What is the projection of #(-i + j + k)# onto # ( i - j + k)#?

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Answer 1 · 2016-05-30T07:53:52

The projection of a vector a onto vector b is given by

#proj_a b=(a*b)/absa^2 *a#

Hence

The dot product of #a=(-1,1,1)# and #b=(1,-1,1)# is

#a*b=-1-1+1=-1#

The magnitude of a is #absa=sqrt(-1^2+1^2+1^2)=sqrt3#

Hence the projection is

#proj_a b=-1/3*(-1,1,1)=(-1/3,1/3,1/3)=1/3*(-i+j+k)#