Question

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

• A. $t=\frac{\pi }{4\omega }$
• B. $t=\frac{\pi }{2\omega }$
• C. $t=\frac{\pi }{\omega }$
• D. t = 0

Answer 1

The correct option is C $t=\frac{\pi }{\omega }$

Given,

$\overrightarrow{A}=(\cos wt\hat{i}+\sin w\hat{j})\cdot$

$\overrightarrow{B}=(\cos \frac{wt}{2}\hat{i}+\sin \frac{wt}{2}\hat{j})$

When they are orthogonal, their dot product will e zero

$\overrightarrow{A}\cdot \overrightarrow{B}=(\cos wt\hat{i}+\sin wt\hat{j})\cdot (\cos \frac{wt}{2}\hat{i}+\sin \frac{wt}{2}j)$

$0=\cos (wt-\frac{wt}{2})$

$\cos (wt-\frac{wt}{2})=0=\frac{\pi }{2}$

$\cos \left(\frac{wt}{2}\right)=\frac{\pi }{2}$

$\frac{wt}{2}=\frac{\pi }{2}$

$t=\frac{\pi }{w}$

Hence, value of time is $\frac{\pi }{\omega }$