
If \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are three non-coplanar vectors, then \[\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\]
A. $0$
B. $2$
C. $3$
D. None of these
Answer
163.2k+ 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 equation 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:
\[\begin{align}
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}\]
Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
Complete step by step solution:It is given that,
\[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are three non-coplanar vectors, then \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\ne 0\].
So,
\[\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}+\dfrac{[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}\text{ }...(1)\]
Since we have a scalar triple product as
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}\]
We have the vector identity
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]\]
On substituting this identity in equation (1), we get
\[\begin{align}
& \dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}+\dfrac{[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]} \\
& \text{ }=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}-\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]} \\
& \text{ }=0 \\
\end{align}\]
Therefore, the value of the given vector equation is 0.
Option ‘A’ 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 equation. I.e., \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]\] and \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}\]. 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:
\[\begin{align}
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
& [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}\]
Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
Complete step by step solution:It is given that,
\[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] are three non-coplanar vectors, then \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\ne 0\].
So,
\[\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}+\dfrac{[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}\text{ }...(1)\]
Since we have a scalar triple product as
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}\]
We have the vector identity
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]\]
On substituting this identity in equation (1), we get
\[\begin{align}
& \dfrac{\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}}{\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}}+\dfrac{\overrightarrow{b}\cdot \overrightarrow{a}\times \overrightarrow{c}}{\overrightarrow{c}\cdot \overrightarrow{a}\times \overrightarrow{b}}=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}+\dfrac{[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]} \\
& \text{ }=\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]}-\dfrac{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}{[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]} \\
& \text{ }=0 \\
\end{align}\]
Therefore, the value of the given vector equation is 0.
Option ‘A’ 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 equation. I.e., \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]\] and \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b}\]. By applying appropriate vector products, the given vector equation is evaluated.
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

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

Instantaneous Velocity - Formula based Examples 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

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
