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Question: Find the volume of the parallelepiped whose co terminal edges are \[4\hat{i}+3\hat{j}+\hat{k},5\hat{...

Find the volume of the parallelepiped whose co terminal edges are 4i^+3j^+k^,5i^+9j^+19k^ and 8i^+6j^+5k^4\hat{i}+3\hat{j}+\hat{k},5\hat{i}+9\hat{j}+19\hat{k}\text{ }and\text{ }8\hat{i}+6\hat{j}+5\hat{k}.

Explanation

Solution

The volume of the parallelepiped is given by, a.(b×c)\vec{a}.\left( \vec{b}\times \vec{c} \right) where a,b,c\vec{a},\vec{b},\vec{c} are the coterminal edges of the parallelepiped. We have already been given the co terminal edges, so we will put a=4i^+3j^+k^,b=5i^+9j^+19k^,c=8i^+6j^+5k^\vec{a}=4\hat{i}+3\hat{j}+\hat{k},\vec{b}=5\hat{i}+9\hat{j}+19\hat{k},\vec{c}=8\hat{i}+6\hat{j}+5\hat{k} and find the volume of the parallelepiped.

Complete step-by-step answer:
It is given in the question that we have to find the volume of the parallelepiped whose co terminal edges are 4i^+3j^+k^,5i^+9j^+19k^ and 8i^+6j^+5k^4\hat{i}+3\hat{j}+\hat{k},5\hat{i}+9\hat{j}+19\hat{k}\text{ }and\text{ }8\hat{i}+6\hat{j}+5\hat{k}. So, before we solve the question, let us understand what a parallelepiped is. It is a solid 3D shape in which all the faces are parallelograms and the opposite faces are of the same dimension. A parallelepiped can be shown as below.

Let us assume that a=4i^+3j^+k^,b=5i^+9j^+19k^,c=8i^+6j^+5k^\vec{a}=4\hat{i}+3\hat{j}+\hat{k},\vec{b}=5\hat{i}+9\hat{j}+19\hat{k},\vec{c}=8\hat{i}+6\hat{j}+5\hat{k}. Then, we know that the volume of a parallelepiped is given by, a.(b×c)\vec{a}.\left( \vec{b}\times \vec{c} \right) where a,b,c\vec{a},\vec{b},\vec{c} are the coterminal edges of the parallelepiped.
So, first, we will find the cross product of b and c\vec{b}\text{ }and\text{ }\vec{c}. In order to find the cross product, we will take determinant of the given vectors. For example, let p=ai^+bj^+ck^ and q=di^+ej^+fk^\vec{p}=a\hat{i}+b\hat{j}+c\hat{k}\text{ }and\text{ }\vec{q}=d\hat{i}+e\hat{j}+f\hat{k}, then their cross product will be, p×q=i^j^k^ abc def \vec{p}\times \vec{q}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ a & b & c \\\ d & e & f \\\ \end{matrix} \right|.
And, the determinant will be, i^(bfec)j^(afdc)+k^(aedb)\hat{i}\left( bf-ec \right)-\hat{j}\left( af-dc \right)+\hat{k}\left( ae-db \right).
So, here, we have to find the cross product of b and c\vec{b}\text{ }and\text{ }\vec{c}, where b=5i^+9j^+19k^,c=8i^+6j^+5k^\vec{b}=5\hat{i}+9\hat{j}+19\hat{k},\vec{c}=8\hat{i}+6\hat{j}+5\hat{k}. So, we will get,
b×c=i^j^k^ 5919 865 \vec{b}\times \vec{c}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 5 & 9 & 19 \\\ 8 & 6 & 5 \\\ \end{matrix} \right|
So, on taking the determinant, we get,

& \hat{i}\left[ \left( 9\times 5 \right)-\left( 6\times 19 \right) \right]-\hat{j}\left[ \left( 5\times 5 \right)-\left( 8\times 19 \right) \right]+\hat{k}\left[ \left( 5\times 6 \right)-\left( 8\times 9 \right) \right] \\\ & \hat{i}\left[ 45-114 \right]-\hat{j}\left[ 25-152 \right]+\hat{k}\left[ 30-72 \right] \\\ & -69\hat{i}+127\hat{j}-42\hat{k} \\\ \end{aligned}$$ So, we get the cross product of $\vec{b}\text{ }and\text{ }\vec{c}$ as $$-69\hat{i}+127\hat{j}-42\hat{k}$$. Now, we will find the dot product of $\vec{a}$ with $\vec{b}\times \vec{c}$. If we consider the vectors, $\vec{p}=a\hat{i}+b\hat{j}+c\hat{k}\text{ }and\text{ }\vec{q}=d\hat{i}+e\hat{j}+f\hat{k}$, then their dot product will be $\vec{p}.\vec{q}=\left( a\times d \right)+\left( b\times e \right)+\left( c\times f \right)$. So, here we have to find the dot product of $\vec{a}$ with $\vec{b}\times \vec{c}$, where $\vec{a}=4\hat{i}+3\hat{j}+\hat{k}$ and $$\vec{b}\times \vec{c}=-69\hat{i}+127\hat{j}-42\hat{k}$$. So, we will get, $$\begin{aligned} & \vec{a}.\left( \vec{b}\times \vec{c} \right)=\left( 4\hat{i}+3\hat{j}+\hat{k} \right)\left( -69\hat{i}+127\hat{j}-42\hat{k} \right) \\\ & =\left( 4\times -69 \right)+\left( 3\times 127 \right)+\left( 1\times -42 \right) \\\ & =-276+381-42 \\\ & =381-18 \\\ & =63\text{ }cubic\text{ }units \\\ \end{aligned}$$ Therefore, the volume of the parallelepiped is 63 cubic units. **Note:** The most common mistake that students make while solving this question is that, while taking the determinant, most of the students put + sign instead of – sign before $\hat{j}$ and as a result, they will get the cross product of $\vec{b}\text{ }and\text{ }\vec{c}$ as $$\vec{b}\times \vec{c}=-69\hat{i}-127\hat{j}-42\hat{k}$$, which is wrong, hence the final answer also will become wrong. So, the students are supposed to do all the calculations step by step in order to avoid any mistakes.