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Q26.

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Found in: Page 646

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

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# The Component of $u$ along $v$Find the component of $u$ along $v$.$\mathbf{u}=⟨-3,5⟩,\mathbf{v}=⟨1/\sqrt{2},1/\sqrt{2}⟩$

The component is $\sqrt{2}$

See the step by step solution

## Step 1. Given information

Vectors are given as-

$\mathbf{u}=⟨-3,5⟩,v=⟨1/\sqrt{2},1/\sqrt{2}⟩$

## Step 2. Concept used

The dot product of two vectors $u$ and $v$ can be defined as-

$u·v={a}_{1}{b}_{1}+{a}_{2}{b}_{2}$

Where $u=\left({a}_{1},{a}_{2}\right)&v=\left({b}_{1},{b}_{2}\right)$

If $u$ and $v$ are two vectors then the component of $u$ along $v$ can be defined as $com{p}_{v}u=\frac{u·v}{|v|}$

Use the formula of components of $u$ along $v$ and the dot product of vectors.

## Step 3. Calculation

Component of $u$ along $v$,

$\begin{array}{l}{\mathrm{comp}}_{v}u=\frac{u\cdot v}{|v|}\\ =\frac{-3\left\{\frac{1}{\sqrt{2}}\right\}+5\left\{\frac{1}{\sqrt{2}}\right\}}{\sqrt{{\left\{\frac{1}{\sqrt{2}}\right\}}^{2}+{\left\{\frac{1}{\sqrt{2}}\right\}}^{2}}}\\ =\frac{\sqrt{2}}{1}\\ =\sqrt{2}\end{array}$

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