Question 1183181
Use the fact that {{{abs(u)^2 = u*u }}} and {{{u*v = abs(u)abs(v)cos(theta)}}}, where {{{abs(u)}}} is the magnitude of the vector {{{u}}}, and {{{u*v}}} is the dot product of the vectors {{{u}}} and {{{v}}}.

Then {{{abs(u+v+w)^2 = (u+v+w)*(u+v+w)

= u*u + v*v + w*w + 2(u*v + v*w + u*w)}}}, by direct application of the distributive, commutative, and associative properties.

Now 

{{{u*u + v*v + w*w + 2(u*v + v*w + u*w) = abs(u)^2 + abs(v)^2 + abs(w)^2 + 2(abs(u)*abs(v)cos(theta[1]) + abs(v)*abs(w)cos(theta[2]) + abs(u)*abs(w)cos(theta[3]))}}}


= {{{abs(u)^2 + abs(v)^2 + abs(w)^2 + 2(abs(u)*abs(v)cos60^0 + abs(v)*abs(w)cos60^0 + abs(u)*abs(w)cos60^0)}}}  since {{{theta[1] = theta[2] = theta[3] = 60^0}}}


={{{abs(u)^2 + abs(v)^2 + abs(w)^2 + abs(u)*abs(v) + abs(v)*abs(w) + abs(u)*abs(w)  }}}, since {{{cos60^0 = 1/2}}}


= 16 + 4 + 36 + 8 + 12 + 24 = 100.


==> {{{abs(u+v+w)^2 = 100}}}  ===> {{{highlight(abs(u+v+w) = 10)}}}