If vectors $\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$ and $\vec{B}=\cos \frac{\omega t}{2

Question

If vectors $\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$ and $\vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}$ are functions of time, then the value of $t$ at which they are orthogonal to each other is\

Professor PC Thomas Classes · Answer

(a) : Two vectors $\vec{A}$ and $\vec{B}$ are orthogonal to each other, if their scalar product is zero i.e. $\vec{A} \cdot \vec{B}=0$\ Here, $\vec{A}=\cos \omega t \hat{i}+\sin \omega t \hat{j}$\ and $\vec{B}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}$\ $	herefore \quad \vec{A} \cdot \vec{B}=(\cos \omega t \hat{i}+\sin \omega t \hat{j}) \cdot\left(\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}ight)$ $$ =\cos \omega t \cos \frac{\omega t}{2}+\sin \omega t \sin \frac{\omega t}{2}=\cos \left(\omega t-\frac{\omega t}{2}ight) $$ But $\vec{A} \cdot \vec{B}=0$ (as $\vec{A}$ and $\vec{B}$ are orthogonal to each other)\ $	herefore \quad \cos \left(\omega t-\frac{\omega t}{2}ight)=0$\ $\cos \left(\omega t-\frac{\omega t}{2}ight)=\cos \frac{\pi}{2}$ or $\omega t-\frac{\omega t}{2}=\frac{\pi}{2}$\ $\frac{\omega t}{2}=\frac{\pi}{2}$ or $t=\frac{\pi}{\omega}$\