Question

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

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

Answer 1

Answer: (A)

2(A${}^2$ + B${}^2$)

Let $\theta$ be angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.$\therefore$ Resultant of $\overrightarrow{A}$ and $\overrightarrow{B}$ is

$P=\sqrt{A^2+B^2+2ABcos\theta }$ ---(i)

When $\overrightarrow{B}$ is reversed, then the angle between $\overrightarrow{A}$ and - $\overrightarrow{B}$ is (180${}^{\circ }$ - $\theta$).Resultant of $\overrightarrow{A}$ and - $\overrightarrow{B}$ is

Q = $\sqrt{A^2+B^2+2ABcos({180}^{\circ }-\theta )}$

Q = $\sqrt{A^2+B^2+2ABcos\theta }$ --(ii)

Squaring and adding (i) and (ii), we get$P^2+Q^2=2(A^2+B^2)$