Question

two vectors of magnitudes A and √3A are perpendicular to each other . what is the angle which their resultant makes with A​

Answer 1

Answer:

$\tan( \alpha ) = \frac{ \sqrt{3}a }{a} = \sqrt{3} \\ \\ \alpha = \tan {}^{ - 1} ( \sqrt{3} ) = \frac{\pi}{6}$

Answer 2

Solution :

✴ Given :-

✈ Two vectors of magnitude A and √3A are perpendicular to each other.

✴ To Find :-

✈ Angle between resultant vector and A

✴ Assumption :-

✈ Let, angle between resultant vector and A be β.

✴ Calculation :-

$\implies\sf\:\tan\beta=\dfrac{\sqrt{3}A\sin90\degree}{A+\sqrt{3}A\cos90\degree}\\ \\ \circ\sf\:\sin90\degree=1\:and\:\cos90\degree=0\\ \\ \implies\sf\:\tan\beta=\dfrac{\sqrt{3}\cancel{A}}{\cancel{A}}\\ \\ \implies\sf\:\tan\beta=\sqrt{3}\\ \\ \implies\sf\:\beta=\tan^{-1}(\sqrt{3})\\ \\ \implies\:\underline{\boxed{\orange{\large{\tt{\beta=60\degree=\dfrac{\pi}{3}\:rad}}}}}$

✴ Additional information :-

• Vector quantity has both magnitude as well as direction.