But, as I said, you don't actually need the coordinates at all.

[hnnhcmmngs]

Homework Statement

Calculate |u+v+w|, knowing that u, v, and w are vectors in space such that
|u|=√2, |v|=√3, u is perpendicular to v, w=u×v.

Homework Equations

|w|=|u×v|=|u|*|v|*sinΘ

The Attempt at a Solution

[/B]
Θ=90°
|w|=(√2)*(√3)*sin(90°)=√(6)

Then I tried to use
u={√2,0,0}
v={0,√3,0}
w={0,0,√6}
and I got that |u+v+w|=√(2+3+6)=√11
but I'm trying to find a way to do this problem where I don't assume that those are the vectors.

 

[Ray Vickson]

Homework Statement

Calculate |u+v+w|, knowing that u, v, and w are vectors in space such that
|u|=√2, |v|=√3, u is perpendicular to v, w=u×v.

Homework Equations

|w|=|u×v|=|u|*|v|*sinΘ

The Attempt at a Solution

[/B]
Θ=90°
|w|=(√2)*(√3)*sin(90°)=√(6)

Then I tried to use
u={√2,0,0}
v={0,√3,0}
w={0,0,√6}
and I got that |u+v+w|=√(2+3+6)=√11
but I'm trying to find a way to do this problem where I don't assume that those are the vectors.

There is no loss of generality in what you did. Assuming that $u,v,w$ are vectors in some Cartesian coordinate system $(x,y,z)$, just change coordinates to a new system $(x', y', z')$ in which $u$ points along the $x'$-axis, $v$ points along the $y'$-axis and $w$ points along the $z'$-axis. That takes you right back to your original calculations.

Note added in edit: that argument assumes that the cross-product $u \times v$ behaves like a vector under a rotation; that is, if ${\cal R}$ is a rotation, then ${\cal R}(u \times v) = {\cal R} u \: \times \: {\cal R} v.$ (Here I mean that ${\cal R} u$ is the vector $u$ re-expressed in a new coordinate system obtained by applying ${\cal R}$ to the original unit coordinate vectors $e_x, e_y, e_z$; it does not mean that the "physical" vector $u$ is rotated.) That "rotation" result is true, but needs to be proved, a task I will leave to you.

 

[Orodruin]

However, it should be noted that you can do it without ever referring to a coordinate system at all, just square and expand the vector sum:
$$
(\vec u + \vec v + \vec w)^2 = \vec u^2 + \vec v^2 + \vec w^2 + 2 (\vec u \cdot \vec v + \vec u \cdot \vec w + \vec v \cdot \vec w).
$$
Now, all of the inner products in the parenthesis are zero because all of the vectors are orthogonal. Furthermore
$$
\vec w^2 = (\vec u \times \vec v)^2 = (\vec u \times \vec v)\cdot(\vec u \times \vec v)
= -\vec v \cdot [\vec u \times (\vec u \times \vec v )].
$$
Apply the BAC-CAB rule:
$$
\vec w^2 = -\vec v \cdot [\vec u (\vec u \cdot \vec v)- \vec v (\vec u^2)] = \vec v^2 \vec u^2 = 6.
$$
It follows that
$$
(\vec u + \vec v + \vec w)^2 = 2+3+6 = 11.
$$