Question

Given that $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=0$. Out of the three vectors, two are equal in magnitude and the magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. Find the angle between the equal magnitude vectors.

Answer 1

Step 1: Given

$\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=0$ . . . . . . (1)
Let:

$A=B$
$C=\sqrt{2}A=\sqrt{2}B$

Step 2: Formula used

$MagnitudeofresultantoftwovectorsAandB=\sqrt{A^2+B^2+2ABcos\theta }$

Step 3: Calculation of angle
Let $\theta$ be the angle between the vectors $\overrightarrow{A}and\overrightarrow{B}$.
We can write equation (1) as $-\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$
Here, $-\overrightarrow{C}$ is the resultant of vectors $\overrightarrow{A}\&\overrightarrow{B}$.
$\therefore |-C|=\sqrt{A^2+B^2+2ABcos\theta }$
$\Rightarrow C=\sqrt{A^2+B^2+2ABcos\theta }$

Using given conditions, we get:
$\Rightarrow \sqrt{2}A=\sqrt{A^2+A^2+2A^{2}cos\theta }$
$\Rightarrow \sqrt{2}A=\sqrt{2A^2+2A^{2}cos\theta }$
$\Rightarrow \sqrt{2}A=\sqrt{2}A\times \sqrt{1+cos\theta }$
$\Rightarrow 1+cos\theta =1$

$\Rightarrow cos\theta =0$

$\Rightarrow cos\theta =cos{90}^0$

$Comparingbothsides,weget$

$\Rightarrow \theta ={90}^{\circ }$

Hence, angle between equal vectors is $\theta ={90}^{\circ }$.