What is the projection of #<6,-4,8 ># onto #<-1,3,-7 >#?

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Answer 1 · 2017-07-21T06:03:43

The vector projection is #=-74/59<-1,3,-7>#
The scalar projection is #=-61/sqrt59#

Let #vecb= <6,-4,8># and #veca= <-1,3,-7>#

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

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

The dot product is

#veca.vecb=<6,-4,8> . <-1,3,-7> =(6*-1)+(-4*3)+(8*-7)#

#=-6-12-56=-74#

The modulus of #veca# is

#||<-1,3,-7>|| =sqrt(1+9+49) = sqrt59#

Therefore,

The vector projection is

#=-74/59<-1,3,-7>#

The scalar projection is

#=(veca.vecb)/(||veca||)=-61/sqrt59#