Question

Resultant of two vectors $\vec{A}$ and $\vec{B}$ is of magnitude $P$. If $\vec{B}$ is reversed, then resultant is of magnitude $Q$. What is the value of $P^2 + Q^2$ ?

• A. $2\left(A^{2}+B^{2}\right)$
• B. $2\left(A^{2}-B^{2}\right)$
• C. $A^{2}-B^{2}$
• D. $A^{2}+B^{2}$

Answer 1

Answer: (A)

Let $\theta$ be angle between $\vec{A}$ and $\vec{B}$.
$\therefore $ Resultant of $\vec{A}$ and $\vec{B}$ is
$P=\sqrt{A^{2}+B^{2}+2AB\,cos\,\theta}\,...\left(i\right)$
When $\vec{B}$ is reversed, then the angle between $\vec{A}$ and $-\vec{B}$ is $\left(180^{\circ}-\theta\right)$.
Resultant of $\vec{A}$ and $\vec{B}$ is
$Q=\sqrt{A^{2}+B^{2}+2AB\,cos\left(180^{\circ}-\theta\right)}$
$Q=\sqrt{A^{2}+B^{2}-2AB\,cos\,\theta}\,...\left(ii\right)$
Squaring and adding $\left(i\right)$ and $\left(ii\right)$, we get
$P^{2}+Q^{2}=2\left(A^{2}+B^{2}\right)$