# Question #eda10

##### 1 Answer
Jul 12, 2016

The volume of a new parallelepiped equals to $80$ cubic units (option 2).

#### Explanation:

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
${V}_{1} = \vec{a} \cdot \left(\vec{b} \times \vec{c}\right)$

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
${V}_{2} = \left(\vec{b} + \vec{c}\right) \cdot \left(\left(\vec{c} + \vec{a}\right) \times \left(\vec{a} + \vec{b}\right)\right) =$
$= \left(\vec{b} + \vec{c}\right) \cdot \left(\vec{c} \times \vec{a} + \vec{a} \times \vec{a} + \vec{c} \times \vec{b} + \vec{a} \times \vec{b}\right) =$
$= \vec{b} \cdot \left(\vec{c} \times \vec{a}\right) + \vec{b} \cdot \left(\vec{a} \times \vec{a}\right) + \vec{b} \cdot \left(\vec{c} \times \vec{b}\right) + \vec{b} \cdot \left(\vec{a} \times \vec{b}\right) +$
$+ \vec{c} \cdot \left(\vec{c} \times \vec{a}\right) + \vec{c} \cdot \left(\vec{a} \times \vec{a}\right) + \vec{c} \cdot \left(\vec{c} \times \vec{b}\right) + \vec{c} \cdot \left(\vec{a} \times \vec{b}\right)$

As we know, vector product of collinear vectors equals to zero-vector. Therefore, $\left(\vec{a} \times \vec{a}\right) = \vec{0}$.
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 $\left(\vec{c} \times \vec{b}\right)$,
$\vec{b} \cdot \left(\vec{c} \times \vec{b}\right) = 0$
Analogously,
$\vec{b} \cdot \left(\vec{a} \times \vec{b}\right) = 0$
$\vec{c} \cdot \left(\vec{c} \times \vec{a}\right) = 0$
$\vec{c} \cdot \left(\vec{c} \times \vec{b}\right) = 0$

The result is:
${V}_{2} = \vec{b} \cdot \left(\vec{c} \times \vec{a}\right) + \vec{c} \cdot \left(\vec{a} \times \vec{b}\right)$

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 ${V}_{1}$:
${V}_{2} = {V}_{1} + {V}_{1} = 40 + 40 = 80$