Question

Let $\overrightarrow{u}$ be a vector on rectangular coordinate system with sloping angle ${60}^{\circ }.$ Suupose that $|\overrightarrow{u}-\hat{i}|$ is geometric mean of $|\overrightarrow{u}|$ and $|\overrightarrow{u}-2\hat{i}|,$ then the value of $2(\sqrt{2}+1)|\overrightarrow{u}|=$

Answer 1

Since angle between $\overrightarrow{u}$ and $\hat{i}$ is ${60}^{\circ }.$
$\Rightarrow \overrightarrow{u}\cdot \hat{i}=|\overrightarrow{u}||\hat{i}|\cos \left({60}^{\circ }\right)=\frac{|\overrightarrow{u}|}{2}\cdots \left(i\right)$
Also $|\overrightarrow{u}-\hat{i}|$ is geometric mean of $|\overrightarrow{u}|$ and $|\overrightarrow{u}-2\hat{i}|$
$\Rightarrow |\overrightarrow{u}-\hat{i}{|}^2=|\overrightarrow{u}|\cdot |\overrightarrow{u}-2\hat{i}|$
Squaring both sides and expanding,
$\Rightarrow {[|\overrightarrow{u}{|}^2+|\hat{i}{|}^2-2\overrightarrow{u}\cdot \hat{i}]}^2=|\overrightarrow{u}{|}^2[|\overrightarrow{u}{|}^2+4|\hat{i}{|}^2-4\overrightarrow{u}\cdot \hat{i}]$
using $\left(i\right),$ we get
$\Rightarrow |\overrightarrow{u}{|}^2+2|\overrightarrow{u}|-1=0\Rightarrow |\overrightarrow{u}|=\frac{-2\pm 2\sqrt{2}}{2}$
So, $|\overrightarrow{u}|=\sqrt{2}-1$