Respuesta :

Ferraz
Good morning.

We see that [tex]\mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}[/tex]

The magnitude(norm, to be precise) can be calculated the following way:

[tex]\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}[/tex]


Now the calculus is trivial:

[tex]\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}[/tex]