
What will be the volume of that parallelopiped whose sides are \[\overrightarrow{a}=\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}\], \[\overrightarrow{b}=\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}\] and \[\overrightarrow{c}=2\overrightarrow{i}-5\overrightarrow{j}+3\overrightarrow{k}\].
A. \[5\] unit
B. \[6\] unit
C. \[7\] unit
D. \[8\] unit
Answer
164.1k+ views
Hint: In this question, the dot and cross products of vectors are applied to find the required vector expression. The dot product is said to be a scalar product and the cross product is said to be a skew product or vector product. By using appropriate formulae, the required vector product 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 parallelopiped with edges \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] is \[V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
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:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
& \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
& \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
& \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
& \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]
Complete step by step solution:It is given that, sides of a parallelopiped are
\[\overrightarrow{a}=\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}\]
\[\overrightarrow{b}=\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}\]
\[\overrightarrow{c}=2\overrightarrow{i}-5\overrightarrow{j}+3\overrightarrow{k}\]
Then, the volume of the given parallelopiped is
\[V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
\[\begin{align}
& \Rightarrow V=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
& \text{ }=\left| \begin{matrix}
1 & -1 & 1 \\
1 & -3 & 4 \\
2 & -5 & 3 \\
\end{matrix} \right| \\
& \text{ }=1(-9+20)+1(3-8)+1(-5+6) \\
& \text{ }=11-5+1 \\
& \text{ }=7 \\
\end{align}\]
Option ‘C’ is correct
Note: Here we may go wrong with the vector identities and scalar triple product. Here are the simple formulae used for solving the given vector. By applying appropriate vector products, the given vector equation is evaluated.
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 parallelopiped with edges \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] is \[V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
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:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
& \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
& \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
& \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
& \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]
Complete step by step solution:It is given that, sides of a parallelopiped are
\[\overrightarrow{a}=\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}\]
\[\overrightarrow{b}=\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}\]
\[\overrightarrow{c}=2\overrightarrow{i}-5\overrightarrow{j}+3\overrightarrow{k}\]
Then, the volume of the given parallelopiped is
\[V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
\[\begin{align}
& \Rightarrow V=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
& \text{ }=\left| \begin{matrix}
1 & -1 & 1 \\
1 & -3 & 4 \\
2 & -5 & 3 \\
\end{matrix} \right| \\
& \text{ }=1(-9+20)+1(3-8)+1(-5+6) \\
& \text{ }=11-5+1 \\
& \text{ }=7 \\
\end{align}\]
Option ‘C’ is correct
Note: Here we may go wrong with the vector identities and scalar triple product. Here are the simple formulae used for solving the given vector. By applying appropriate vector products, the given vector equation is evaluated.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
