Question

The volume of the parallelepiped whose sides is given by $OA=2\hat{i}-2\hat{j}$, $OB=\hat{i}+\hat{j}-\hat{k}$ and $OC=3\hat{i}-\hat{k}$ is A.$\dfrac{4}{13}$
B.4 C.$\dfrac{2}{7}$
D.2 $$$$

Answer 1

We know that the parallelepiped whose sides are given as vectors say $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are given the scalar triple product of the vectors$\overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)$. We denote $OA=\overrightarrow{a},OB=\overrightarrow{b},OC=\overrightarrow{c}$ and then take the determinant of the components in orthogonal unit vectors to find the scalar product. $$$$

Complete step by step answer:
We know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta$ where $\theta$ is the angle between the vectors $\vec{a}$ and$\vec{b}$. We also know that $\hat{i}$,$\hat{j}$ and $\hat{k}$ are unit vectors(vectors with magnitude 1) along $x,y$ and $z$ axes respectively. So the magnitude of these vectors is$\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=\left| {\hat{k}} \right|=1$. The vectors just like their axes are perpendicular to each other which means any angle among$\hat{i}$,$\hat{j}$ and $\hat{k}$is ${{90}^{\circ }}.$ So $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\cdot 1\cdot \cos {{0}^{\circ }}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{j}=\hat{i}\cdot \hat{k}=\hat{k}\cdot \hat{i}=1\cdot 1\cdot \cos {{90}^{\circ }}=0$.
The dot product of two vectors written in components of unit orthogonal vectors$\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ can also be given by
$\overrightarrow{a}\cdot \overrightarrow{b}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}$
The cross product of two vectors written in components of unit orthogonal vectors$\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ can also be given by

{\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \right|$$ The scalar triple product of three 3 dimensional vectors is obtained by taking the dot product of a vector with cross product of two other vectors. If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors in three dimension written in components of unit orthogonal vectors$\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$, $\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ and $\overrightarrow{c}={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}$ then the scalar triple product is defined as $$\overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right|$$ The geometrical interpretation of scalar triple product is the volume of the parallelepiped whose sides are represented by vectors$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$. We are given in the question that sides of the parallelepiped are given by $OA=2\hat{i}-2\hat{j}$ $OB=\hat{i}+\hat{j}-\hat{k}$ and $OC=3\hat{i}-\hat{k}$. Let us denote $OA=\overrightarrow{a},OB=\overrightarrow{b},OC=\overrightarrow{c}$.  The volume of the parallelepiped can be obtained by taking the scalar triple product of vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$. So we have, $$\begin{aligned} & \overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \right| \\\ & \Rightarrow \overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left| \begin{matrix} 2 & -2 & 0 \\\ 1 & 1 & -1 \\\ 3 & 0 & -1 \\\ \end{matrix} \right| \\\ \end{aligned}$$ We expand by first row to have, $$\overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)=2\left( -1-0 \right)-\left( -2 \right)\left( -1-\left( -3 \right) \right)=-2+2\times 2=2$$ **So, the correct answer is “Option D”.** **Note:** We note that the scalar triple product follows circular shift in operands which means $\overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\overrightarrow{b}\cdot \left( \overrightarrow{c}\times \overrightarrow{a} \right)=\overrightarrow{c}\cdot \left( \overrightarrow{a}\times \overrightarrow{b} \right)$ and also allows swapping operation once without changing the value $\overrightarrow{a}\cdot \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\cdot c$. This is why we can assign $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ with $OA,OB,OC$ randomly and find the volume. A parallelepiped with perpendicular edges is called cuboid and a cuboid with equal sides is called cube.