Question 130: Motion in 2D & Vectors

Question

If $\sqrt{A^2+B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$, then the angle between two vectors is

Accepted Answer

As we know that the magnitude of the resultant of two vectors $\mathrm{X}$ and $\mathbf{Y}$,
$$R^2=X^2+Y^2+2 X Y \cos 	heta\quad 	ext{.... (i)}$$
where, $	heta$ is the angle between $X$ and $Y$.
Putting,
$$\begin{aligned}
& X=(A+B) \
& Y=(A-B)
\end{aligned}$$
$$\begin{aligned}
&	ext { and } \quad R=\sqrt{A^2+B^2} 	ext { in Eq. (i), we get }\
&\begin{aligned}
& \begin{aligned}
& A^2+B^2=(A+B)^2+(A-B)^2+2(A+B)(A-B) \cos 	heta \
& \Rightarrow \quad A^2+B^2=A^2+B^2+2 A B+A^2 \
&+B^2-2 A B+2\left(A^2-B^2ight) \cos 	heta
\end{aligned}
\end{aligned}\
&\begin{aligned}
& \Rightarrow \frac{-\left(A^2+B^2ight)}{2\left(A^2-B^2ight)}=\cos 	heta \
& 	ext { we get, } \quad 	heta=\cos ^{-1}\left[-\frac{\left(A^2+B^2ight)}{2\left(A^2-B^2ight)}ight]
\end{aligned}
\end{aligned}$$

Answer

$\cos ^{-1}\left[-\frac{2\left(A^2-B^2\right)}{\left(A^2+B^2\right)}\right]$

Answer

$\cos ^{-1}\left[-\frac{A^2-B^2}{A^2 B^2}\right]$

Answer

$\cos ^{-1}\left[-\frac{\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}\right]$

Answer

$\cos ^{-1}\left[-\frac{\left(A^2-B^2\right)}{A^2+B^2}\right]$

Motion in 2D & Vectors

If \sqrt{A^2+B^2} represents the magnitude of resultant of two vectors (\mathbf{A}+\mathbf{B}) and (\mathbf{A}-\mathbf{B}), then the angle between two vectors is

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