I'm at quite a loss as to how to answer this question, and I'd really appreciate some help.
The question is as follows:
If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three vectors such that |$\vec{a}$| = 3, $\vec{b}$ = 4 and $\vec{c}$ = 5 and each one of them is perpendicular to the sum of the other two, find |$\vec{a}$ + $\vec{b}$ + $\vec{c}$|.
I'm familiar with vectors as explained in Thomas' Calculus, Twelfth Edition.
On
Hint:
the three vectors are the sides of a rectangula cuboid, and their sum is the space diagonal.
On
Probably the piece you are missing is the fact that, because each vector is perpendicular to the other two, the following three equations must be true:
$$ (\vec{a}+\vec{b})\cdot\vec{c} = 0, \\ (\vec{a}+\vec{c})\cdot\vec{b} = 0, \\ (\vec{b}+\vec{c})\cdot\vec{a} = 0. \\ $$
This is because, whenever two vectors are perpendicular in Euclidean geometry, their dot products must be 0. Combine this with the relationship between dot product and vector length:
$$ \vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta,\\ \hbox{where }\theta\hbox{ is the angle between the vectors, and therefore} \\ \vec{v}\cdot\vec{v} = |\vec{v}|^2, $$
and should be able to deduce the solution. Of course, if you're looking for a geometric intuition rather than a means of proving the solution, Emilio Novati's answer gives a good hint.
Fleshing out nben's answer above:
because $$ \vec{v}\cdot\vec{v} = |\vec{v}|^2 $$ we have: $$ |\vec{a}+\vec{b}+\vec{c}|^2 = (\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = $$ distributing $\cdot$ over $+$: $$ [\vec{a}\cdot\vec{a}+(\vec{b}+\vec{c})\cdot\vec{a}]+[\vec{b}\cdot\vec{b}+(\vec{a}+\vec{c})\cdot\vec{b}]+[\vec{c}\cdot\vec{c}+(\vec{a}+\vec{b})\cdot\vec{c}] = $$ by the perpendicularity property given in the problem (see nben's equations above) $$ [\vec{a}\cdot\vec{a}+0]+[\vec{b}\cdot\vec{b}+0]+[\vec{c}\cdot\vec{c}+0] =\\ |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 =\\ 9+16+25=\\ 50 $$ thus $$ |\vec{a}+\vec{b}+\vec{c}| = \sqrt{50} $$