Question

If the three vectors $\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{c}+\overrightarrow{a}$ are coterminous edges of a parallelepiped then its volume is _.
A) $3\left[\overrightarrow{a}\overrightarrow{c}\overrightarrow{b}\right]$
B) $0$
C) $2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
D) $4\left[\overrightarrow{b}\overrightarrow{a}\overrightarrow{c}\right]$

Answer 1

Hint: Scalar triple product can directly be applied on the given sides of parallelepiped. Drawing sketches for parallelepiped with given coterminous edges might lead to an error as they are given in the form of sum of two vectors.

Complete step-by-step answer:
Here, we have a parallelepiped with $\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{c}+\overrightarrow{a}$ as their coterminous edges.
And coterminous edges mean the edges of a figure having or sharing the same boundaries.
Now, volume of a parallelepiped with their edges as let’s say $\overrightarrow{x},\overrightarrow{y}$ and $\overrightarrow{z}$ is defined as the area of the base times the height., $V=\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]$,
 $$\begin{array}{rl}& \Rightarrow V=\left[\overrightarrow{x}\overrightarrow{y}\overrightarrow{z}\right]\\ & \Rightarrow V=\overrightarrow{x}\cdot (\overrightarrow{y}\times \overrightarrow{z})=(\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}...\left(1\right)\end{array}$$
Which is also known as the scalar-triple product and it is further defined as,
$$\Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=\left|\begin{array}{ccc}{z}_1& z_2& z_3\\ x_1& x_2& x_3\\ y_1& y_2& y_3\end{array}\right|...\left(2\right)$$
where $\overrightarrow{x}=(x_1,x_2,x_3)$, $\overrightarrow{y}=(y_1,y_2,y_3)$ and $$\overrightarrow{z}=(z_1,z_2,z_3)$$ defined in vector form.
Thus, from given conditions, we have $\overrightarrow{x}=\overrightarrow{a}+\overrightarrow{b}$, $\overrightarrow{y}=\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{z}=\overrightarrow{c}+\overrightarrow{a}$.
Substituting these values in equation (2), we get
$$\Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=((\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{b}+\overrightarrow{c}))\cdot (\overrightarrow{c}+\overrightarrow{a})$$
Applying the product of vectors, we get
$$\begin{array}{rl}& \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})+\overrightarrow{b}\times (\overrightarrow{b}+\overrightarrow{c}))\cdot (\overrightarrow{c}+\overrightarrow{a})\\ & \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}+\overrightarrow{a})\end{array}$$
Now, on applying properties of cross product of two parallel vectors, i.e.,
$$\Rightarrow \overrightarrow{b}\times \overrightarrow{b}=\left|\overrightarrow{b}\right|\left|\overrightarrow{b}\right|\sin 0^{\circ }=0$$
Substituting this value in above equation, we get
$$\begin{array}{rl}& \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}+\overrightarrow{a})\\ & \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}+\overrightarrow{a})\end{array}$$
Now, applying scalar product of vectors, we get
$$\begin{array}{rl}& \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}+\overrightarrow{a})\\ & \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{c}+(\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}+0+\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{a}\\ & \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}+(\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{c}+0\cdot \overrightarrow{c}+(\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{c}+(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{a}+0\cdot \overrightarrow{a}+(\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{a}\end{array}$$
Using properties of scalar triple product of vectors, we have
$$\Rightarrow (\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$$
And, $$\Rightarrow (\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{c}=|\overrightarrow{a}\times \overrightarrow{c}|\left|\overrightarrow{c}\right|\cos {90}^{\circ }=0$$
Thus, from above equation, we have
$$\begin{array}{rl}& \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}+(\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{c}+0\cdot \overrightarrow{c}+(\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{c}+(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{a}+0\cdot \overrightarrow{a}+(\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{a}\\ & \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+0+0+0+0+0+0+\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}\right]\end{array}$$
Also, from properties of scalar triple product, we have $$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}\right]$$
$$\begin{array}{rl}& \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+0+0+0+0+0+0+\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{a}\right]\\ & \Rightarrow (\overrightarrow{x}\times \overrightarrow{y})\cdot \overrightarrow{z}=2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\end{array}$$
Hence, the volume of parallelepiped = $V=2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$, thus option [C] is correct.

Note: As per question, the edges of parallelepiped are given as the sum of two vectors. So, calculation of scalar triple product for volume of parallelepiped becomes quite complex. Keeping the angle between the vectors in mind might ease the calculations, in a product.