Test Index

Vector Algebra

Section: Physics

Question : 32 of 47

Marks: +1, -0

Solution:

In the given figure, the magnitude of vectors OA, OB and OC areequal. Let it be m.

From figure, we can write

OA = m(\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j})

= m\left(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}\right)

OB = m(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})

= m\left(\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}\right)

OC = m\left(\cos 45^{\circ}(-\hat{i}) + \sin 45^{\circ} \hat{j}\right)

= m\left\{-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}\right\}

According to the question,

OA + OB - OC = m\left(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}\right) + m\left(\frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j}\right)

-m\left(-\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}\right)

= m \hat{i} \left(\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}\right) + m \hat{j} \left(\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\right)

∴ Angle with X - axis is given by

\tan \theta = \frac{y}{x} = \frac{m\left(\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\right)}{m\left(\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}}\right)}

\Rightarrow \tan \theta = \frac{\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{\sqrt{2}}{2}}

\Rightarrow \tan \theta = \frac{1 - \sqrt{3} - \sqrt{2}}{\sqrt{3} + 1 + \sqrt{2}}

\Rightarrow \theta = \tan^{-1}\left(\frac{1 - \sqrt{3} - \sqrt{2}}{1 + \sqrt{3} + \sqrt{2}}\right)

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