Notes: `u = cos(pi/4)i + sin(pi/4)j, v = cos(pi/2)i + sin(pi/2)j` Find the angle theta between the vectors.

Tuesday, 12 August 2014

$\displaystyle{u}={\cos{{\left(\frac{\pi}{{4}}\right)}}}{i}+{\sin{{\left(\frac{\pi}{{4}}\right)}}}{j},{v}={\cos{{\left(\frac{\pi}{{2}}\right)}}}{i}+{\sin{{\left(\frac{\pi}{{2}}\right)}}}{j}$ 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:

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

$\displaystyle{u}\cdot{v}={\cos{{\left(\frac{\pi}{{4}}\right)}}}\cdot{\cos{{\left(\frac{\pi}{{2}}\right)}}}+{\sin{{\left(\frac{\pi}{{4}}\right)}}}\cdot{\sin{{\left(\frac{\pi}{{2}}\right)}}}$

$\displaystyle{u}\cdot{v}={\cos{{\left(\frac{\pi}{{2}}-\frac{\pi}{{4}}\right)}}}=\ldots$

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:

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

$\displaystyle{u}\cdot{v}={\cos{{\left(\frac{\pi}{{4}}\right)}}}\cdot{\cos{{\left(\frac{\pi}{{2}}\right)}}}+{\sin{{\left(\frac{\pi}{{4}}\right)}}}\cdot{\sin{{\left(\frac{\pi}{{2}}\right)}}}$

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

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

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

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

$\displaystyle{\cos{\theta}}=\frac{{\frac{\sqrt{{2}}}{{2}}}}{{{1}\cdot{1}}}\Rightarrow{\cos{\theta}}=\frac{\sqrt{{2}}}{{2}}\Rightarrow\theta=\frac{\pi}{{4}}$

Hence, the cosine of the angle between the vectors u and v is $\displaystyle{\cos{\theta}}=\frac{\sqrt{{2}}}{{2}}$ , so, $\displaystyle\theta=\frac{\pi}{{4}}.$