Question

Given, $|A+B|=P,|A-B|=Q$. The value of $P^2+Q^2$ is

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

Answer 1

Answer: (A)

According to the question, representation of vectors $A$ and $B$ can be shown as follows

Given, $|A+B|=P$
$\Rightarrow |A+B{|}^2=P^2$
$P^2=A^2+B^2+2AB\cos \theta$....(i)
Also, $|A-B|=Q\Rightarrow |A-B{|}^2=Q^2$
$\Rightarrow A^2+B^2+2AB\cos \left(180^{\circ }-\theta \right)=Q^2$
$\Rightarrow A^2+B^2-2AB\cos \theta =Q^2$....(ii)
Adding Eqs. (i) and (ii), we get
$P^2+Q^2=2\left(A^2+B^2\right)$