Question

Let $\overrightarrow{u}$ be a vector on rectangular system with sloping angle ${60}^{\circ }$. Suppose that $|\overrightarrow{u}-\hat{i}|$ is geometric mean of $\left|\overrightarrow{u}\right|$ and $|\overrightarrow{u}-2\hat{i}|$, where $\overrightarrow{i}$ is the unit vector along $x$-axis, then $\left|\overrightarrow{u}\right|$ has the value equal to $\sqrt{a}-\sqrt{b}$, where $a,b$ $\in$ $N$. Find the value $\frac{{(a+b)}^3+{(a-b)}^3}{7}$

Answer 1

Let $\left|\overrightarrow{u}\right|=\lambda \overrightarrow{u}=\frac{\lambda }{2}(\hat{i}+\sqrt{3}\hat{j})$

Given, $|\overrightarrow{u}-\overrightarrow{i}|=\sqrt{|\overrightarrow{u}|\cdot |\overrightarrow{u}-2\overrightarrow{i}|}$

$\Rightarrow {|\frac{\lambda }{2}(\hat{i}+\sqrt{3}\hat{j})-\hat{i}|}^2=\lambda |\frac{\lambda }{2}(\hat{i}+\sqrt{3}\hat{j})-2\hat{i}|$

$\Rightarrow {({(\frac{\lambda }{2}-1)}^2+\frac{3{\lambda }^2}{4})}^2$ $={\lambda }^2({\left(\frac{\lambda -4}{2}\right)}^2+\frac{3{\lambda }^2}{4})$

$\Rightarrow {(4{\lambda }^2-4\lambda +4)}^2$$=16{\lambda }^2({\lambda }^2-2\lambda +4)$

$\Rightarrow {({\lambda }^2-\lambda +1)}^2={\lambda }^2({\lambda }^2-2\lambda +4)$

On solving, we get

$\lambda =\frac{-2\pm \sqrt{4+4}}{2}=-1\pm \sqrt{2}$
But $\lambda >0\Rightarrow \lambda =\sqrt{2}-1$

$\Rightarrow a=2,b=1$

$\Rightarrow {(a+b)}^3+{(a-b)}^3=3^3+1^3=27+1=28$

$\therefore \frac{(a+b{)}^3+(a-b{)}^3}{7}=4$