# What is the projection of <3,4,-1 > onto <-1,3,-6 >?

Jul 12, 2016

The projection is $\left(- \frac{15}{\sqrt{46}} , \frac{45}{\sqrt{46}} , - \frac{90}{\sqrt{46}}\right)$.

#### Explanation:

Given two vectors $\vec{V}$ and $\vec{W}$, the projection of $\vec{V}$ onto $\vec{W}$ is given by:

$\left(\vec{V} \cdot \vec{W}\right) \frac{\vec{W}}{|} | \vec{W} | |$

The inner product gives the component of $V e c V$ in the direction of $V e c W$, and the fraction performs the same function as multiplying this magnitude by a unit vector in the direction of $\vec{W}$.

So plugging in given values:
$\vec{V} = \left(3 , 4 , - 1\right)$
$\vec{W} = \left(- 1 , 3 , - 6\right)$
$| | \vec{W} | | = \sqrt{1 + 9 + 36} = \sqrt{46}$

$\vec{V} \cdot \vec{W} = \left(3 \cdot - 1\right) + \left(4 \cdot 3\right) + \left(- 1 \cdot - 6\right)$
$= - 3 + 12 + 6 = 15$

So plugging everything in, the projection is:
$\frac{15 \cdot \left(- 1 , 3 , - 6\right)}{\sqrt{46}} = \left(- \frac{15}{\sqrt{46}} , \frac{45}{\sqrt{46}} , - \frac{90}{\sqrt{46}}\right)$