Answer
77.7k+ views
Hint: When two vectors are perpendicular to each other, their scalar product (dot product) vanishes. Use this to find a relation between the two forces.
Complete step by step solution
Let the two forces be $\overrightarrow {{F_1}} $ and $\overrightarrow {{F_2}} $.
The vector sum of the two forces will be${\overrightarrow F _{sum}} = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} $.
The vector difference of the two forces will be${\overrightarrow F _{difference}} = \overrightarrow {{F_1}} - \overrightarrow {{F_2}} $.
In the question, we are given that the vector sum of the two forces and their vector difference are perpendicular to each other. This implies that their scalar product or their dot product, must vanish.
$ \Rightarrow \left( {{{\overrightarrow F }_{sum}}} \right) \cdot \left( {{{\overrightarrow F }_{difference}}} \right) = 0$
Substituting the values of ${\overrightarrow F _{sum}}$and ${\overrightarrow F _{difference}}$in the above equation yields,
$ \Rightarrow \left( {{{\overrightarrow F }_1} + {{\overrightarrow F }_2}} \right) \cdot \left( {{{\overrightarrow F }_1} - {{\overrightarrow F }_2}} \right) = 0$
Now, we will use distributive law of dot product and open the brackets in the above equation,
\[
\Rightarrow {\overrightarrow F _1} \cdot \left( {{{\overrightarrow F }_1} - {{\overrightarrow F }_2}} \right) + {\overrightarrow F _2} \cdot \left( {{{\overrightarrow F }_1} - {{\overrightarrow F }_2}} \right) = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} - {\overrightarrow F _1} \cdot {\overrightarrow F _2} + {\overrightarrow F _2} \cdot {\overrightarrow F _1} - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\]
We know that the order of multiplication does not matter in a dot product, i.e.\[{\overrightarrow F _2}.{\overrightarrow F _1} = {\overrightarrow F _1}.{\overrightarrow F _2}\], substituting this in above equation,
\[
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} - {\overrightarrow F _1} \cdot {\overrightarrow F _2} + {\overrightarrow F _1} \cdot {\overrightarrow F _2} - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} + ( - {\overrightarrow F _1} \cdot {\overrightarrow F _2} + {\overrightarrow F _1} \cdot {\overrightarrow F _2}) - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} + (0) - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} = {\overrightarrow F _2} \cdot {\overrightarrow F _2} \\
\]
Now we can substitute \[{\overrightarrow F _1} \cdot {\overrightarrow F _1}\]as \[{\left| {{{\overrightarrow F }_1}} \right|^2}\]and \[{\overrightarrow F _2} \cdot {\overrightarrow F _2}\]as \[{\left| {{{\overrightarrow F }_2}} \right|^2}\]in the above equation,
\[
\Rightarrow {\left| {{{\overrightarrow F }_1}} \right|^2} = {\left| {{{\overrightarrow F }_2}} \right|^2} \\
\Rightarrow \left| {{{\overrightarrow F }_1}} \right| = \left| {{{\overrightarrow F }_2}} \right| \\
\]
Therefore, the magnitude of the two forces, \[{\overrightarrow F _1}\]and \[{\overrightarrow F _2}\]are coming out to be equal.
Option (C) is correct.
Note: We have used that the scalar product of two mutually perpendicular vectors is zero. This can be understood from the definition of scalar product. The scalar product (also called as dot product) of two vectors $\overrightarrow A $ and $\overrightarrow B $ is given by: $\left| {\overrightarrow A \cdot \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $ where, $\theta $ is the angle between vectors $\overrightarrow A $ and $\overrightarrow B $. If the vectors $\overrightarrow A $ and $\overrightarrow B $ are mutually perpendicular, the angle $\theta$ between them will be $90^\circ $.
$
\Rightarrow \left| {\overrightarrow A \cdot \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos 90^\circ \\
\Rightarrow \left| {\overrightarrow A \cdot \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|(0) \\
\Rightarrow \left| {\overrightarrow A \cdot \overrightarrow B } \right| = 0 \\
$
Therefore, the dot product of two mutually perpendicular vectors is always zero.
Complete step by step solution
Let the two forces be $\overrightarrow {{F_1}} $ and $\overrightarrow {{F_2}} $.
The vector sum of the two forces will be${\overrightarrow F _{sum}} = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} $.
The vector difference of the two forces will be${\overrightarrow F _{difference}} = \overrightarrow {{F_1}} - \overrightarrow {{F_2}} $.
In the question, we are given that the vector sum of the two forces and their vector difference are perpendicular to each other. This implies that their scalar product or their dot product, must vanish.
$ \Rightarrow \left( {{{\overrightarrow F }_{sum}}} \right) \cdot \left( {{{\overrightarrow F }_{difference}}} \right) = 0$
Substituting the values of ${\overrightarrow F _{sum}}$and ${\overrightarrow F _{difference}}$in the above equation yields,
$ \Rightarrow \left( {{{\overrightarrow F }_1} + {{\overrightarrow F }_2}} \right) \cdot \left( {{{\overrightarrow F }_1} - {{\overrightarrow F }_2}} \right) = 0$
Now, we will use distributive law of dot product and open the brackets in the above equation,
\[
\Rightarrow {\overrightarrow F _1} \cdot \left( {{{\overrightarrow F }_1} - {{\overrightarrow F }_2}} \right) + {\overrightarrow F _2} \cdot \left( {{{\overrightarrow F }_1} - {{\overrightarrow F }_2}} \right) = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} - {\overrightarrow F _1} \cdot {\overrightarrow F _2} + {\overrightarrow F _2} \cdot {\overrightarrow F _1} - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\]
We know that the order of multiplication does not matter in a dot product, i.e.\[{\overrightarrow F _2}.{\overrightarrow F _1} = {\overrightarrow F _1}.{\overrightarrow F _2}\], substituting this in above equation,
\[
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} - {\overrightarrow F _1} \cdot {\overrightarrow F _2} + {\overrightarrow F _1} \cdot {\overrightarrow F _2} - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} + ( - {\overrightarrow F _1} \cdot {\overrightarrow F _2} + {\overrightarrow F _1} \cdot {\overrightarrow F _2}) - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} + (0) - {\overrightarrow F _2} \cdot {\overrightarrow F _2} = 0 \\
\Rightarrow {\overrightarrow F _{1.}} \cdot {\overrightarrow F _1} = {\overrightarrow F _2} \cdot {\overrightarrow F _2} \\
\]
Now we can substitute \[{\overrightarrow F _1} \cdot {\overrightarrow F _1}\]as \[{\left| {{{\overrightarrow F }_1}} \right|^2}\]and \[{\overrightarrow F _2} \cdot {\overrightarrow F _2}\]as \[{\left| {{{\overrightarrow F }_2}} \right|^2}\]in the above equation,
\[
\Rightarrow {\left| {{{\overrightarrow F }_1}} \right|^2} = {\left| {{{\overrightarrow F }_2}} \right|^2} \\
\Rightarrow \left| {{{\overrightarrow F }_1}} \right| = \left| {{{\overrightarrow F }_2}} \right| \\
\]
Therefore, the magnitude of the two forces, \[{\overrightarrow F _1}\]and \[{\overrightarrow F _2}\]are coming out to be equal.
Option (C) is correct.
Note: We have used that the scalar product of two mutually perpendicular vectors is zero. This can be understood from the definition of scalar product. The scalar product (also called as dot product) of two vectors $\overrightarrow A $ and $\overrightarrow B $ is given by: $\left| {\overrightarrow A \cdot \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $ where, $\theta $ is the angle between vectors $\overrightarrow A $ and $\overrightarrow B $. If the vectors $\overrightarrow A $ and $\overrightarrow B $ are mutually perpendicular, the angle $\theta$ between them will be $90^\circ $.
$
\Rightarrow \left| {\overrightarrow A \cdot \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos 90^\circ \\
\Rightarrow \left| {\overrightarrow A \cdot \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|(0) \\
\Rightarrow \left| {\overrightarrow A \cdot \overrightarrow B } \right| = 0 \\
$
Therefore, the dot product of two mutually perpendicular vectors is always zero.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)