Question

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
(A) Cannot be predicted
(B) Are equal to each other
(C) Are equal to each other in magnitude
(D) Are not equal to each other in magnitude

Answer 1

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.