# What is the projection of <4,-1,6 > onto <1,3,-2 >?

Sep 16, 2016

$= - \frac{11}{\sqrt{14}}$

#### Explanation:

$\vec{A} \cdot \vec{B} = \left\mid \vec{A} \right\mid \left\mid \vec{B} \right\mid \cos \phi$, where $\phi$ is the angle between the vectors

then you insist that $\vec{B}$ be the unit vector $\hat{B} = \frac{\vec{B}}{\left\mid \vec{B} \right\mid}$ so that

$\left\mid \hat{B} \right\mid = 1$

and

$\vec{A} \cdot \hat{B} = \left\mid \vec{A} \right\mid \left(1\right) \cos \phi = \left\mid \vec{A} \right\mid \cos \phi$ which is the length of $\vec{A}$ in the $\setminus \hat{B}$ direction

so that we can actually say that

$\left(\vec{A} \cdot \frac{\vec{B}}{\left\mid \vec{B} \right\mid} = \vec{A} \cdot \hat{B} = \left\mid \vec{A} \right\mid \cos \phi\right) \setminus \hat{B}$

so with $\vec{A} = < 4 , - 1 , 6 >$

and $\vec{B} = < 1 , 3 , - 2 >$ meaning $\hat{B} = \frac{1}{\sqrt{14}} < 1 , 3 , - 2 >$

$\left(\vec{A} \cdot \hat{B} = \frac{1}{\sqrt{14}} \left(< 4 , - 1 , 6 > \cdot < 1 , 3 , - 2 >\right)\right) \hat{B}$

$= \frac{1}{\sqrt{14}} \left(4 - 3 - 12\right) \hat{B}$

$= - \frac{11}{\sqrt{14}} \hat{B}$