How do you find the projection of u onto v given #u=<3, 15># and #<v=-1, 5>#?

1 Answer
Feb 4, 2017

Answer:

The vector projection is #=36/13<-1,5>#
The scalar projection is #=72/sqrt26#

Explanation:

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

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

The dot product is

#vecu.vecv=<3,15>.<-1,5>=-3+75=72#

The modulus of #vecv# is

#||vecv||=||<-1,5>||=sqrt(1+25)=sqrt26#

The vector projection is

#=72/26<-1,5> = 36/13<-1,5>#

The scalar projection is

#(vecu.vecv)/(||vecv||)=72/sqrt26#