Trying to prove: Show that there are no vectors u and v such that ∥u∥=1, ∥v∥=2, and u*v=3.

Melina Barber 2022-09-19 Answered
Trying to prove:
Show that there are no vectors u and v such that $‖u‖=1$, $‖v‖=2$, and $u\cdot v=3$
Don't know where to go from here:
$‖u‖=2⟹‖u{‖}^{2}=4⟹u\cdot u=1$
$‖v‖=1⟹‖v{‖}^{2}=1⟹v\cdot v=4$
Not sure if this is the right direction to take, but we have:
$\begin{array}{r}u\cdot v=v\cdot v-u\cdot u=3\\ =\left(v+u\right)\cdot \left(v-u\right)\end{array}$
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Santiago Collier
$u\cdot v=‖u‖‖v‖\mathrm{cos}\theta$
So
$3=\left(1\right)\left(2\right)\mathrm{cos}\theta$
which implies
$\mathrm{cos}\theta =\frac{3}{2}$
which is not possible.

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easternerjx
Another way to realize the answer:
$\begin{array}{rl}‖u+v{‖}^{2}& =⟨u+v,u+v⟩\\ \\ & =⟨u,u⟩+2⟨u,v⟩+⟨v,v⟩\\ \\ & =‖u{‖}^{2}+2⟨u,v⟩+‖v{‖}^{2}\\ \\ & ={1}^{2}+2×3+{2}^{2}\\ \\ & =1+6+4\\ \\ & >\left(1+2{\right)}^{2}\\ \\ & =\left(‖u‖+‖v‖{\right)}^{2}\end{array}$
which violates the triangle inequality.

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