[Math] geometric mean and vectors

Question

Let $\bar{u}$ be a vector on rectangular coordinate system with sloping angle
$60^\circ$. Suppose that $\lvert \bar{u}-\bar{i} \rvert$ is the geometric mean of $\lvert \bar{u} \rvert$ and $\lvert \bar{u}-2\bar{i} \rvert$ where
$\bar{i}$
is the unit vector along the $x$-axis then find the value of
$\lvert \bar{u}\rvert$.

I encountered this problem in a vector worksheet, but I have no idea on how to use vectors to solve this. Please help

Answer 1

You got the geometric mean equation $$ \lvert \bar{u}-\bar{i} \rvert = \sqrt{\lvert \bar{u} \rvert \lvert \bar{u}-2\bar{i} \rvert} \iff \\ \lvert \bar{u}-\bar{i} \rvert^4 = ((\bar{u}-\bar{i})\cdot (\bar{u}-\bar{i}))^2 = \lvert\bar{u} \rvert^2 \lvert \bar{u}-2\bar{i} \rvert^2 = (\bar{u}\cdot \bar{u}) ((\bar{u}-2\bar{i})\cdot (\bar{u}-2\bar{i})) \iff \\ (\bar{u}^2 + \bar{i}^2 - 2 \bar{u}\cdot \bar{i})^2 = \bar{u}^2 (\bar{u}^2 + 4\bar{i}^2 -4 \bar{u}\cdot \bar{i}) \iff \\ (\bar{u}^2 + 1 - 2 \bar{u}\cdot \bar{i})^2 = \bar{u}^2 (\bar{u}^2 + 4 -4 \bar{u}\cdot \bar{i}) \iff \\ (\bar{u}^2 + 1)^2 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}^2 + 1)(\bar{u}\cdot \bar{i}) = (\bar{u}^2)^2 + 4\bar{u}^2 -4 \bar{u}^2 (\bar{u}\cdot \bar{i}) \iff \\ (\bar{u}^2)^2 + 1 + 2 \bar{u}^2 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}^2 + 1)(\bar{u}\cdot \bar{i}) = (\bar{u}^2)^2 + 4\bar{u}^2 -4 \bar{u}^2 (\bar{u}\cdot \bar{i}) \iff \\ 1 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}^2 + 1)(\bar{u}\cdot \bar{i}) = 2\bar{u}^2 -4 \bar{u}^2 (\bar{u}\cdot \bar{i}) \iff \\ 1 + 4 (\bar{u}\cdot \bar{i})^2 + 4 (\bar{u}^2-(\bar{u}^2 + 1))(\bar{u}\cdot \bar{i}) = 2\bar{u}^2 \iff \\ 1 + 4 (\bar{u}\cdot \bar{i})^2 - 4 (\bar{u}\cdot \bar{i}) = 2\bar{u}^2 $$ with $$ \bar{u}\cdot \bar{i} = \lvert \bar{u} \rvert \cos 60^\circ = (1/2) \lvert \bar{u} \rvert $$ we end up with the equation $$ 1 + \lvert \bar{u} \rvert^2 - 2 \lvert \bar{u} \rvert = 2\lvert \bar{u} \rvert^2 \iff \\ \lvert \bar{u} \rvert^2 + 2 \lvert \bar{u} \rvert - 1 = 0 \iff \\ (\lvert \bar{u} \rvert + 1)^2 = 2 \iff \\ \lvert \bar{u} \rvert = \sqrt{2} - 1 $$ I talked with my friend Ruby to check the result:

irb(main):007:0> ul = Math.sqrt(2) - 1
=> 0.41421356237309515
irb(main):008:0> ux = ul / 2
=> 0.20710678118654757
irb(main):009:0> uy = Math.sin(60*Math::PI/180.0) * ul
=> 0.35871946760715046
irb(main):015:0> vx = ux - 1
=> -0.7928932188134524
irb(main):016:0> vy = uy
=> 0.35871946760715046
irb(main):017:0> vl = Math.sqrt(vx**2+vy**2)
=> 0.8702639328851419
irb(main):019:0> wx = ux - 2
=> -1.7928932188134525
irb(main):020:0> wy = uy
=> 0.35871946760715046
irb(main):021:0> wl = Math.sqrt(wx**2 + wy**2)
=> 1.8284271247461903
irb(main):022:0> Math.sqrt(ul*wl)
=> 0.8702639328851421
irb(main):023:0> vl
=> 0.8702639328851419