Example - Volume of Parallelepiped

Question : 

Find the volume of parallelepiped spanned by (non-coplanar) vectors :

\(\vec{a} = 1\hat{i} + 1\hat{j} + 4\hat{k}\),

\(\vec{b} = 2\hat{i} + 3\hat{j} - 1\hat{k}\) and

\(\vec{c} = -1\hat{i} + 1\hat{j} + 3\hat{k}\).

Solution : 

“Spanned by vectors” here means that the vectors are along the three adjacent sides of the parallelepiped. They have magnitudes equal to the lengths of the corresponding edges.

For finding the volume of parallelepiped, first we want to select its base. Any face can be selected as its base. 

For the chosen base in the figure, \(\vec{b}\) and \(\vec{c}\) are the vectors along its adjacent sides(shown separately). So for the volume, we first find \(\vec{b} \times \vec{c}\)… and ‘dot’ the result  with \(\vec{a}\).

Volume of parallelepiped \(= |\vec{a} \cdot (\vec{b} \times \vec{c})|\)

If a different base was selected, say the one facing us in the figure with adjacent sides as \(\vec{a}\) and \(\vec{b}\), then the volume will be \(|\vec{c} \cdot (\vec{a} \times \vec{b})|\). 

Both are very different in their appearances, however both will give the same final answer! 

\(\vec{b} \times \vec{c} = \Biggl|\matrix{\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 1 & 3}\Biggl|\)

\( = (9 + 1)\hat{i} - (6 - 1)\hat{j} + (2 + 3)\hat{k}\)

\( = 10\hat{i} - 5\hat{j} + 5\hat{k}\)

\(|\vec{a} \cdot (\vec{b} \times \vec{c})| = |<1, 1, 4> \cdot <10, -5, 5>|\)

                              \( = |10 - 5 + 20|\)

                              \( = |25| = 25\)

Therefore, the volume of the parallelepiped is 25 cubic units.

It is important to enclose the dot product inside the modulus sign, because for certain combinations of vectors, the dot product may output a negative answer. The volume cannot be negative! 

(For instance, try finding the volume with the base that has \(\vec{a}\) and \(\vec{c}\) as its adjacent sides.)