Question

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 1

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.