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

##### 2 Answers

It is <-3/14, -1/7, -1/14>.

#### Explanation:

The length of the projection of a vector

The dot product is the sum of the products of the components:

The direction of the vector is the one of

The length is calculated again with the scalar product, this time of

So the unitary vector in the direction of

This unitary vector gives the direction, while the length and direction was given by the initial dot product that was

So the final projection is

I got

**PROJECTION OF** **ONTO**

Let

The projection of

#\mathbf("proj"_(vecb) veca = (vecacdotvecb)/(||vecb||cdot||vecb||)vecb)#

This is drawn as:

For this, we should define a few more things.

- The
**dot product**of two#n# -long vectors#vecu# and#vecv# :

#\mathbf(vecucdotvecv)#

#= << u_1,u_2, . . . ,u_N >> cdot << v_1,v_2, . . . ,v_N >>#

#= \mathbf(u_1v_1 + u_2v_2 + . . . + u_Nv_n)#

- The
**norm**of#vecv# :

#\mathbf(||vecv||)#

#= sqrt(vecvcdotvecv)#

#= sqrt(v_1v_1 + v_2v_2 + . . . + v_Nv_N)#

#= \mathbf(sqrt(v_1^2 + v_2^2 + . . . + v_N^2))#

**COMPONENTS OF THE PROJECTION**

So, what we now have to do is:

#color(green)(vecacdotvecb)#

#= << 3,-6,2 >>cdot<< 3,2,1 >>#

#= 3*3 + (-6*2) + 2*1#

#= 9 - 12 + 2#

#= color(green)(-1)#

#color(green)(||vecb||)#

#= sqrt(<< 3,2,1 >>cdot << 3,2,1 >>)#

#= sqrt(3*3 + 2*2 + 1*1)#

#= sqrt(9 + 4 + 1)#

#= color(green)(sqrt14)#

**FINAL CALCULATION**

Finally, put it all back in the first equation.

#color(blue)("proj"_(vecb) veca) = (vecacdotvecb)/(||vecb||cdot||vecb||)vecb#

#= (-1)/(sqrt14*sqrt14)<< 3,2,1 >>#

#= color(blue)(<< -3/14,-1/7,-1/14 >>)#