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