Question

Suppose u = <2,2,4and v = <-3,5,-2>. Then:
The projection of u along v is ?
The projection along u orthogonal to v is ?

Answer 1

you just need to do the dot product of the two vectors u and v, and divide this by the length of v to get the magnitude of projection of u along v. We multiply this magnitude by a unit vector in the direction of vector v to get get a new vector in the direction of v:\[proj _{v}u=\frac{u*v}{\left| v \right|}\frac{v}{\left| v \right|}\]We get:\[proj _{v}u=\frac{-6+10-8}{\sqrt{38}}\frac{<-3,5,-2>}{\sqrt{38}}\]So your projection of u onto v is\[proj _{v}u=\frac{1}{38}<12,-20,8>\]

Answer 2

Now for the second part. I take this to mean the component of u perpendicular to v, as in the picture linked below.
http://www.ltcconline.net/greenl/courses/107/vectors/img4.gif
To find the vector s, we note the following vector sum\[proj _{v}u+s=u\]Thus,\[s=<\frac{6}{19},\frac{-10}{19},\frac{4}{19}>-<2,2,4>\]We get,\[s=<\frac{-32}{19},\frac{-48}{19},\frac{-72}{19}>\]Double check my arithmetic :)

Answer 3

ooops, did that last part backward. Should be \[s=u-proj _{v}u=<2,2,4>-<\frac{6}{19},\frac{-10}{19},\frac{4}{19}>\]

Answer 4

This gives,\[s=<\frac{32}{19},\frac{48}{19},\frac{72}{19}>\]