
If \[\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\], \[\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}\] and \[\overrightarrow{c}=\widehat{i}-\widehat{j}+2\widehat{k}\], then $\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=$
A. \[6~\]
B. \[10\]
C. \[12\]
D. \[24\]
Answer
163.2k+ views
Hint: In the above question, the dot and cross products are applied to find the required vector product.
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
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
Complete step by step solution: Given that,
\[\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\]
\[\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}\]
\[\overrightarrow{c}=\widehat{i}-\widehat{j}+2\widehat{k}\]
Then,
$\begin{align}
& \overrightarrow{b}\times \overrightarrow{c}=\left| \begin{matrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
1 & 2 & 1 \\
1 & -1 & 2 \\
\end{matrix} \right| \\
& =\widehat{i}(4+1)-\widehat{j}(2-1)+\widehat{k}(-1-2) \\
& =5\widehat{i}-\widehat{j}-3\widehat{k} \\
\end{align}$
Then, the required scalar triple product is
\[\begin{align}
& \overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})=(2\widehat{i}+\widehat{j}-\widehat{k})\cdot (5\widehat{i}-\widehat{j}-3\widehat{k}) \\
& \text{ }=2\times 5+1\times -1+1\times 3 \\
& \text{ }=10-1+3 \\
& \text{ }=12 \\
\end{align}\]
Thus, Option (C) is the correct value.
Additional Information: 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 product identities (dot and cross).
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
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
Complete step by step solution: Given that,
\[\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\]
\[\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}\]
\[\overrightarrow{c}=\widehat{i}-\widehat{j}+2\widehat{k}\]
Then,
$\begin{align}
& \overrightarrow{b}\times \overrightarrow{c}=\left| \begin{matrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
1 & 2 & 1 \\
1 & -1 & 2 \\
\end{matrix} \right| \\
& =\widehat{i}(4+1)-\widehat{j}(2-1)+\widehat{k}(-1-2) \\
& =5\widehat{i}-\widehat{j}-3\widehat{k} \\
\end{align}$
Then, the required scalar triple product is
\[\begin{align}
& \overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})=(2\widehat{i}+\widehat{j}-\widehat{k})\cdot (5\widehat{i}-\widehat{j}-3\widehat{k}) \\
& \text{ }=2\times 5+1\times -1+1\times 3 \\
& \text{ }=10-1+3 \\
& \text{ }=12 \\
\end{align}\]
Thus, Option (C) is the correct value.
Additional Information: 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 product identities (dot and cross).
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

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

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
