Question

At what angle the two vectors of magnitudes $(A+B)$ and $(A-B)$ must act, so that resultant is $\sqrt{A^2+B^2}?$

• A. ${\cos }^{-1}\frac{(A+B)}{A-B}$
• B. ${\cos }^{-1}\left(\frac{\left(A^2+B^2\right)}{2\left(B^2-A^2\right)}\right)$
• C. ${\cos }^{-1}\left(\frac{A^2+B^2}{A^2-B^2}\right)$
• D. None of these

Answer 1

Answer: (B)

$R=\sqrt{P^2+Q^2+2PQ\cos \theta }$
$$\begin{gathered} R=\sqrt{A^2+B^2},P=(A+B),Q=(A-B) \\ <br/><br/>\sqrt{A^2+B^2}=\sqrt{(A+B{)}^2+\left(A^2-B^2\right)+2(A+B)(A-B)\cos \theta } \end{gathered}$$
Squaring both sides
$A^2+B^2=(A+B{)}^2+(A-B{)}^2+2\left(A^2-B^2\right)\cos \theta$
$A^2+B^2=A^2+B^2+2AB+A^2+B^2-2AB+2\left(A^2-B^2\right)\cos \theta$
$-A^2-B^2=2\left(A^2-B^2\right)\cos \theta$
$-\left(A^2+B^2\right)=2\left(A^2-B^2\right)\cos \theta$
Multiply by (-)
$A^2+B^2=2\left(B^2-A^2\right)\cos \theta$
${\cos }^{-1}\left(\frac{A^2+B^2}{2\left(B^2-A^2\right)}\right)=\theta$