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
$P=\sqrt{A^2+B^2+2ABcos\theta }...\left(i\right)$
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
$Q=\sqrt{A^2+B^2+2ABcos\left(180^{\circ }-\theta \right)}$
$Q=\sqrt{A^2+B^2-2ABcos\theta }...\left(ii\right)$
Squaring and adding $\left(i\right)$ and $\left(ii\right)$, we get
$P^2+Q^2=2\left(A^2+B^2\right)$