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Expert-verified Found in: Page 897 ### Calculus

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 $pro{j}_{\mathbit{u}}\mathbit{v}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{u}\mathbf{.}\mathbf{v}}{|\left|u\right||}\mathbf{,}\mathbf{}whereu\ne 0.$

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

See the step by step solution

## Step 1. Given Information.

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

## Step 2. Dot product.

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

## Step 3. Proof

$Letsupposewehaveavector\mathbit{v}andanothervector\mathbit{u}andthesevectors\phantom{\rule{0ex}{0ex}}makeanangleof\theta betweenthem,\phantom{\rule{0ex}{0ex}}Now,thewetakethecomponentofthe\mathbit{v}alongthe\mathbit{u}weget\phantom{\rule{0ex}{0ex}}com{p}_{\mathbf{u}}\mathbit{v}=\left|\mathbit{v}\right|\mathrm{cos}\theta ,\phantom{\rule{0ex}{0ex}}Nowtheprojectionisthesameasthecomponentwecanwrite,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathbit{v}=com{p}_{\mathbf{u}}\mathbit{v}=\left|\mathbit{v}\right|\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}Nowmultiplydividebythemagnitudeof\mathbit{u}vectorinthenumeratorand\phantom{\rule{0ex}{0ex}}denominatorintheRHS,weget,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathbit{v}=\frac{\left|\mathbit{v}\right|\left|\mathbit{u}\right|\mathrm{cos}\theta }{\left|\mathbit{u}\right|},\phantom{\rule{0ex}{0ex}}Nowfromdotproductweget,\phantom{\rule{0ex}{0ex}}pro{j}_{\mathbf{u}}\mathbit{v}=\frac{\mathbit{u}\mathbf{.}\mathbit{v}}{\left|\left|\mathbit{u}\right|\right|}.$ ### Want to see more solutions like these? 