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Question Number 39172 by Rio Mike last updated on 03/Jul/18

$$findtheanglebetween$$$$3i-4jandi-j$$

Commented by math khazana by abdo last updated on 03/Jul/18

$$let\overrightarrow{u}=3i-4j\Rightarrow u(3,-4)$$$$\overrightarrow{v}=i-j?\Rightarrow \overrightarrow{v}(1,-1)$$$$cos(\overrightarrow{u},\overrightarrow{v})=\frac{\overrightarrow{u}.\overrightarrow{v}}{\mid \mid \overrightarrow{u}\mid \mid .\mid \mid \overrightarrow{v}\mid \mid }=\frac{3\times 1)+\left(4\right)}{5\sqrt{2}}=\frac{7}{5\sqrt{2}}$$$$sin(\overrightarrow{u},\overrightarrow{v})=\frac{det(u,v)}{\mid \mid u\mid \mid .\mid \mid v\mid \mid }=\frac{\left|\begin{array}{c}31\\ -4-1\end{array}\right|}{5\sqrt{2}}=\frac{1}{5\sqrt{2}}\Rightarrow$$$$tan(u,v)=\frac{1}{7}\Rightarrow \theta =arctan\left(\frac{1}{7}\right).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

$$vector3i-4jmakeangle\alpha withxaxis$$$$so{m}_1=tan\alpha =\frac{-4}{3}$$$$m_2=tan\beta =\frac{-1}{1}$$$$tan\theta =\frac{m_1\sim m_2}{1+m_1{m}_2}=\frac{-1+\frac{4}{3}}{1+\frac{4}{3}}=\frac{\frac{1}{3}}{\frac{7}{3}}=\frac{1}{7}$$$$\theta ={tan}^{-1}\left(\frac{1}{7}\right)$$$$cos\theta =\frac{7}{\sqrt{50}}$$$$$$$$anotherapproach...$$$$cos\theta =\frac{3\times 1+(-4)\times (-1)}{\sqrt{3^2+{(-4)}^2}\times \sqrt{1^2+{(-1)}^2}}$$$$=\frac{7}{\sqrt{50}}$$

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