# What is the projection of <0, 1, 3> onto <3, 2, 1>?

##### 1 Answer
Apr 22, 2017

The vector projection, $m a t h b f p$, of $m a t h b f a$ on $m a t h b f b$ is:

$m a t h b f p = \left(m a t h b f a \cdot m a t h b f \left(\hat{b}\right)\right) \setminus m a t h b f \left(\hat{b}\right) = \left(\left\mid m a t h b f a \right\mid \left\mid m a t h b f \left(\hat{b}\right) \right\mid \cos \theta\right) \setminus m a t h b f \left(\hat{b}\right)$

implies mathbf p = (abs(mathbf a) cos theta) \ mathbf (hatb

Now:

$m a t h b f a \cdot m a t h b f b = \left\mid m a t h b f a \right\mid \left\mid m a t h b f b \right\mid \cos \theta$

$\implies \cos \theta = \frac{m a t h b f a \cdot m a t h b f b}{\left\mid m a t h b f a \right\mid \left\mid m a t h b f b \right\mid}$

implies mathbf p = abs(mathbf a) (mathbf a cdot mathbf b)/( abs(mathbf a) abs( mathbf b)) mathbf (hatb

$= \frac{m a t h b f a \cdot m a t h b f b}{\left\mid m a t h b f b \right\mid} \frac{m a t h b f b}{\left\mid m a t h b f b \right\mid}$

$= \frac{m a t h b f a \cdot m a t h b f b}{{\left\mid m a t h b f b \right\mid}^{2}} m a t h b f b$

$= \frac{\left\langle 0 , 1 , 3\right\rangle \cdot \left\langle 3 , 2 , 1\right\rangle}{{3}^{2} + {2}^{2} + {1}^{2}} \left\langle 3 , 2 , 1\right\rangle$

$= \frac{5}{14} \left\langle 3 , 2 , 1\right\rangle$