# if vector r1= 4m along north east and r2 vector = 2m due north, find vector mgnitude r1 + r2 and r1 - r2.

r1 is along north east and r2 is along north, angle between r1 and r2 is 45 degree

$\left|r1+r2\right|=\sqrt{{\left|r1\right|}^{2}+{\left|r2\right|}^{2}+2\left|r1\right|\left|r2\right|\mathrm{cos}\left(\theta \right)}\left(\theta istheanglebetweenr1andr2\right)\phantom{\rule{0ex}{0ex}}\left|r1+r2\right|=\sqrt{{\left(4m\right)}^{2}+{\left(2m\right)}^{2}+2\left(4m\right)\left(2m\right)\mathrm{cos}\left(45°\right)}\phantom{\rule{0ex}{0ex}}\left|r1+r2\right|=\sqrt{20{m}^{2}+16{m}^{2}\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{20+8\sqrt{2}}\phantom{\rule{0ex}{0ex}}\left|r1+r2\right|=2\sqrt{5+2\sqrt{2}}m\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left|r1-r2\right|=\sqrt{{\left|r1\right|}^{2}+{\left|r2\right|}^{2}-2\left|r1\right|\left|r2\right|\mathrm{cos}\left(\theta \right)}\phantom{\rule{0ex}{0ex}}\left|r1-r2\right|=\sqrt{{\left(4m\right)}^{2}+{\left(2m\right)}^{2}-2\left(4m\right)\left(2m\right)\mathrm{cos}\left(45°\right)}\phantom{\rule{0ex}{0ex}}\left|r1-r2\right|=\sqrt{20{m}^{2}-16{m}^{2}\left(\frac{1}{\sqrt{2}}\right)}=\sqrt{20-8\sqrt{2}}\phantom{\rule{0ex}{0ex}}\left|r1-r2\right|=2\sqrt{5-2\sqrt{2}}m$

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