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Position Vector and Direction
Concepts tested here
- Slope of a straight line path
All Questions
2007 AIPMT-PRE 1 question
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A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are $(\sqrt{3}, 3)$. The path of the particle makes with the x-axis an angle ofLet $\theta$ be the angle which the path of the particle makes with the x-axis.
The particle moves along the straight line from (0, 0) to $(\sqrt{3}, 3)$.
$\tan\theta = \dfrac{y}{x} = \dfrac{3}{\sqrt{3}}$
$\tan\theta = \sqrt{3}$
$\theta = \tan^{-1}(\sqrt{3}) = 60^\circ$
