Motion In A Plane Ques 3
3. The vectors $\vec{A}$ and $\vec{B}$ are such that
$|\vec{A}+\vec{B}|=|\vec{A}-\bar{B}|$
The angle between the two vectors is
[2006, 2001, 1996, 1991]
(a) $60^{\circ}$
(b) $75^{\circ}$
(c) $45^{\circ}$
(d) $90^{\circ}$
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Answer:
Correct Answer: 3.(d)
Solution: (d) $|\vec{A}+\vec{B}|^2=|\vec{A}-\vec{B}|^2$
$ \begin{aligned} & =|\vec{A}|^2+|\vec{B}|^2+2 \vec{A} \cdot \vec{B}=A^2+B^2+2 A B \cos \theta \\ & =|\vec{A}-\vec{B}|^2=|\vec{A}|^2+|\vec{B}|^2-2 \vec{A} \cdot \vec{B} \\ & =A^2+B^2-2 A B \cos \theta \end{aligned} $
$ \begin{aligned} & \text { So, } A^2+B^2+2 A B \cos \theta \\ & =A^2+B^2-2 A B \cos \theta \\ & 4 A B \cos \theta=0 \Rightarrow \cos \theta=0 \\ & \theta=90^{\circ} \end{aligned} $
So, angle between $A$ & $B$ is $90^{\circ}$.
BETA
