The volume of a parallelepiped based on another.

Question

The volume of the parallelepiped determined by the three-dimensional vectors $\textbf{a}$, $\textbf{b}$, and $\textbf{c}$ is 11. Find the volume of the parallelepiped determined by the vectors $\textbf{a} + 2 \textbf{b}$, $\textbf{b} + 2 \textbf{c}$, and $\textbf{c} + 2 \textbf{a}$.

I know that you can take the determinant and alter it, and I know that the parallelepiped volume is determined by a determinant. However, I can't figure this out. Any solutions?

Answer 1

The volume of the first parallelepiped is $$\mathbf{a}.\mathbf{b}\times\mathbf{c}=11$$ Bearing in mind

• dot products of perpendicular vectors are zero
• cross-products of parallel vectors are zero
• otherwise cross products are perpendicular to both vectors being 'crossed'
• a scalar triple product is invariant to cyclic permutations of the vectors (e.g. $\mathbf{a}\,.\mathbf{b}\times\mathbf{c}=\mathbf{c}\,.\mathbf{a}\times\mathbf{b})$

then the volume of the new parallelepiped is $$(\mathbf{a}+2\mathbf{b}).(\mathbf{b}+2\mathbf{c})\times(\mathbf{c}+2\mathbf{a})$$ $$=(\mathbf{a}+2\mathbf{b}).[\mathbf{b}\times\mathbf{c}+\mathbf{b}\times2\mathbf{a}+2\mathbf{c}\times\mathbf{c}+2\mathbf{c}\times2\mathbf{a}]$$ $$=\mathbf{a}.\mathbf{b}\times\mathbf{c}+2\mathbf{b}.2\mathbf{c}\times2\mathbf{a}$$ $$=\mathbf{a}.\mathbf{b}\times\mathbf{c}+8\mathbf{a}.\mathbf{b}\times\mathbf{c}$$ $$=9\mathbf{a}.\mathbf{b}\times\mathbf{c}=9\times 11=99$$

Answer 2

It suffices to consider that this operation is rendered by matrix product :

$$\begin{pmatrix}a_x+2b_x&b_x+2c_x&c_x+2a_x\\a_y+2b_y&b_y+2c_y&c_y+2a_y\\a_z+2b_z&b_z+2c_z&c_z+2a_z\end{pmatrix}=\begin{pmatrix}a_x&b_x&c_x\\a_y&b_y&c_y\\a_z&b_z&c_z\end{pmatrix}\begin{pmatrix}1&0&2\\2&1&0\\0&2&1\end{pmatrix}$$

It suffices to take the determinants of the LHS and RHS: the determinant of the second matrix is 9, thus the volume of the second parallelepided (or more exactly rhomboid) is

9 times the volume of the initial parallelepiped, i.e., 99.