
Question
Assertion
\[\begin{array}{*{20}{c}}
\tau & = &{\overrightarrow r \times \overrightarrow F }
\end{array}\]and \[\begin{array}{*{20}{c}}
\tau & \ne &{\overrightarrow F \times \overrightarrow r }
\end{array}\]
Reason
Cross product of vectors is commutative.
\[\begin{array}{*{20}{c}}
\tau & = &{\overrightarrow r \times \overrightarrow F }
\end{array}\] and\[\begin{array}{*{20}{c}}
\tau & = &{\overrightarrow F \times \overrightarrow r }
\end{array}\]
A) Both assertion and reason are correct and Reason is the correct explanation for the assertion.
B) Both assertion and reason are correct and the reason is not the correct explanation for the assertion.
C) Assertion is correct but Reason is incorrect.
D) Both assertion and reason are incorrect.
Answer
162k+ views
Hint:
First of all, we will determine the cross-product of the two vectors. The Curl of two vectors is normal to the plane. And then we will change the order of the vectors and again find the cross product of the vectors. After that, we will check the correct option.
Complete step by step solution:
The direction of the curl of two vectors is set by the Right-Hand rule.
Let us assume that there is a body. A force F is acting at the corner of the body. And r is the perpendicular distance from the line of action of force F, then the vector field\[\begin{array}{*{20}{c}}
{\overrightarrow F }& = &{{f_1}\widehat i + {f_2}\widehat j + {f_3}\widehat k}
\end{array}\] and the position vector is\[\begin{array}{*{20}{c}}
{\overrightarrow r }& = &{{r_1}\widehat i + {r_2}\widehat j + {r_3}\widehat k}
\end{array}\].
Now we will determine the cross product of the vector \[\overrightarrow F \]and \[\overrightarrow r \]. Therefore, we can write
\[ \Rightarrow \overrightarrow F \times \begin{array}{*{20}{c}}
{\overrightarrow r }& = &{\left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
{{f_1}}&{{f_2}}&{{f_3}} \\
{{r_1}}&{{r_2}}&{{r_3}}
\end{array}} \right|}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow F \times \overrightarrow r }& = &{\left( {{f_2}{r_3} - {f_3}{r_2}} \right)\widehat i - \left( {{f_1}{r_3} - {f_3}{r_1}} \right)\widehat j + \left( {{f_1}{r_2} - {f_2}{r_1}} \right)\widehat k}
\end{array}\]…………. (1).
Now we will determine the cross product of the vector \[\overrightarrow r \]and \[\overrightarrow F \]. Then we will get,
\[ \Rightarrow \overrightarrow r \times \begin{array}{*{20}{c}}
{\overrightarrow F }& = &{\left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
{{r_1}}&{{r_2}}&{{r_3}} \\
{{f_1}}&{{f_2}}&{{f_3}}
\end{array}} \right|}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow r \times \overrightarrow F }& = &{\left( {{f_3}{r_2} - {f_2}{r_3}} \right)\widehat i - \left( {{f_3}{r_1} - {f_1}{r_3}} \right)\widehat j + \left( {{f_2}{r_1} - {f_1}{r_2}} \right)\widehat k}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow r \times \overrightarrow F }& = &{ - \left[ {\left( {{f_2}{r_3} - {f_3}{r_2}} \right)\widehat i - \left( {{f_1}{r_3} - {f_3}{r_1}} \right)\widehat j + \left( {{f_1}{r_2} - {f_2}{r_1}} \right)\widehat k} \right]}
\end{array}\]
Now from equation 1, we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow r \times \overrightarrow F }& = &{ - \overrightarrow F \times \overrightarrow r }
\end{array}\]
Now from the above result, we can conclude that the assertion is true but the reason is false.
Therefore, the correct option is C.
Note:
The first point is to keep in mind that the cross-product of the two vectors is not commutative.
First of all, we will determine the cross-product of the two vectors. The Curl of two vectors is normal to the plane. And then we will change the order of the vectors and again find the cross product of the vectors. After that, we will check the correct option.
Complete step by step solution:
The direction of the curl of two vectors is set by the Right-Hand rule.
Let us assume that there is a body. A force F is acting at the corner of the body. And r is the perpendicular distance from the line of action of force F, then the vector field\[\begin{array}{*{20}{c}}
{\overrightarrow F }& = &{{f_1}\widehat i + {f_2}\widehat j + {f_3}\widehat k}
\end{array}\] and the position vector is\[\begin{array}{*{20}{c}}
{\overrightarrow r }& = &{{r_1}\widehat i + {r_2}\widehat j + {r_3}\widehat k}
\end{array}\].
Now we will determine the cross product of the vector \[\overrightarrow F \]and \[\overrightarrow r \]. Therefore, we can write
\[ \Rightarrow \overrightarrow F \times \begin{array}{*{20}{c}}
{\overrightarrow r }& = &{\left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
{{f_1}}&{{f_2}}&{{f_3}} \\
{{r_1}}&{{r_2}}&{{r_3}}
\end{array}} \right|}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow F \times \overrightarrow r }& = &{\left( {{f_2}{r_3} - {f_3}{r_2}} \right)\widehat i - \left( {{f_1}{r_3} - {f_3}{r_1}} \right)\widehat j + \left( {{f_1}{r_2} - {f_2}{r_1}} \right)\widehat k}
\end{array}\]…………. (1).
Now we will determine the cross product of the vector \[\overrightarrow r \]and \[\overrightarrow F \]. Then we will get,
\[ \Rightarrow \overrightarrow r \times \begin{array}{*{20}{c}}
{\overrightarrow F }& = &{\left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
{{r_1}}&{{r_2}}&{{r_3}} \\
{{f_1}}&{{f_2}}&{{f_3}}
\end{array}} \right|}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow r \times \overrightarrow F }& = &{\left( {{f_3}{r_2} - {f_2}{r_3}} \right)\widehat i - \left( {{f_3}{r_1} - {f_1}{r_3}} \right)\widehat j + \left( {{f_2}{r_1} - {f_1}{r_2}} \right)\widehat k}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow r \times \overrightarrow F }& = &{ - \left[ {\left( {{f_2}{r_3} - {f_3}{r_2}} \right)\widehat i - \left( {{f_1}{r_3} - {f_3}{r_1}} \right)\widehat j + \left( {{f_1}{r_2} - {f_2}{r_1}} \right)\widehat k} \right]}
\end{array}\]
Now from equation 1, we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow \overrightarrow r \times \overrightarrow F }& = &{ - \overrightarrow F \times \overrightarrow r }
\end{array}\]
Now from the above result, we can conclude that the assertion is true but the reason is false.
Therefore, the correct option is C.
Note:
The first point is to keep in mind that the cross-product of the two vectors is not commutative.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Class 11 JEE Main Physics Mock Test 2025

Differentiate between audible and inaudible sounds class 11 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
