Question

Two equal vectors have a resultant equal to either of them, then the angle between them will be

• A. $120^{\circ }$
• B. $110^{\circ }$
• C. $60^{\circ }$
• D. $150^{\circ }$

Answer 1

Answer: (A)

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
$\vec{R}\cdot \vec{R}=(\vec{A}+\vec{B})\cdot (\vec{A}+\vec{B})$
$=\vec{A}\cdot \vec{A}+2\vec{A}\cdot \vec{B}+\vec{B}\cdot \vec{B}$
$=A^2+2AB\cos \theta +B^2$
where $\theta$ is angle between $\vec{A}$ and $\vec{B}$
When $\vec{R}=\vec{A}=\vec{B}$, then we have
$A^2=A^2+2A^2\cos \theta +A^2$
$\Rightarrow 2A^2\cos \theta =-A^2$
$\Rightarrow \cos \theta =-\frac{1}{2}$
$\Rightarrow \theta =120^{\circ }$
In this condition angle between given vectors should be $120^{\circ }$.