# Find the values of x such that the angle between

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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Hector Roberts
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
$$\displaystyle{x}={\frac{{-{\left(-{4}\right)}\pm\sqrt{{{\left(-{4}\right)}^{{{2}}}-{4}{\left({2}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({2}\right)}}}}={\frac{{{4}\pm\sqrt{{{24}}}}}{{{2}{\left({2}\right)}}}}={\frac{{{2}\pm\sqrt{{{6}}}}}{{{2}}}}$$
###### Not exactly what youâ€™re looking for?
sukljama2
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",
$$\displaystyle{2}{x}^{{{2}}}-{4}{x}-{1}={0}$$
$$\displaystyle{x}={\frac{{-{d}\pm\sqrt{{{d}^{{{2}}}-{4}{a}{c}}}}}{{{2}{a}}}}$$
$$\displaystyle{x}={\frac{{-{\left(-{4}\right)}\pm\sqrt{{{\left(-{4}\right)}^{{{2}}}-{4}\cdot{2}{\left(-{1}\right)}}}}}{{{2}\cdot{2}}}}$$
$$\displaystyle{x}={\frac{{{2}+\sqrt{{{6}}}}}{{{2}}}},{\frac{{{2}-\sqrt{{{6}}}}}{{{2}}}}$$