Question

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

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

Answer 1

Answer: (A)

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

Answer 2

Two vectors $\overrightarrow{A}$ and $\overrightarrow{ B}$ are orthogonal to each other, if their scalar product is zero i.e. $\overrightarrow{A} . \overrightarrow{ B} = 0$.
Here, $\overrightarrow{ A} = \cos \omega t \widehat{i} + \sin \omega t \widehat{j}$
and $\overrightarrow{ B} = \cos \frac{ \omega t}{ 2} \widehat{i} + \sin \frac{ \omega t }{ 2} \widehat{j}$
$\therefore \overrightarrow{A} \cdot \overrightarrow{ B} = ( \cos \, \omega t \widehat{i} + \sin \, \omega t \widehat{j} ) \cdot \bigg( \cos \frac{ \omega t}{ 2} \widehat{i} + \sin \frac{ \omega t }{ 2} \widehat{j} \bigg)$
$= \cos \omega t \cos \frac{ \omega t }{ 2} + \sin \omega t \sin \frac{ \omega t }{ 2}$
$( \because \widehat{i} \cdot \widehat{i} = \widehat{j} \cdot \widehat{j} = 1 \, and \, \widehat{i} \cdot \widehat{j} = \widehat{j} \cdot \widehat{i} = 0 )$
$= \cos \bigg( \omega t - \frac{ \omega t }{ 2} \bigg)$ $( \because \cos (A - B) = cos A \cos B + sin A sin B)$
But $\overrightarrow{A} \cdot \overrightarrow{B} = 0$ ( as $\overrightarrow{A}$ and $\overrightarrow{B}$ are orthogonal to each other).
$\therefore \cos \bigg( \omega t - \frac{ \omega t}{ 2} \bigg) = 0$
$cos \bigg( \omega t - \frac{ \omega t}{ 2} \bigg) = \cos \frac{ \pi}{ 2}$ or $\omega t - \frac{ \omega t}{ 2} = \frac{ \pi}{ 2}$
$\frac{ \omega t}{ 2} = \frac{ \pi} { 2}$ or $t = \frac{ \pi}{ \omega}$