How do you find the projection of u onto v given #u=<0, 3># and #v=<2, 15>#?

1 Answer
Feb 26, 2017

Answer:

The vector projection is #=45/sqrt229<2,15>#
The scalar projection is #=45/sqrt229#

Explanation:

The vector projection of #vecu# onto #vecv# is

#=(vecu.vecv)/(||vecv||^2)vecv#

The dot product is

#vecu.vecv=<0,3>,<2,15>=0+45=45#

The modulus of #vecv# is

#=||vecv||=||vecv||= ||<2,15||#

#=sqrt(4+225)=sqrt229#

The vector projection is

#=45/sqrt229<2,15>#

The scalar projection is

#=(vecu.vecv)/(||vecv||)#

#=45/sqrt229#