If A and B are two non-zero vectors having eq | vedclass

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(D) Let the magnitude of vectors $A$ and $B$ be $|A| = |B| = a$. Let the angle between $A$ and $B$ be $	heta$.Consider the vector $R = A - B$. This c

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dependent on the orientation of $A$ and $B$

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(D) Let the magnitude of vectors $A$ and $B$ be $|A| = |B| = a$. Let the angle between $A$ and $B$ be $	heta$.Consider the vector $R = A - B$. This can be written as $R = A + (-B)$.The magnitude of vector $-B$ is equal to the magnitude of vector $B$,which is $a$.The angle between $A$ and $-B$ is $(180^{\circ} - 	heta)$.Using the law of vector addition,the resultant vector $R$ forms a triangle with $A$ and $-B$.Since $|A| = |-B| = a$,the triangle formed by $A$,$-B$,and $R$ is an isosceles triangle.The angle $\alpha$ between $A$ and $R$ (which is $A - B$) is given by the base angle of this isosceles triangle.The sum of angles in a triangle is $180^{\circ}$. The angle between $A$ and $-B$ is $(180^{\circ} - 	heta)$.Thus,$2\alpha + (180^{\circ} - 	heta) = 180^{\circ}$.Solving for $\alpha$,we get $2\alpha = 	heta$,or $\alpha = \frac{	heta}{2}$.Since the angle $\alpha$ depends on $	heta$ (the angle between $A$ and $B$),the correct option is $(d)$.

If $A$ and $B$ are two non-zero vectors having equal magnitude,the angle between the vectors $A$ and $A - B$ is

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