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Q.
Two equal vectors have a resultant equal to either of them, then the angle between them will be
Bihar CECEBihar CECE 2007Motion in a Plane
Solution:
Two vectors (inclined at any angle) and their sum vector form a triangle.
It is given that two vectors have a resultant equal to either of them, hence these three vectors form an equilateral triangle each angle of $60^{\circ}$.
In the figure $\vec{A}$ and $\vec{B}$ are two vector $(\vec{A}=\vec{B})$ having their sum vectors $\vec{R}$ such that.
$\vec{ R }=\vec{ A }=\vec{ B }$
Thus, the vectors $\vec{ A }$ and $\vec{ B }$ of same magnitude have the resultant vectors $\vec{ R }$ of the same magnitude.
In this case angle between $\vec{ A }$ and $\vec{ B }$ is $120^{\circ}$
Alternative : Let there be two vectors $\vec{ A }$ and $\vec{ B }$
where, $A=B$
Their sum is
$\vec{ R }=\vec{ A }+\vec{ B }$
Taking self product of both sides, we get
$\vec{ R } \cdot \vec{ R } =(\vec{ A }+\vec{ B }) \cdot(\vec{ A }+\vec{ B }) $
$=\vec{ A } \cdot \vec{ A }+2 \vec{ A } \cdot \vec{ B }+\vec{ B } \cdot \vec{ B } $
$=A^{2}+2 A B \cos \theta+B^{2}$
where $\theta$ is angle between $\vec{ A }$ and $\vec{ B }$
When $\vec{ R }=\vec{ A }=\vec{ B }$, then we have
$A^{2}=A^{2}+2 A^{2} \cos \theta+A^{2} $
$\Rightarrow 2 A^{2} \cos \theta=-A^{2}$
$\Rightarrow \cos \theta=-\frac{1}{2}$
$\Rightarrow \theta=120^{\circ}$
In this condition angle between given vectors should be $120^{\circ}$.
