Given that → A + → B + → C = 0 . Out of three vectors, two are...

Let, \left|\overset{→}{A}\right| = \left|\overset{→}{B}\right| = \text{x}$<br/>Then \left|\overset{→}{C}\right| = \sqrt{2} \text{x}$
Now $\overset{→}{A} + \overset{→}{B} = - \overset{→}{C}$
\left(\overset{→}{A} + \overset{→}{B}\right) \cdot \left(\overset{→}{A} + \overset{→}{B}\right) = \left(- \overset{→}{C}\right) \cdot \left(- \overset{→}{C}\right)
$2 \left(\text{A.B}\right) +$A^{2} + B^{2} = C^{2}$<br/>$2$\text{ABcos}\theta +$x^{2} + x^{2} = 2x^{2}<br/>\text\cos \theta =0$<br/>\theta =$9 {0}^{o}$<br/>$⇒$Angle between$A$and$B$is${90}^{0}$<br/>Again$\overset{→}{A} + \overset{→}{C} = - \overset{→}{B}$<br/>$\left(\overset{→}{A} + \overset{→}{C}\right) \cdot $\left(\overset{→}{A} + \overset{→}{C}\right) = \left(- \overset{→}{B}\right) \left(- \overset{→}{B}\right)<br/>$A^{2} + C^{2} + 2 $\text{Acos} \theta =$B^{2}$<br/>$3 x^{2} + 2 \text{.} \text{x} \sqrt{2} \text{x} \text\cos \theta =x^{2}<br/>\text\cos \theta = \dfrac{- 2}{2 \sqrt{2}} = \dfrac{- 1}{\sqrt{2}}<br/>\theta =1 3 {5}^{o}$<br/>i.e Angle between$A$and$C$is${135}^{o}$<br/>Similarly, angle between$B$and$C$is${θ}_{2}$<br/>$\overset{→}{B} + \overset{→}{C} = - \overset{→}{A}$<br/>$B^{2} + C^{2} + 2 $\text{Bcos} \theta =$A^{2}<br/>\text\cos \theta = \dfrac{- 2}{2 \sqrt{2}} = \left(\dfrac{- 1}{\sqrt{2}}\right)<br/>{θ}_{2} = 1 3 {5}^{o}