a and b are two unit vectors such that |2 a +3 b |=|3 a -2 b |. Find the angle between a and b.

\[\begin{aligned} & |2 \vec{a}+3 \vec{b}|=|3 \vec{a}-2 \vec{b}| \\ & \Rightarrow|2 \vec{a}+3 \vec{b}|^{2}=|3 \vec{a}-2 \vec{b}|^{2} \\ & \Rightarrow 4|\vec{a}|^{2}+12 \vec{a} \cdot \vec{b}+9|\vec{b}|^{2}=9|\vec{a}|^{2}-12 \vec{a} \cdot \vec{b}+4|\vec{b}|^{2} \end{aligned} \] As \(\displaystyle |\vec{a}|=|\vec{b}|=1\) \(\displaystyle \therefore 24 \vec{a} \cdot \vec{b}=5|\vec{a}|^{2}-5|\vec{b}|^{2}=0 \Longrightarrow \vec{a} \cdot \vec{b}=0\) So, \(\displaystyle \vec{a} \perp \vec{b}\) or Angle between them is \(\displaystyle \frac{\pi}{2}\)