What is the projection of #< 3 , -7, 0># onto #< -1, -4 , 6 >#?

Question

1378 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-05T16:07:04

The vector projection is #=<-25/53,-100/53,150/53>#

The vector projection of #vecb# onto #veca# is

#proj_(veca)vecb=(veca.vecb)/(||veca||)^2*veca#

#veca= <-1,-4,6>#

#vecb= <3,-7,0>#

The dot product is

#veca.vecb= <-1,-4,6> . <3, -7,0> = (-1)*(3)+(-4)*(-7)+(6)*(0)#

#=-3+28+0=25#

The modulus of #veca# is

#||veca|| = ||<-1,-4,6>|| = sqrt((-1)^2+(-4)^2+(6)^2)#

#= sqrt(1+16+36)=sqrt(53)#

Therefore,

#proj_(veca)vecb=(25)/(sqrt(53))^2* <-1,-4,6>#

#=25/53<-1,-4,6>#

#= <-25/53,-100/53,150/53>#