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Q12E

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Found in: Page 556

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# To find a dot product $$u \cdot v$$ and $$u \cdot w$$.

The dot product of$${\rm{u}} \cdot {\rm{v}}$$ is $$0.5$$ and $${\rm{u}} \cdot {\rm{w}}$$ is $$0$$.

See the step by step solution

## Step 1: Concept of Dot Product

Formula Used:

Write the expression to find $${\rm{u}} \cdot {\rm{v}}$$ in terms of $$\theta$$.

$${\rm{u}} \cdot {\rm{v}} = |{\rm{u}}||{\rm{v}}|\cos \theta$$

Here,

$$|u|$$is the magnitude of vector $$u$$,

$$|{\rm{v}}|$$ is the magnitude of vector $$v$$, and

$$\theta$$ is the angle between vectors $$u$$ and $$v$$.

## Step 2: Calculation of the dot product of$${\rm{u}} \cdot {\rm{v}}$$

Consider the right angle triangle formed by $${\bf{u}}$$ and $$v$$.

Write the expression to find magnitude of $$v(|v|)$$.

$$|v| = |u|\cos \theta$$

As $${\bf{u}}$$, and $${\bf{w}}$$ are the unit vectors, the value of $$|u|$$and $$|{\bf{w}}|$$ is$$1$$.

The angle between $${\bf{u}}$$ and $$v$$ is $${45^\circ }$$.

In equation of$$|v|$$, substitute$$1$$ for $$|{\rm{u}}|$$ and $${45^\circ }$$ for $$\theta$$.

\begin{aligned}{l}|v| &= (1)\cos \left( {{{45}^\circ }} \right)\\|v| &= (1)(0.707)\\|v| &= 0.707\end{aligned}

## Step 3: Calculation of the dot product of$${\rm{u}} \cdot {\rm{v}}$$

In equation of$${\rm{u}} \cdot {\rm{v}}$$, substitute$$1$$for $$|{\rm{u}}|,0.707$$ for $$|{\rm{v}}|$$ and $${45^\circ }$$ for $$\theta$$.

\begin{aligned}{l}{\rm{u}} \cdot {\rm{v}} &= (1)(0.707)\cos \left( {{{45}^\circ }} \right)\\{\rm{u}} \cdot {\rm{v}} &= (0.707)(0.707)\\{\rm{u}} \cdot {\rm{v}} &= 0.49\\{\rm{u}} \cdot {\rm{v}} \cong 0.5\end{aligned}

Thus, $${\rm{u}} \cdot {\rm{v}}$$ is $$0.5.$$

## Step 4: Calculation of the dot product$${\rm{u}} \cdot w$$

Rewrite equation of$${\rm{u}} \cdot {\rm{v}}$$.

$${\rm{u}} \cdot {\rm{w}} = |{\rm{u}}||{\rm{w}}|\cos \theta$$

Substitute $$1$$ for $$|{\rm{u}}|,1$$ for $$|{\rm{w}}|$$ and $${90^\circ }$$ for $$\theta$$.

\begin{aligned}{l}{\rm{u}} \cdot {\rm{w}} &= (1)(1)\cos \left( {{{90}^\circ }} \right)\\{\rm{u}} \cdot {\rm{w}} &= (1)(0)\\{\rm{u}} \cdot {\rm{w}} &= 0\end{aligned}

Thus, $${\rm{u}} \cdot {\rm{w}}$$ is$$0$$ .