Question

Determinar um vetor de módulo 5 paralelo ao vetor v=(1,-1,2)

Answer 1

$\vec{u}=k\vec{v}\\ \\ \vec{u}=k\left(1,-1,2\right)\\ \\ \vec{u}=(k\cdot 1,\ k\cdot \left(-1\right),\ k\cdot 2)\ \Rightarrow \vec{u}=(k,\ -k,\ 2k)\\ \\ \\ |\vec{u}|=\sqrt{k^{2}+\left(-k\right)^{2}+\left(2k\right)^{2}}=5\\ \\ \\ \sqrt{k^{2}+k^{2}+4k^{2}}=5\\ \\ \\ \sqrt{6k^{2}}=5\\ \\ \\ \sqrt{6}\cdot \sqrt{k^{2}}=5\\ \\ \\ \sqrt{6}\cdot \left(k^{2}\right)^{\frac{1}{2}}=5\\ \\ \\ \sqrt{6}\cdot \left(k\right)^{\not 2\cdot \frac{1}{\not 2}}=5$


$\sqrt{6}\cdot k=5\\ \\ \\ k=\dfrac{5}{\sqrt{6}}\\ \\ \\ \vec{u}=k\left(1,-1,2\right)\\ \\ \\ \vec{u}=\dfrac{5}{\sqrt{6}}\left(1,-1,2\right)\\ \\ \\ \vec{u}=\left(\dfrac{5}{\sqrt{6}},\ -\dfrac{5}{\sqrt{6}},\ \dfrac{10}{\sqrt{6}}\right)$