Question

Two vectors $\vec{A}$ and $\vec{B}$ inclined at an angle $\theta$ have a resultant $\vec{R}$ which makes an angle $\alpha$ with $\vec{A}$ and angle $\beta$ with $\vec{B}$ . Let the magnitudes of the vectors $\vec{A}$, $\vec{B}$ and $\vec{R}$ be represented by $A$, $B$ and $R$ respectively. Which of the following relations is not correct?

• A. $\frac{R}{sin\left(\alpha+\beta\right)}=\frac{A}{sin\,\alpha}=\frac{B}{sin\,\beta}$
• B. $Rsin\alpha=Bsin\left(\alpha+\beta\right)$
• C. $Asin\alpha=Bsin\beta$
• D. $Rsin\beta=Asin\left(\alpha+\beta\right)$

Answer 1

The Correct Option is A

Let $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ represent two vectors $\vec{A}$ and $\vec{B}$ making an angle $\left(\alpha+\beta\right)$. Using the parallelogram method of vector addition, Resultant vector, $\vec{R}=\vec{A}+\vec{B}$ $SN$ is normal to $OP$ and $PM$ is normal to $OS$. From the geometry of the figure, $OS^2 = ON^2 + SN^2 = (OP + PN)^2 + SN^2$ $=\left(A+Bcos\left(\alpha+\beta\right)\right)^{2}+\left(Bsin\left(\alpha+\beta\right)\right)^{2}$ $R^{2}=A^{2}+B^{2}+2ABcos\left(\alpha+\beta\right)$ In $\Delta OSN$, $SN = OS ,S sin\alpha = Rsin\alpha$ and in $\Delta PSN$, $SN=PSsin\left(\alpha+\beta\right)=Bsin\left(\alpha+\beta\right)$ $\therefore Rsin\alpha=Bsin\left(\alpha+\beta\right)$ or $\frac{R}{sin\left(\alpha+\beta\right)}=\frac{B}{sin\,\alpha}\,...\left(i\right)$ Similarly, $PM = Asin\alpha = Bsin\beta$ $\frac{A}{sin\,\beta}=\frac{B}{sin\,\alpha}\,...\left(ii\right)$ Combining (i) and (ii), we get $\frac{R}{sin\left(\alpha+\beta\right)}=\frac{A}{sin\,\beta}=\frac{B}{sin\,\alpha}\,...\left(iii\right)$ From eqn. (iii), $Rsin\beta=A\,sin\left(\alpha+\beta\right)$