looking for but it has happened in the past that for some reason a mistake that I made got me the right answer. Also if this is true for a 3D plane, does that mean that it holds through for all dimensions?
Thanks for any and all help
Your proof looks okay except that I would make the following change
$$ |s\vec{\mathbf{v}}| = \sqrt{(sx)^2 + (sy)^2 + (sz)^2} $$ $$ |s\vec{\mathbf{v}}| = \sqrt{s^2x^2 + s^2y^2 + s^2z^2} $$ $$ |s\vec{\mathbf{v}}| = \sqrt{s^2(x^2 + y^2 + z^2)} $$ $$ |s\vec{\mathbf{v}}| = \sqrt{s^2} \sqrt{x^2 + y^2 + z^2} $$ $$ |s\vec{\mathbf{v}}| = |s| \sqrt{x^2 + y^2 + z^2} $$
Remembering that $\sqrt{s^2} = |s|$. By placing the $|s|$ within the square root in your solution, you have already to some degree presupposed that $|s\vec{\mathbf{v}}| = |s| |\vec{\mathbf{v}}|$.
The proof extends without a problem to more dimensions as long as you use a more generic formulation of the Euclidean norm:
$$|\vec{\mathbf{v}}| := \sqrt{v_1^2 + \cdots + v_n^2}$$