Question

$\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C},|\overrightarrow{A}|=|\overrightarrow{B}|=|\overrightarrow{C}|$, then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is:

• A. 45${}^0$
• B. 60${}^0$
• C. 90${}^0$
• D. 120${}^0$

Answer 1

Answer: (D)

120${}^0$

Let $\theta$ be the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.

Now we have:

$(\overrightarrow{A}+\overrightarrow{B})=\overrightarrow{C}$

$(\overrightarrow{A}+\overrightarrow{B})\cdot (\overrightarrow{A}+\overrightarrow{B})$=$\overrightarrow{C}\cdot \overrightarrow{C}$

$\overrightarrow{A}\cdot \overrightarrow{A}+\overrightarrow{B}\cdot \overrightarrow{B}$+$\overrightarrow{A}\cdot \overrightarrow{B}$+$\overrightarrow{B}\cdot \overrightarrow{A}$=$\overrightarrow{C}\cdot \overrightarrow{C}$

We know that scalar product of a vector by itself is equal to the square of the magnitude of that vector therefore

$A^2+B^2+2\overrightarrow{A}\cdot \overrightarrow{B}$ = $C^2$

Since given that the magnitude of three vectors is equal so we can put $A=B=C$ in above equation.

Now, $C^2+C^2+2AB\cos \theta$=$C^2$

$2C^2\cos \theta$= $-C^2$

$\cos \theta =-\frac{1}{2}$

$\cos \theta =\cos {120}^o$

$\theta ={120}^o$