Notes: `u = -6i - 3j, v = -8i + 4j` Find the angle theta between the vectors.

Tuesday, 22 December 2015

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;

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

$\displaystyle{u}=-{6}{i}-{3}{j}$

$\displaystyle{v}=-{8}{i}+{4}{j}$

$\displaystyle{u}.{v}={\left(-{6}\right)}\times{\left(-{8}\right)}+{\left(-{3}\right)}\times{4}={36}$

$\displaystyle{\left|{u}\right|}=\sqrt{{{{\left(-{6}\right)}}^{{2}}+{{\left(-{3}\right)}}^{{2}}}}=\sqrt{{{45}}}$

$\displaystyle{\left|{v}\right|}=\sqrt{{{{\left(-{8}\right)}}^{{2}}+{{4}}^{{2}}}}=\sqrt{{80}}$

The angle between...

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;

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

$\displaystyle{u}=-{6}{i}-{3}{j}$

$\displaystyle{v}=-{8}{i}+{4}{j}$

$\displaystyle{u}.{v}={\left(-{6}\right)}\times{\left(-{8}\right)}+{\left(-{3}\right)}\times{4}={36}$

$\displaystyle{\left|{u}\right|}=\sqrt{{{{\left(-{6}\right)}}^{{2}}+{{\left(-{3}\right)}}^{{2}}}}=\sqrt{{{45}}}$

$\displaystyle{\left|{v}\right|}=\sqrt{{{{\left(-{8}\right)}}^{{2}}+{{4}}^{{2}}}}=\sqrt{{80}}$

The angle between vectors is given by;

$\displaystyle{\cos{\theta}}=\frac{{36}}{{{\left(\sqrt{{45}}\right)}{\left(\sqrt{{80}}\right)}}}$

$\displaystyle\theta={{\cos}}^{{-{1}}}{\left(\frac{{36}}{\sqrt{{{3600}}}}\right)}$

$\displaystyle\theta={53.13}{d}{e}{g{}}$

So the angle between two vectors is 53.13 deg

Notes