Notes: `u = cos(pi/3)i + sin(pi/3)j, v = cos((3pi)/4)i + sin((3pi)/4)j` Find the angle theta between the vectors.

Thursday, 9 February 2017

$\displaystyle{u}={\cos{{\left(\frac{\pi}{{3}}\right)}}}{i}+{\sin{{\left(\frac{\pi}{{3}}\right)}}}{j},{v}={\cos{{\left(\frac{{{3}\pi}}{{4}}\right)}}}{i}+{\sin{{\left(\frac{{{3}\pi}}{{4}}\right)}}}{j}$ Find the angle theta between the vectors.

You need to use the formula of dot product to find the angle between two vectors, $\displaystyle{u}={u}_{{x}}\cdot{i}+{u}_{{y}}\cdot{j},{v}={v}_{{x}}\cdot{i}+{v}_{{y}}\cdot{j},$ such that:

$\displaystyle{u}\cdot{v}={\left|{u}\right|}\cdot{\left|{v}\right|}\cdot{\cos{{\left(\theta\right)}}}$

The angle between the vectors u and v is theta.

$\displaystyle{\cos{\theta}}=\frac{{{u}\cdot{v}}}{{{\left|{u}\right|}\cdot{\left|{v}\right|}}}$

First, you need to evaluate the product of the vectors u and v, such that:

$\displaystyle{u}\cdot{v}={u}_{{x}}\cdot{v}_{{x}}+{u}_{{y}}\cdot{v}_{{y}}$

$\displaystyle{u}\cdot{v}={\cos{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\cos{{\left({3}\frac{\pi}{{4}}\right)}}}+{\sin{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\sin{{\left({3}\frac{\pi}{{4}}\right)}}}$

$\displaystyle{\sin{{\left({3}\frac{\pi}{{4}}\right)}}}={\sin{{\left(\frac{\pi}{{2}}+\frac{\pi}{{4}}\right)}}}=\ldots$

You need to use the formula of dot product to find the angle between two vectors, $\displaystyle{u}={u}_{{x}}\cdot{i}+{u}_{{y}}\cdot{j},{v}={v}_{{x}}\cdot{i}+{v}_{{y}}\cdot{j},$ such that:

$\displaystyle{u}\cdot{v}={\left|{u}\right|}\cdot{\left|{v}\right|}\cdot{\cos{{\left(\theta\right)}}}$

The angle between the vectors u and v is theta.

$\displaystyle{\cos{\theta}}=\frac{{{u}\cdot{v}}}{{{\left|{u}\right|}\cdot{\left|{v}\right|}}}$

First, you need to evaluate the product of the vectors u and v, such that:

$\displaystyle{u}\cdot{v}={u}_{{x}}\cdot{v}_{{x}}+{u}_{{y}}\cdot{v}_{{y}}$

$\displaystyle{u}\cdot{v}={\cos{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\cos{{\left({3}\frac{\pi}{{4}}\right)}}}+{\sin{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\sin{{\left({3}\frac{\pi}{{4}}\right)}}}$

$\displaystyle{\sin{{\left({3}\frac{\pi}{{4}}\right)}}}={\sin{{\left(\frac{\pi}{{2}}+\frac{\pi}{{4}}\right)}}}={\cos{{\left(\frac{\pi}{{4}}\right)}}}=\frac{\sqrt{{2}}}{{2}}$

$\displaystyle{\cos{{\left({3}\frac{\pi}{{4}}\right)}}}={\cos{{\left(\frac{\pi}{{2}}+\frac{\pi}{{4}}\right)}}}=-{\sin{{\left(\frac{\pi}{{4}}\right)}}}=-\frac{\sqrt{{2}}}{{2}}$

$\displaystyle{u}\cdot{v}=-{\cos{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\sin{{\left(\frac{\pi}{{4}}\right)}}}+{\sin{{\left(\frac{\pi}{{3}}\right)}}}\cdot{\sin{{\left(\frac{\pi}{{4}}\right)}}}$

$\displaystyle{u}\cdot{v}=\frac{\sqrt{{2}}}{{2}}\cdot{\left(\frac{\sqrt{{3}}}{{2}}-\frac{{1}}{{2}}\right)}$

$\displaystyle{u}\cdot{v}={\cos{{\left({3}\frac{\pi}{{4}}-\frac{\pi}{{3}}\right)}}}={\cos{{\left({5}\frac{\pi}{{12}}\right)}}}=\frac{{\sqrt{{2}}\cdot{\left(\sqrt{{3}}-{1}\right)}}}{{4}}$

You need to evaluate the magnitudes |u| and |v|, such that:

$\displaystyle{\left|{u}\right|}=\sqrt{{{{\cos}}^{{2}}{\left(\frac{\pi}{{3}}\right)}+{{\sin}}^{{2}}{\left(\frac{\pi}{{3}}\right)}}}\Rightarrow{\left|{u}\right|}=\sqrt{{{1}}}\Rightarrow{\left|{u}\right|}={1}$

$\displaystyle{\left|{v}\right|}=\sqrt{{{{\cos}}^{{2}}{\left({3}\frac{\pi}{{4}}\right)}+{{\sin}}^{{2}}{\left({3}\frac{\pi}{{4}}\right)}}}\Rightarrow{\left|{v}\right|}=\sqrt{{{1}}}\Rightarrow{\left|{v}\right|}={1}$

$\displaystyle{\cos{\theta}}=\frac{{{\cos{{\left({5}\frac{\pi}{{12}}\right)}}}}}{{{1}\cdot{1}}}\Rightarrow{\cos{\theta}}={\cos{{\left({5}\frac{\pi}{{12}}\right)}}}\Rightarrow\theta={5}\frac{\pi}{{12}}$

Hence, the cosine of the angle between the vectors u and v is $\displaystyle{\cos{\theta}}={\cos{{\left({5}\frac{\pi}{{12}}\right)}}}$ , so, $\displaystyle\theta={5}\frac{\pi}{{12}}.$