Motion in a Plane - Result Question 6
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7. $\vec{A}$ and $\vec{B}$ are two vectors and $\theta$ is the angle between them, if $|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})$, the value of $\theta$ is
======= ####7. $\vec{A}$ and $\vec{B}$ are two vectors and $\theta$ is the angle between them, if $|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})$, the value of $\theta$ is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/motion-in-a-plane/motion-in-a-plane—result-question-6.md (a) $45^{\circ}$
(b) $30^{\circ}$
(c) $90^{\circ}$
(d) $60^{\circ}$
[2007]
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Answer:
Correct Answer: 7. (d)
Solution:
(d) $|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B})$
$\Rightarrow A B \sin \theta=\sqrt{3} A B \cos \theta$
$\Rightarrow \tan \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}$
8
(d) $|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2}$
$=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}=A^{2}+B^{2}+2 A B \cos \theta$
$=|\vec{A}-\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}-2 \vec{A} \cdot \vec{B}$
$=A^{2}+B^{2}-2 A B \cos \theta$
So, $A^{2}+B^{2}+2 A B \cos \theta$
$=A^{2}+B^{2}-2 A B \cos \theta$
$4 A B \cos \theta=0 \Rightarrow \cos \theta=0$
$\theta=90^{\circ}$
So, angle between $A$ & $B$ is $90^{\circ}$.
