Question

A unit vector in the xy-plane that makes an angle of $\frac{\pi }{3}$ with the vector $3i-4j$ is

• A. $i$
• B. $\frac{1}{\sqrt{2}}(i+j)$
• C. $\frac{i}{\sqrt{3}}(i+j+k)$
• D. Noe of the above

Answer 1

The correct option is D Noe of the above

$\because Threquiredvectorisinx-yplane,$

$\therefore Lettheunitvectorbe\left(\cos \theta \right)\stackrel{\wedge }{i}+\left(\sin \theta \right)\stackrel{\wedge }{j}$

$Thisvectormakes\frac{\pi }{3}with3\stackrel{\wedge }{i}-4\stackrel{\wedge }{j}$

$⟹\left(\right(\cos \theta )\stackrel{\wedge }{i}+(\sin \theta \left)\stackrel{\wedge }{j}\right).(3\stackrel{\wedge }{i}-4\stackrel{\wedge }{j})$

$=\left(\sqrt{\cos {}^2\theta +\sin {}^2\theta }\right)\left(\sqrt{3^2+4^2}\right)\left(\cos \frac{\pi }{3}\right)$

$⟹3\cos \theta -4\sin \theta =5\cos \frac{\pi }{3}$

$⟹3\cos \theta -4\sin \theta =\frac{5}{2}$

$⟹\frac{3}{5}\cos \theta -\frac{4}{5}\sin \theta =\frac{1}{2}$

$⟹\cos (\theta +{53}^{\circ })=\frac{1}{2}$

$⟹\cos (\theta +{53}^{\circ })=\cos \left({60}^{\circ }\right)$

$⟹\theta +53={60}^{\circ }$

$⟹\theta =7^{\circ }$

$\therefore Therequiredvectoris$

$\cos \left(7^{\circ }\right)\stackrel{\wedge }{i}+\sin \left(7^{\circ }\right)\stackrel{\wedge }{j}$

$D)Noneoftheabove.$