What is the projection of #<4,-1,6 ># onto #<1,3,-2 >#?

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Answer 1 · 2016-09-16T20:10:44

#= - (11)/sqrt(14) #

well you start with the scalar dot product

#vec A * vec B = abs vec A abs vec B cos phi#, where #phi# is the angle between the vectors

then you insist that #vec B # be the unit vector #hat B = (vec B)/(abs vec B)# so that

#abs hat B = 1#

and

#vec A * hat B = abs vec A (1) cos phi = abs vec A cos phi # which is the length of #vec A# in the #\hat B# direction

so that we can actually say that

# (vec A * (vec B)/(abs vec B) = vec A * hat B = abs vec A cos phi )\ hat B #

so with #vec A = <4,-1,6 >#

and #vec B = <1,3,-2 ># meaning #hat B = 1/sqrt(14) <1,3,-2 >#

# (vec A * hat B = 1/sqrt(14) ( <4,-1,6 > * <1,3,-2 > ) )hat B#

#= 1/sqrt(14) (4 - 3 - 12 ) hat B#

#= - (11)/sqrt(14) hat B#