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

Answer: (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,
$O{S}^2=O{N}^2+S{N}^2=(OP+PN{)}^2+S{N}^2$
$={\left(A+Bcos\left(\alpha +\beta \right)\right)}^2+{\left(Bsin\left(\alpha +\beta \right)\right)}^2$
$R^2=A^2+B^2+2ABcos\left(\alpha +\beta \right)$
In $\Delta OSN$, $SN=OS,Ssin\alpha =Rsin\alpha$
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), $Rsin\beta =Asin\left(\alpha +\beta \right)$