Question 174: Motion in 2D & Vectors

Question

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

Accepted Answer

Two vectors $$\overline A $$ and $$\overline B $$ are orthogonal to
each other, if their scalar product is zero
i.e. $$\overline A $$.$$\overline B $$ = 0.
Here, $$\overline A  = \cos \omega t\widehat i + \sin \omega t\widehat j$$
and $$\overline B  = \cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j$$
$$ 	herefore \overline A .\overline B  = \left( {\cos \omega t\widehat i + \sin \omega t\widehat j} ight)\left( {\cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j} ight)$$
= $$\cos \omega t\cos {{\omega t} \over 2} + \sin \omega t\sin {{\omega t} \over 2}$$
$$  \left( \because {\widehat i.\widehat i = \widehat j.\widehat j = 1\,and\,\widehat i.\widehat j = \widehat j.\widehat i = 0} ight)$$
= $$\cos \left( {\omega t - {{\omega t} \over 2}} ight)$$
($$ \because \cos (A - B) = cosAcosB + sinAsinB$$)
But $$\overline A .\overline B  = 0$$ (as $$\overline A $$ and $$\overline B $$ are orthogonal to each
other)
$$ 	herefore \cos \left( {\omega t - {{\omega t} \over 2}} ight) = 0$$
$$\cos \left( {\omega t - {{\omega t} \over 2}} ight) = \cos {\pi  \over 2}\,\,or\,\omega t - {{\omega t} \over 2} = {\pi  \over 2}$$
$${{\omega t} \over 2} = {\pi  \over 2}\,or\,t = {\pi  \over \omega }$$

Answer

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

Answer

t $$=$$ 0

Answer

$$t = {\pi  \over {4\omega }}$$

Answer

$$t = {\pi  \over {2\omega }}$$

Motion in 2D & Vectors

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

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