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Find the angle between unit vectors \overrightarrow{a} and\overrightarrow{b} so that \sqrt{3}\overrightarrow{a}-\overrightarrow{b} is also a unit vector

 

 

Answers (1)

Given:

|\overrightarrow{a}|=1

|\overrightarrow{b}|=1

|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}|=1

Assume angle between \overrightarrow{a} and \overrightarrow{b} is  \theta 

\\ |\sqrt{3} \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=1\\\\ (\sqrt{3} \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})\cdot(\sqrt{3} \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) = 1 \\\\ 3 |\overrightarrow{\mathrm{a}}|^2 +| \overrightarrow{\mathrm{b}}|^2 -2\sqrt{3}\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} =1 \\\\ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{\sqrt{3}}{2}\\\\ |\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta =\frac{\sqrt{3}}{2} \\\\ \cos \theta =\frac{\sqrt{3}}{2} \\\\ \theta = \frac{\pi}{6}

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Safeer PP

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