Q80 / 110 · VITEEE 2024 · Physics

The resultant of $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{R}_1$. On reversing the vector $\mathbf{B}$, the resultant becomes $\mathbf{R}_2$. What is the value of $R_1^2+R_2^2$ ?

  • A
    $A^2+B^2$
  • B
    $A^2-B^2$
  • Correct answer
    C
    $2\left(A^2+B^2\right)$
  • D
    $2\left(A^2-B^2\right)$

Solution

$\begin{aligned} & R_1^2=A^2+B^2+2 A B \cos \theta \\ & R_2^2=A^2+B^2-2 A B \cos \theta \\ & \therefore R_1^2+R_2^2=2\left(A^2+B^2\right)\end{aligned}$

Hence, the correct answer is option: 'C'.