Notes: `u = 2i - j ,v = 6i + 4j` Find the angle theta between the vectors.

Sunday, 15 June 2014

The angle between two vectors u and v is given by;

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

u.v represent the vector dot product and |u| and |v| represents the magnitude of vectors.

We know that in unit vectors;

$\displaystyle{i}\times{i}={j}\times{j}={1}$ and $\displaystyle{i}\times{j}={j}\times{i}={0}$

$\displaystyle{u}={2}{i}-{j}$

$\displaystyle{v}={6}{i}+{4}{j}$

$\displaystyle{u}.{v}={2}\times{6}-{1}\times{4}={8}$

The magnitude of the vectors is given by;

$\displaystyle{\left|{u}\right|}=\sqrt{{{{2}}^{{2}}+{{\left(-{1}\right)}}^{{2}}}}=\sqrt{{5}}$

$\displaystyle{\left|{v}\right|}\ldots$

The angle between two vectors u and v is given by;

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

u.v represent the vector dot product and |u| and |v| represents the magnitude of vectors.

We know that in unit vectors;

$\displaystyle{i}\times{i}={j}\times{j}={1}$ and $\displaystyle{i}\times{j}={j}\times{i}={0}$

$\displaystyle{u}={2}{i}-{j}$

$\displaystyle{v}={6}{i}+{4}{j}$

$\displaystyle{u}.{v}={2}\times{6}-{1}\times{4}={8}$

The magnitude of the vectors is given by;

$\displaystyle{\left|{u}\right|}=\sqrt{{{{2}}^{{2}}+{{\left(-{1}\right)}}^{{2}}}}=\sqrt{{5}}$

$\displaystyle{\left|{v}\right|}=\sqrt{{{{6}}^{{2}}+{{4}}^{{2}}}}=\sqrt{{52}}$

$\displaystyle{\cos{\theta}}=\frac{{8}}{{\sqrt{{5}}\times\sqrt{{52}}}}$

$\displaystyle\theta={{\cos}}^{{-{1}}}{\left(\frac{{8}}{\sqrt{{260}}}\right)}$

$\displaystyle\theta={60.255}{d}{e}{g{}}$

So the angle theta between two vectors is 60.255 deg

Notes