Question

If$\stackrel{\rightharpoonup }{a}$ and $\stackrel{\rightharpoonup }{b}$are unit vectors, then the greatest value of$\sqrt{3}\left|\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right|+\left|\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}\right|$ is

Answer 1

Finding the greatest value of$\sqrt{3}\left|\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right|+\left|\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}\right|$:

we know,

$\begin{array}{rcl}& & \left|\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right|\end{array}=\sqrt{\left|a\right|+\left|b\right|+2ab\cos \theta },\begin{array}{rcl}& & \left|\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}\right|\end{array}=\sqrt{\left|a\right|+\left|b\right|-2ab\cos \theta }$

Then,

$\begin{array}{rcl}\sqrt{3}\left|\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}\right|+\left|\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}\right|& =& \sqrt{3}\left(\sqrt{(2+2\cos \theta )}\right)+\sqrt{(2-2\cos \theta )}\\ & =& \sqrt{6}\sqrt{(1+\cos \theta })+\sqrt{2}(1-\cos \theta )\\ & =& \sqrt{6}\sqrt{\left(1+2{\cos }^2\left(\frac{\theta }{2}\right)-1\right)}+\sqrt{2}\sqrt{\left(1-1+2{\sin }^2\frac{\theta }{2}\right)}\\ & =& \sqrt{6}\sqrt{\left(2{\cos }^2\frac{\theta }{2}\right)}+\sqrt{2}\left(\sqrt{2{\sin }^2\frac{\theta }{2}}\right)\end{array}$

The greatest value of a trigonometric function $a\cos \theta +b\sin \theta \le \surd (a^2+b^2)$

$$\begin{gathered} 2\sqrt{3}\cos \left(\frac{\theta }{2}\right)+2\sin \left(\frac{\theta }{2}\right)\le \sqrt{{\left(2\sqrt{3}\right)}^2+{\left(2\right)}^2} \\ =4 \end{gathered}$$

Hence, the correct answer is $4$.