let u, v,w be vectors such that u+v+w=0 u=3, v=4, w=5 find 47 costhita and u.v+v.w+w.a

2- if a= 2i-3j+4k ,b= 3i+2j-4k, c= 4i-3j+5k which is meaningful and how can we evaluate this (a.b)crossc, a cross(b crossc), a.(bcrossc) how we can solve this type of prob

u, v, w be vectors such that u + v + w = 0  and  |u| = 3, |v| = 4, |w| = 5
So u.( u + v + w) = 0 [ as u + v + w = 0]
So u.u + u.v + u.w = 0
Or |u|² + u.v + u.w = 0  [as u.u = |u|² ]
Or u.v + u.w = - |u|² = -(3)² = -9  (1)

And v.(u + v + w )  = 0 [as u + v + w = 0 ]
v.u + v.v + v.w = 0
Or v.u +|v|² + v.w = 0
Or v.u + v.w = -|v|² = -16  (2)

And w.(u + v + w ) = 0
So w.u + w.v + w.w = 0
Or w.u + w.v = -|w|² = -25 (3)

So adding (1) (2) and (3) , we get
u.v + u.w + w.u + w.v + v.u + v.w = -16 -25 -9 = -50
So 2( u.v + v.w + w.v ) = -50  [ as using A.B = B.A]
So ( u.v + v.w + w.v ) = -50/2 = -25


47 cos(theta) is not clear, what do you want to ask , as theta can be a angle between two vectors, not three.


2) a = 2i - 3j + 4k , b = 3i + 2j - 4k, c = 4i -3j +5k
So a.(b x c) is scalar triple product and can be calculated by evaluating the determinant of a ,b and c
So a.(b x c) = $\left|\begin{array}{ccc}2& -3& 4\\ 3& 2& -4\\ 4& -3& 5\end{array}\right|=2(2\times 5 - (-3)\times (-4\left)\right) - (-3) ( 3\times 5 - 4\times (-4\left)\right) + 4 (3\times -3 -4\times 2)$
= 2( 10 -12) + 3(15 +16) + 4(-9 -8) = -4 +93 -68 = 21

(a.b) x c  is not meaningful as the bracket gives you a scalar quantity, and cross product of two vectors is only possible.

a x (b x c)  here take the cross product of b and c first, then take the cross product of a and solution of b x c.
Try it.