Kinematics 6 Question 2
2. Two vectors $\mathbf{A}$ and $\mathbf{B}$ have equal magnitudes. The magnitude of $(\mathbf{A}+\mathbf{B})$ is ’ $n$ ’ times the magnitude of $(\mathbf{A}-\mathbf{B})$. The angle between $\mathbf{A}$ and $\mathbf{B}$ is
(2019 Main, 10 Jan II)
(a) $\sin ^{-1} \frac{n^{2}-1}{n^{2}+1}$
(b) $\sin ^{-1} \frac{n-1}{n+1}$
(c) $\cos ^{-1} \frac{n^{2}-1}{n^{2}+1}$
(d) $\cos ^{-1} \frac{n-1}{n+1}$
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Answer:
Correct Answer: 2. (c)
Solution:
- Given, $|\mathbf{A}|=\mid \mathbf{B}$
$$ \text { or } \quad A=B $$
Let magnitude of $(\mathbf{A}+\mathbf{B})$ is $R$ and for $(\mathbf{A}-\mathbf{B})$ is $R^{\prime}$.
Now, $\quad \mathbf{R}=\mathbf{A}+\mathbf{B}$
and $R^{2}=A^{2}+B^{2}+2 A B \cos \theta$
$$ R^{2}=2 A^{2}+2 A^{2} \cos \theta $$
$$ \begin{array}{rlrl} \text { Again, } & & & \mathbf{R}^{\prime}=\mathbf{A}-\mathbf{B} \\ \Rightarrow & & R^{\prime 2}=A^{2}+B^{2}-2 A B \cos \theta \\ & & R^{\prime 2}=2 A^{2}-2 A^{2} \cos \theta \end{array} $$
$[\because$ using Eq. (i) $]$
Given,
$$ R=n R^{\prime} \text { or } \quad{\frac{R}{R^{\prime}}}^{2}=n^{2} $$
Dividing Eq. (ii) with Eq. (iii), we get
$$ \frac{n^{2}}{1}=\frac{1+\cos \theta}{1-\cos \theta} $$
$$ \begin{aligned} & \text { or } \frac{n^{2}-1}{n^{2}+1}=\frac{(1+\cos \theta)-(1-\cos \theta)}{(1+\cos \theta)+(1-\cos \theta)} \\ & \Rightarrow \quad \frac{n^{2}-1}{n^{2}+1}=\frac{2 \cos \theta}{2}=\cos \theta \text { or } \theta=\cos ^{-1} \frac{n^{2}-1}{n^{2}+1} \end{aligned} $$
