Find the values of x such that the angle between

tearstreakdl 2021-12-14 Answered
Find the values of x such that the angle between the vectors (2, 1, -1), and (1, x, 0) is \(\displaystyle{45}^{{\circ}}\).

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Expert Answer

Hector Roberts
Answered 2021-12-15 Author has 3535 answers
Theorem 3 from this section states that:
\(\displaystyle{a}\cdot{b}={m}{i}{d}{a}{p}{a}{r}{a}{l}\le{l}{b}{m}{i}{d}{\cos{\theta}}\)
where \theta is the angle between the vectors a,b
\(\displaystyle{\left\langle{2},{1},-{1}\right\rangle}\cdot{\left\langle{1},{x},{0}\right\rangle}=\sqrt{{{2}^{{{2}}}+{1}^{{{2}}}+{\left(-{1}\right)}^{{{2}}}}}\sqrt{{{1}^{{{2}}}+{x}^{{{2}}}+{0}^{{\lbrace}}}}{{\cos{{45}}}^{{\circ}}}\)
Remember that:
\(\displaystyle{\left\langle{a}_{{{1}}},{b}_{{{1}}},{c}_{{{1}}}\right\rangle}\cdot{\left\langle{a}_{{{2}}},{b}_{{{2}}},{c}_{{{2}}}\right\rangle}={a}_{{{1}}}{a}_{{{2}}}+{b}_{{{1}}}{b}_{{{2}}}+{c}_{{{1}}}{c}_{{{2}}}\)
This is because when it comes to the unit vectors i,j,k, their dot product with themselves is 1 and their dot product with the other two is 0.
\(\displaystyle{\left({2}\right)}{\left({1}\right)}+{\left({1}\right)}{\left({x}\right)}+{\left(-{1}\right)}{\left({0}\right)}=\sqrt{{{4}+{1}+{1}}}\sqrt{{{1}+{x}^{{{2}}}+{0}}}{\frac{{{1}}}{{\sqrt{{{2}}}}}}\)
\(\displaystyle{2}+{x}+{0}=\sqrt{{{6}}}\sqrt{{{1}+{x}^{{{2}}}}}{\frac{{{1}}}{{\sqrt{{{2}}}}}}\)
\(\displaystyle{2}+{x}=\sqrt{{{3}}}\sqrt{{{1}+{x}^{{{2}}}}}\)
Square both sides
\(\displaystyle{\left({2}+{x}\right)}^{{{2}}}={3}{\left({1}+{x}^{{{2}}}\right)}\)
\(\displaystyle{4}+{4}{x}+{x}^{{{2}}}={3}+{3}{x}^{{{2}}}\)
Rewrite the quadratic equation in the standard form
\(\displaystyle{2}{x}^{{{2}}}-{4}{x}-{1}={0}\)
Using the quadratic formula, we can write
\(\displaystyle{x}={\frac{{-{\left(-{4}\right)}\pm\sqrt{{{\left(-{4}\right)}^{{{2}}}-{4}{\left({2}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({2}\right)}}}}={\frac{{{4}\pm\sqrt{{{24}}}}}{{{2}{\left({2}\right)}}}}={\frac{{{2}\pm\sqrt{{{6}}}}}{{{2}}}}\)
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sukljama2
Answered 2021-12-16 Author has 3829 answers
Step 1
Consider the vectors:
\(\displaystyle{v}={\left\langle{2},{1},-{1}\right\rangle}\)
\(\displaystyle{w}={\left\langle{1},{x},{0}\right\rangle}\)
\(\displaystyle\theta={45}^{{\circ}}\)
\(\displaystyle\vec{{{v}}}{2}{i}+{j}-{k}\)
\(\displaystyle\vec{{{w}}}={i}+{x}{j}\)
\(\displaystyle\vec{{{v}}}\cdot\vec{{{w}}}={\left({2}{i}+{j}-{k}\right)}\cdot{\left({i}+{k}\right)}\)
\(\displaystyle={\left({2}\right)}{\left({1}\right)}+{\left({1}\right)}{\left({x}\right)}+{\left(-{1}\right)}{\left({0}\right)}\)
\(\displaystyle={2}+{x}\)
Step 2
Find the magnitude each vector:
\(\displaystyle{m}{i}{d}\vec{{{v}}}{m}{i}{d}=\sqrt{{{2}^{{{2}}}+{1}^{{{2}}}+{\left({1}\right)}^{{{2}}}}}\)
\(\displaystyle=\sqrt{{{4}+{1}+{1}}}\)
\(\displaystyle=\sqrt{{{6}}}\)
\(\displaystyle{m}{i}{d}\vec{{{w}}}{m}{i}{d}=\sqrt{{{1}^{{{2}}}+{\left({x}\right)}^{{{2}}}+{\left({0}\right)}^{{{2}}}}}\)
\(\displaystyle=\sqrt{{{1}+{x}^{{{2}}}+{0}}}\)
\(\displaystyle=\sqrt{{{1}+{x}^{{{2}}}}}\)
Step 3 Find the cosine angle between two vectors:
\(\displaystyle\vec{{{v}}}\cdot\vec{{{w}}}={m}{i}{d}\vec{{{v}}}{p}{a}{r}{a}{l}\le{l}\vec{{{w}}}{m}{i}{d}{\cos{\theta}}\)
\(\displaystyle{2}+{x}={\left(\sqrt{{{6}}}\right)}{\left(\sqrt{{{1}+{x}^{{{2}}}}}\right\rbrace}{{\cos{{45}}}^{{\circ}}}\)
\[2+x=(\sqrt{6})(\sqrt{1+x^{2})}(\begin{array}{c}\frac{1}{\sqrt{2}}\end{array})\]
\[2+x=(\sqrt{6})(\sqrt{1+x^{2})}(\begin{array}{c}\frac{\sqrt{2}}{2}\end{array})\]
\(\displaystyle{2}+{x}={\frac{{\sqrt{{{12}}}}}{{{2}}}}{\left(\sqrt{{{1}+{x}^{{{2}}}}}\right\rbrace}\)
\(\displaystyle{\frac{{\sqrt{{{12}}}}}{{{2}}}}{\left({2}+{x}\right)}=\sqrt{{{1}+{x}^{{{2}}}}}\)
\(\displaystyle{\frac{{{4}}}{{{12}}}}{\left({2}+{x}\right)}^{{{2}}}={1}+{x}^{{{2}}}\)
\(\displaystyle{\frac{{{4}}}{{{12}}}}{\left({4}+{x}^{{{2}}}+{4}{x}\right)}={1}+{x}^{{{2}}}\)
\(\displaystyle{\frac{{{1}}}{{{3}}}}{\left({4}+{x}^{{{2}}}+{4}{x}\right)}={1}+{x}^{{{2}}}\)
\(\displaystyle{4}+{x}^{{{2}}}+{4}{x}={3}+{3}{x}^{{{2}}}\)
\(\displaystyle{4}+{4}{x}={3}+{3}{x}^{{{2}}}\)
\(\displaystyle{0}={3}+{2}{x}^{{{2}}}-{4}{x}-{4}\)
\(\displaystyle{2}{x}^{{{2}}}-{4}{x}-{1}={0}\)
Step 4 Find"x",
\(\displaystyle{2}{x}^{{{2}}}-{4}{x}-{1}={0}\)
\(\displaystyle{x}={\frac{{-{d}\pm\sqrt{{{d}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}\)
\(\displaystyle{x}={\frac{{-{\left(-{4}\right)}\pm\sqrt{{{\left(-{4}\right)}^{{{2}}}-{4}\cdot{2}{\left(-{1}\right)}}}}}{{{2}\cdot{2}}}}\)
\(\displaystyle{x}={\frac{{{2}+\sqrt{{{6}}}}}{{{2}}}},{\frac{{{2}-\sqrt{{{6}}}}}{{{2}}}}\)
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