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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 the values of $$x$$.

The values of $$x$$ is $$v$$

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

The given vectors are,

$${\rm{a}} = \langle 2,1, - 1\rangle ,{\rm{b}} = \langle 1,x,0\rangle$$ and $$\theta = {45^\circ }$$

Consider a general expression to find the dot product between two three-dimensional vectors.

\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= \left\langle {{a_1},{a_2},{a_2}} \right\rangle \cdot \left\langle {{b_1},{b_2},{b_2}} \right\rangle \\{\rm{a}} \cdot {\rm{b}} &= {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\end{aligned}

In the above equation, substitute $$2$$ for $${a_1},1$$ for $${a_2}, - 1$$ for $${a_3},1$$ for $${b_1},x$$ for $${b_2}$$ and $$0$$ for $${b_3}$$.

\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (2)(1) + (1)(x) + ( - 1)(0)\\{\rm{a}} \cdot {\rm{b}} &= 2 + x + 0\\{\rm{a}} \cdot {\rm{b}} &= 2 + x\end{aligned}

## Step 3: Calculation of the magnitude of the vector$$a$$and$$b$$

Consider a general expression to find the magnitude of a three-dimensional vector that is$$a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle$$.

$$|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2}$$

In the above equation, substitute $$2$$ for $${a_1},1$$ for $${a_2}$$ and $$- 1$$ for $${a_3}$$.

\begin{aligned}{l}|a| &= \sqrt {{{(2)}^2} + {{(1)}^2} + {{( - 1)}^2}} \\|a| &= \sqrt {4 + 1 + 1} \\|a| &= \sqrt 6 \end{aligned}

Similarly, consider a general expression to find the magnitude of a three-dimensional vector that is

$${\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle$$

$$|{\rm{b}}| = \sqrt {b_1^2 + b_2^2 + b_3^2}$$

In the above equation, substitute $$1$$ for $${b_1},x$$ for $${b_2}$$ and $$0$$ for $${b_3}$$.

\begin{aligned}{l}|{\rm{b}}| &= \sqrt {{{(1)}^2} + {{(x)}^2} + {{(0)}^2}} \\|{\rm{b}}| &= \sqrt {1 + {x^2}} \end{aligned}

## Step 4: Calculation for the value of$$x$$

In the formula, substitute $$2 + x$$ for $$a \cdot b,\sqrt 6$$ for $$|{\rm{a}}|,\sqrt {1 + {x^2}}$$ for $$|{\rm{b}}|$$ and $${45^\circ }$$ for $$\theta$$.

\begin{aligned}{l}2 + x &= (\sqrt 6 )\left( {\sqrt {1 + {x^2}} } \right)\cos \left( {{{45}^\circ }} \right)\\2 + x &= (\sqrt 6 )\left( {\sqrt {1 + {x^2}} } \right)\left( {\frac{1}{{\sqrt 2 }}} \right)\\2 + x &= (\sqrt 3 )\left( {\sqrt {1 + {x^2}} } \right)\end{aligned}

Take square on both sides.

## Step 5: Simplification of the above equation for$$x$$

Solve this second order equation and find the values of $$x$$ as follows.

\begin{aligned}{l}x &= \frac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(2)( - 1)} }}{{2(2)}}\\x &= \frac{{4 \pm \sqrt {16 + 8} }}{4}\\x &= \frac{{4 \pm \sqrt {24} }}{4}\\x &= \frac{{4 \pm 2\sqrt 6 }}{4}\end{aligned}

Expand the equation.

\begin{aligned}{l}x &= \frac{4}{4} \pm \frac{{2\sqrt 6 }}{4}\\x &= 1 \pm \frac{{\sqrt 6 }}{2}\\x &= 1 + \frac{{\sqrt 6 }}{2},1 - \frac{{\sqrt 6 }}{2}\end{aligned}

Thus, values of $$x$$ are $$1 + \frac{{\sqrt 6 }}{2}$$ and $$1 - \frac{{\sqrt 6 }}{2}$$.