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
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 >>)#