Mathematics - Vector Question with Solution | TestHub

Given that, $\left|\frac{\hat{\mathbf{u}}+\hat{\mathbf{v}}}{2}+\hat{\mathbf{u}}\times \hat{\mathbf{v}}\right|=1$ $$\begin{array}{ll}\Rightarrow & {\left|\frac{\hat{\mathbf{u}}+\hat{\mathbf{v}}}{2}+\hat{\mathbf{u}}\times \hat{\mathbf{v}}\right|}^2=1\\ \Rightarrow & \frac{2+2\cos \theta }{4}+{\sin }^2\theta =1\\ & [\because \hat{\mathbf{u}}\cdot (\hat{\mathbf{u}}\times \hat{\mathbf{v}})=\hat{\mathbf{v}}(\hat{\mathbf{u}}\times \hat{\mathbf{v}})=0]\\ \Rightarrow & {\cos }^2\frac{\theta }{2}=\cos 2\theta \\ \Rightarrow & \theta =n\pi \pm \frac{\theta }{2},n\in z\\ \Rightarrow & \theta =\frac{2\pi }{3}\end{array}$$ $\Rightarrow |\hat{\mathbf{u}}\times \hat{\mathbf{v}}|=\sin \frac{2\pi }{3}=\sin \frac{\pi }{3}=\left|\frac{\hat{\mathbf{u}}-\hat{\mathbf{v}}}{2}\right|$