$u=\u27e82,-4\u27e9\text{and}v=\u27e85,-3\u27e9$. Find the angle between them to the nearest tenth of a degree

Edward Hendricks
2022-08-12
Answered

$u=\u27e82,-4\u27e9\text{and}v=\u27e85,-3\u27e9$. Find the angle between them to the nearest tenth of a degree

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Melany Flynn

Answered 2022-08-13
Author has **12** answers

Given vectors u=2i-4j, v=5i-3j

Angle between the two given vectrors,

$\mathrm{cos}\theta =\frac{u\cdot v}{|u||v|}\dots (1)\phantom{\rule{0ex}{0ex}}u\cdot v=(2)(5)+(-4)(-3)=10+12=22\phantom{\rule{0ex}{0ex}}|u|=\sqrt{(2{)}^{2}+(-4{)}^{2}}\phantom{\rule{0ex}{0ex}}|u|=\sqrt{4+16}\phantom{\rule{0ex}{0ex}}|u|=\sqrt{20}=2\sqrt{5}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{{5}^{2}+(-3{)}^{2}}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{25+9}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{34}$

Substituting in the equation (1),

$\mathrm{cos}\theta =\frac{u\cdot v}{|u||v|}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{22}{(2\sqrt{5})(\sqrt{34})}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{11}{(4.472)(5.830)}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{11}{26.071}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =0.4219\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}(0.4219)\phantom{\rule{0ex}{0ex}}\theta ={65}^{\circ}$

Angle between the two given vectrors,

$\mathrm{cos}\theta =\frac{u\cdot v}{|u||v|}\dots (1)\phantom{\rule{0ex}{0ex}}u\cdot v=(2)(5)+(-4)(-3)=10+12=22\phantom{\rule{0ex}{0ex}}|u|=\sqrt{(2{)}^{2}+(-4{)}^{2}}\phantom{\rule{0ex}{0ex}}|u|=\sqrt{4+16}\phantom{\rule{0ex}{0ex}}|u|=\sqrt{20}=2\sqrt{5}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{{5}^{2}+(-3{)}^{2}}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{25+9}\phantom{\rule{0ex}{0ex}}|v|=\sqrt{34}$

Substituting in the equation (1),

$\mathrm{cos}\theta =\frac{u\cdot v}{|u||v|}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{22}{(2\sqrt{5})(\sqrt{34})}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{11}{(4.472)(5.830)}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{11}{26.071}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =0.4219\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{cos}}^{-1}(0.4219)\phantom{\rule{0ex}{0ex}}\theta ={65}^{\circ}$

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