Question

If a + b, b + c, c + a are coterminus edges of parallelopiped, then its volume is

• A. 0
• B. 4 [b a c]
• C. 3 [a c b]
• D. 2 [a b c]

Answer 1

Answer: (D)

2[a b c]

Answer 2

Let the coterminal edges be

$\vec{u} = \vec{a}+\vec{b}, \quad \vec{v} = \vec{b}+\vec{c}, \quad \vec{w} = \vec{c}+\vec{a}$.

The volume $V$ of the parallelepiped is given by:

$V = \left|[\vec{u}, \vec{v}, \vec{w}]\right| = \left|[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]\right|$.

Using the linearity of the scalar triple product (i.e. the determinant function), we have:

$[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = [\vec{a}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] + [\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]$.

Expanding the first term:

$[\vec{a}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = [\vec{a}, \vec{b}, \vec{c}+\vec{a}] + [\vec{a}, \vec{c}, \vec{c}+\vec{a}]$.

Now, applying linearity again:

$[\vec{a}, \vec{b}, \vec{c}+\vec{a}] = [\vec{a},\vec{b},\vec{c}] + [\vec{a},\vec{b},\vec{a}]$ and $[\vec{a},\vec{b},\vec{a}] = 0$.

Similarly,

$[\vec{a}, \vec{c}, \vec{c}+\vec{a}] = [\vec{a},\vec{c},\vec{c}] + [\vec{a},\vec{c},\vec{a}] = 0$.

Thus the first term equals $[\vec{a}, \vec{b}, \vec{c}]$.

Now, for the second term:

$[\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = [\vec{b},\vec{b}, \vec{c}+\vec{a}] + [\vec{b},\vec{c}, \vec{c}+\vec{a}]$.

The first part $[\vec{b},\vec{b}, \vec{c}+\vec{a}]=0$ and

$[\vec{b},\vec{c}, \vec{c}+\vec{a}] = [\vec{b},\vec{c},\vec{c}] + [\vec{b},\vec{c},\vec{a}] = [\vec{b}, \vec{c}, \vec{a}]$.

Since the scalar triple product is cyclic, we have:

$[\vec{b},\vec{c},\vec{a}] = [\vec{a},\vec{b},\vec{c}]$.

Thus, the second term equals $[\vec{a},\vec{b},\vec{c}]$.

Adding both terms:

$[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = [\vec{a},\vec{b},\vec{c}] + [\vec{a}, \vec{b}, \vec{c}] = 2[\vec{a},\vec{b},\vec{c}]$.

Hence, the volume is:

$V = \left| 2[\vec{a},\vec{b},\vec{c}] \right| = 2\left| [\vec{a},\vec{b},\vec{c}] \right|$.