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Question
If \[\vec{a} \text{ and } \vec{b}\] are unit vectors, then the greatest value of \[\sqrt{3}\left| \vec{a} + \vec{b} \right| + \left| \vec{a} - \vec{b} \right|\]
Options
(a) 2
(b) \[2\sqrt{2}\]
(c) 4
(d) None of these
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Solution
(c) 4
\[\text{ We have }\]
\[\sqrt{3}\left| \vec{a} + \vec{b} \right| + \left| \vec{a} - \vec{b} \right|\]
\[ = \sqrt{3} \times \sqrt{\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2\left| \vec{a} \right|\left| \vec{b} \right|\cos\theta} + \sqrt{\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 - 2\left| \vec{a} \right|\left| \vec{b} \right|\cos\theta}\]
\[ = \sqrt{3} \times \sqrt{1^2 + 1^2 + 2 \times 1 \times 1\cos\theta} + \sqrt{1^2 + 1^2 - 2 \times 1 \times 1 \cos\theta} \left( As \vec{a} and \vec{b} unit vectors \right)\]
\[ = \sqrt{3} \times \sqrt{2 + 2 \cos\theta} + \sqrt{2 - 2 \cos\theta}\]
\[ = \sqrt{3} \times \sqrt{2\left( 1 + \cos\theta \right)} + \sqrt{2\left( 1 - \cos\theta \right)}\]
\[ = \sqrt{3} \times \sqrt{2 \times 2 \cos^2 \frac{\theta}{2}} + \sqrt{2 \times 2 \sin^2 \frac{\theta}{2}}\]
\[ = 2\sqrt{3} \cos\frac{\theta}{2} + 2\sin \frac{\theta}{2}\]
\[ = 2\left( \sqrt{3} \cos\frac{\theta}{2} + \sin \frac{\theta}{2} \right)\]
\[ = 2 \times 2\left( \frac{\sqrt{3}}{2} \cos\frac{\theta}{2} + \frac{1}{2}\sin \frac{\theta}{2} \right)\]
\[ = 2 \times 2\left( \sin\frac{\pi}{3} \cos\frac{\theta}{2} + \cos\frac{\pi}{3}\sin \frac{\theta}{2} \right)\]
\[ = 4 \sin\left( \frac{\pi}{3} + \frac{\theta}{2} \right)\]
\[\text{ Now, maximum value of } \sin\alpha = 1\]
\[ \Rightarrow \text{ Maximum value of } \sin\left( \frac{\pi}{3} + \frac{\theta}{2} \right) = 1\]
\[ \Rightarrow \text{ Maximum value of 4 }\sin\left( \frac{\pi}{3} + \frac{\theta}{2} \right) = 4\]
\[ \therefore \text{ Maximum value of } \sqrt{3}\left| \vec{a} + \vec{b} \right| + \left| \vec{a} - \vec{b} \right| = 4\]
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