Question

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

• A. are not equal to each other in magnitude.
• B. cannot be predicted.
• C. are equal to each other.
  
• D. are equal to each other in magnitude.

Answer 1

The correct option is D are equal to each other in magnitude.

Formula Used: $\overrightarrow{A}\cdot \overrightarrow{B}=A\cdot B\cos \theta$

Given:

Vector sum of two forces is perpendicular to their diffrences $i.e.\theta ={90}^{\circ }$

As we know,
$\overrightarrow{A}\cdot \overrightarrow{B}=A\cdot B\cos \theta$

$\cos {90}^{\circ }=0$

Therefor, as per the question, if two vectors are perpendicular, then
$(\overrightarrow{A}+\overrightarrow{B})\cdot (\overrightarrow{A}-\overrightarrow{B})=|\overrightarrow{A}+\overrightarrow{B}||\overrightarrow{A}-\overrightarrow{B}|\cos {90}^{\circ }$

$(\overrightarrow{A}+\overrightarrow{B})\cdot (\overrightarrow{A}-\overrightarrow{B})=0$

Hence,

$(\overrightarrow{A}+\overrightarrow{B})\cdot (\overrightarrow{A}-\overrightarrow{B})=0$
$\overrightarrow{A}\cdot \overrightarrow{A}-\overrightarrow{A}\cdot \overrightarrow{B}+\overrightarrow{B}\cdot \overrightarrow{A}-\overrightarrow{B}\cdot \overrightarrow{B}=0$
$A^2-B^2=0$
$A^2=B^2$
Therefore,
$A=B$ $i.e.,$two vectors are equal to each other in magnitude.

Final answer: (a)