The volume of a parallelepiped with vec(a), vec(b) and vec(c) as vectors coinciding with three edges sharing the same vertex equals to
V_1 = vec(a) * (vec(b) xx vec(c))
Let's examine what happens if we will use vectors vec(b)+vec(c), vec(c)+vec(a) and vec(a)+vec(b) instead.
The volume of a new parallelepiped will be
V_2 = (vec(b)+vec(c)) * ((vec(c)+vec(a)) xx (vec(a)+vec(b))) =
= (vec(b)+vec(c)) * (vec(c) xx vec(a)+vec(a) xx vec(a)+vec(c) xx vec(b)+vec(a) xx vec(b)) =
= vec(b) * (vec(c) xx vec(a))+vec(b) * (vec(a) xx vec(a)) + vec(b) * (vec(c) xx vec(b))+vec(b) * (vec(a) xx vec(b))+
+ vec(c) * (vec(c) xx vec(a))+vec(c) * (vec(a) xx vec(a)) + vec(c) * (vec(c) xx vec(b))+vec(c) * (vec(a) xx vec(b))
As we know, vector product of collinear vectors equals to zero-vector. Therefore, (vec(a) xx vec(a)) = vec(0).
Subsequent scalar product of any vector by zero-vector is zero.
Also, a scalar product of perpendicular vectors is equal to zero. Since vec(b) is perpendicular to (vec(c) xx vec(b)),
vec(b) * (vec(c) xx vec(b)) = 0
Analogously,
vec(b) * (vec(a) xx vec(b)) = 0
vec(c) * (vec(c) xx vec(a)) = 0
vec(c) * (vec(c) xx vec(b)) = 0
The result is:
V_2 = vec(b) * (vec(c) xx vec(a))+vec(c) * (vec(a) xx vec(b))
Each component of this sum equals to original volume of a parallelepiped since they are just cyclical permutation of vectors that preserves their order. Therefore, each is equal to V_1:
V_2 = V_1 + V_1 = 40+40=80
