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(D) The magnitude of the resultant $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given

Vedclass · Accepted Answer

$120$

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(D) The magnitude of the resultant $R$ of two vectors $\vec{A}$ and $\vec{B}$ is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given that the magnitudes of the two vectors are equal,let $A = B = x$.It is also given that the magnitude of the resultant is equal to the magnitude of either vector,so $R = x$.Substituting these values into the formula: $x = \sqrt{x^2 + x^2 + 2x^2 \cos 	heta}$.Squaring both sides: $x^2 = 2x^2 + 2x^2 \cos 	heta$.Dividing by $x^2$ (assuming $x 
eq 0$): $1 = 2 + 2 \cos 	heta$.Rearranging the terms: $2 \cos 	heta = 1 - 2 = -1$.Therefore,$\cos 	heta = -\frac{1}{2}$.This corresponds to an angle of $	heta = 120^{\circ}$.

The resultant magnitude of two vectors of same magnitude is equal to the magnitude of either. The angle between the two vectors is (in $^{\circ}$)

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