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Q 26.

Expert-verifiedFound in: Page 812

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

In Exercises 24-27, find ${\mathrm{comp}}_{\mathbf{u}}\mathbf{v}\mathbf{},{\mathrm{proj}}_{\mathbf{u}}\mathbf{v}$ and the component of **v** orthogonal to** u**.

$\mathit{u}=>">2,0,-5$

The values are ${\mathrm{comp}}_{\mathbf{u}}\mathbf{v}=\frac{-1}{\sqrt{29}},{\mathrm{proj}}_{\mathbf{u}}\mathbf{}\mathbf{v}=\frac{-1}{29}>">2,0,-5$ and the component of **v** orthogonal to** u** is $>">\frac{-85}{29},7,\frac{-3}{29}$.

The given vectors are:

$\mathit{u}=>">2,0,-5$

${\mathrm{comp}}_{\mathbf{u}}\mathbf{v}=\frac{\mathit{u}\mathbf{\xb7}\mathit{v}}{||\mathit{u}||}\phantom{\rule{0ex}{0ex}}\mathit{u}=>">2,0,-5$

Then,

${\mathrm{comp}}_{\mathbf{u}}\mathbf{v}=\frac{\mathit{u}\mathbf{\xb7}\mathit{v}}{||\mathit{u}||}\phantom{\rule{0ex}{0ex}}=\frac{-1}{\sqrt{29}}$

Therefore, localid="1649675490199" ${\mathrm{comp}}_{\mathbf{u}}\mathbf{v}=\frac{-1}{\sqrt{29}}$.

${\mathrm{proj}}_{\mathbf{u}}\mathbf{}\mathbf{v}=\frac{\mathit{u}\mathbf{\xb7}\mathit{v}}{{||\mathit{u}||}^{2}}\mathit{u}\phantom{\rule{0ex}{0ex}}=\frac{-1}{{\left(\sqrt{29}\right)}^{2}}>">2,0,-5$

Therefore, the vector projection of **v** onto **u **is $\frac{-1}{29}>">2,0,-5$.

The component of **v** orthogonal to** u** is** $\mathbf{v}-{\mathrm{proj}}_{\mathbf{u}}\mathbf{v}$.**

$\mathbf{v}-{\mathrm{proj}}_{\mathbf{u}}\mathbf{v}=>">-3,7,-1$

Therefore, the component of v orthogonal to u is $>">\frac{-85}{29},7,\frac{-3}{29}$.

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