Question

To find the values of \(x\).

Short answer

The values of \(x\) is \(v\)

Step-by-step solution

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\(v\).

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 \({b_3}\).

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

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 \({b_3}\).

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

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.

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 \(1 - \frac{{\sqrt 6 }}{2}\).