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)

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

Answer 2

Let $\theta$be angle between $\overset{\rightarrow}{A}$and $\overset{\rightarrow}{B}$is

$\therefore$Resultant of $\overset{\rightarrow}{A}$and $\overset{\rightarrow}{B}$is

$P = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$ ….(i)

When $\overset{\rightarrow}{B}$ is reversed, then the angle between $\overset{\rightarrow}{A}$ and $- \overset{\rightarrow}{B}$ is ($180{^\circ} - \theta$)

Resultant of $\overset{\rightarrow}{A}$ and $- \overset{\rightarrow}{B}$is

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

$Q = \sqrt{A^{2} + B^{2} - 2AB\cos\theta}$ ….. (ii)

Squaring and adding (i) and (ii), we get

$P^{2} + Q^{2} = 2(A^{2} + B^{2})$