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Q. 1

Expert-verifiedFound in: Page 897

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Projecting one vector onto another: Show that the formula for the projection of a vector **v** onto a nonzero vector **u** is given by $pro{j}_{\mathit{u}}\mathit{v}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{u}\mathbf{.}\mathbf{v}}{\left|\left|u\right|\right|}\mathbf{,}\mathbf{}whereu\ne 0.$

$Dotproduct:\phantom{\rule{0ex}{0ex}}\mathit{u}\mathbf{.}\mathit{v}=\left|\mathit{u}\right|\left|\mathit{v}\right|\mathrm{cos}\theta ,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathit{v}=\left|\mathit{v}\right|\mathrm{cos}\theta ,\phantom{\rule{0ex}{0ex}}Solvingabovetwoequationsweget,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathit{v}=\frac{\mathit{u}\mathbf{.}\mathit{v}}{\left|\left|\mathit{u}\right|\right|}.$

Given two vectors **u **and **v **where u is a nonzero vector.

$\mathit{u}\mathbf{.}\mathit{v}=\left|u\right|.\left|v\right|\mathrm{cos}\theta $

$Letsupposewehaveavector\mathit{v}andanothervector\mathit{u}andthesevectors\phantom{\rule{0ex}{0ex}}makeanangleof\theta betweenthem,\phantom{\rule{0ex}{0ex}}Now,thewetakethecomponentofthe\mathit{v}alongthe\mathit{u}weget\phantom{\rule{0ex}{0ex}}com{p}_{\mathbf{u}}\mathit{v}=\left|\mathit{v}\right|\mathrm{cos}\theta ,\phantom{\rule{0ex}{0ex}}Nowtheprojectionisthesameasthecomponentwecanwrite,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathit{v}=com{p}_{\mathbf{u}}\mathit{v}=\left|\mathit{v}\right|\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}Nowmultiplydividebythemagnitudeof\mathit{u}vectorinthenumeratorand\phantom{\rule{0ex}{0ex}}denominatorintheRHS,weget,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathit{v}=\frac{\left|\mathit{v}\right|\left|\mathit{u}\right|\mathrm{cos}\theta}{\left|\mathit{u}\right|},\phantom{\rule{0ex}{0ex}}Nowfromdotproductweget,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathit{v}=\frac{\mathit{u}\mathbf{.}\mathit{v}}{\left|\left|\mathit{u}\right|\right|}.$

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