
If three vectors \[\overrightarrow{a}=12\widehat{i}+4\widehat{j}+3\widehat{k}\], \[\overrightarrow{b}=8\widehat{i}-12\widehat{j}-9\widehat{k}\] and \[\overrightarrow{c}=33\widehat{i}-4\widehat{j}-24\widehat{k}\] represents a cube, then its volume will be
A. \[616\]
B. \[308\]
C. \[154\]
D. None of these
Answer
162.3k+ views
Hint: In the above question we are to find the volume of the cube which is represented by three vectors. By using appropriate formulae (i.e. scalar triple product), the required solution is calculated.
Formula used: The dot product of two vectors is
$\overrightarrow{a}\cdot \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\overrightarrow{a},\overrightarrow{b})$
The cross-product of two vectors is
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
Scalar triple product of three vectors:
We have the vectors \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] as
\[\begin{align}
& \overrightarrow{a}={{a}_{1}}\overrightarrow{i}+{{a}_{2}}\overrightarrow{j}+{{a}_{3}}\overrightarrow{k} \\
& \overrightarrow{b}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}+{{b}_{3}}\overrightarrow{k} \\
& \overrightarrow{c}={{c}_{1}}\overrightarrow{i}+{{c}_{2}}\overrightarrow{j}+{{c}_{3}}\overrightarrow{k} \\
\end{align}\]
Then, the triple product is calculated by,
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]
Thus, the volume of a 3D structure is calculated by $V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]$
Complete step by step solution: Here, we are given the three vectors that represent the three axes that form a cube when produced in their respective planes in both directions.
The given vectors are
\[\overrightarrow{a}=12\widehat{i}+4\widehat{j}+3\widehat{k}\]
\[\overrightarrow{b}=8\widehat{i}-12\widehat{j}-9\widehat{k}\]
\[\overrightarrow{c}=33\widehat{i}-4\widehat{j}-24\widehat{k}\]
We know that the concept of calculating the volume of any 3D structure which has sides or faces is the scalar triple product of its sides or faces. I.e.,
\[\begin{align}
& V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}] \\
& \text{ }=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
On substituting,
\[\begin{align}
& V=\left| \begin{matrix}
12 & 4 & 3 \\
8 & -12 & -9 \\
33 & -4 & -24 \\
\end{matrix} \right| \\
& \text{ }=12(12\times 24-4\times 9)-4(8\times -24+9\times 33)+3(8\times -4+12\times 33) \\
& \text{ }=12(288-36)-4(-192+297)+3(-32+396) \\
& \text{ }=12\times 252-4\times 105+3\times 364 \\
& \text{ }=3024-420+1092 \\
& \text{ }=3096 \\
\end{align}\]
Thus, Option (D) is correct.
Additional Information: In vector triple product is cross and dot products are interchangeable. i.e.,
\[\begin{align}
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}\]
Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
& \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
& \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
& \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
& \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]
Note: Here we may go wrong with the vector identities and scalar triple product. Here, the scalar triple product is used for calculating the volume. By calculating the determinant of the matrix formed by the given vectors and their scalar product, the volume of the cube is obtained.
Formula used: The dot product of two vectors is
$\overrightarrow{a}\cdot \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\overrightarrow{a},\overrightarrow{b})$
The cross-product of two vectors is
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
Scalar triple product of three vectors:
We have the vectors \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] as
\[\begin{align}
& \overrightarrow{a}={{a}_{1}}\overrightarrow{i}+{{a}_{2}}\overrightarrow{j}+{{a}_{3}}\overrightarrow{k} \\
& \overrightarrow{b}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}+{{b}_{3}}\overrightarrow{k} \\
& \overrightarrow{c}={{c}_{1}}\overrightarrow{i}+{{c}_{2}}\overrightarrow{j}+{{c}_{3}}\overrightarrow{k} \\
\end{align}\]
Then, the triple product is calculated by,
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]
Thus, the volume of a 3D structure is calculated by $V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]$
Complete step by step solution: Here, we are given the three vectors that represent the three axes that form a cube when produced in their respective planes in both directions.
The given vectors are
\[\overrightarrow{a}=12\widehat{i}+4\widehat{j}+3\widehat{k}\]
\[\overrightarrow{b}=8\widehat{i}-12\widehat{j}-9\widehat{k}\]
\[\overrightarrow{c}=33\widehat{i}-4\widehat{j}-24\widehat{k}\]
We know that the concept of calculating the volume of any 3D structure which has sides or faces is the scalar triple product of its sides or faces. I.e.,
\[\begin{align}
& V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}] \\
& \text{ }=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
\end{align}\]
On substituting,
\[\begin{align}
& V=\left| \begin{matrix}
12 & 4 & 3 \\
8 & -12 & -9 \\
33 & -4 & -24 \\
\end{matrix} \right| \\
& \text{ }=12(12\times 24-4\times 9)-4(8\times -24+9\times 33)+3(8\times -4+12\times 33) \\
& \text{ }=12(288-36)-4(-192+297)+3(-32+396) \\
& \text{ }=12\times 252-4\times 105+3\times 364 \\
& \text{ }=3024-420+1092 \\
& \text{ }=3096 \\
\end{align}\]
Thus, Option (D) is correct.
Additional Information: In vector triple product is cross and dot products are interchangeable. i.e.,
\[\begin{align}
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}\]
Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
& \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
& \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
& \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
& \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]
Note: Here we may go wrong with the vector identities and scalar triple product. Here, the scalar triple product is used for calculating the volume. By calculating the determinant of the matrix formed by the given vectors and their scalar product, the volume of the cube is obtained.
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