What is the projection of #<2,-7,1 ># onto #<4,-5,9 >#?

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Answer 1 · 2018-04-22T12:01:10

The projection is #=7/sqrt122<4, -5, 9>#

The projection of #vecv# onto #vecu# is

#proj_(vecu)(vecv)= (< vecu, vecv >)/ (< vecu, vecu >) vecu#

#vecu = <4, -5, 9>#

#vecv= <2, -7,1>#

The dot product is

#< vecu, vecv > = <4, -5, 9> .<2, -7,1> #

#=(4xx2)+(-5xx2)+(9xx1)#

#=8-10+9#

#=7#

The magnitude of #vecu# is

#< vecu, vecu > = ||<4, -5, 9>|| =sqrt(4^2+(-5)^2+9^2)#

#=sqrt(16+25+81)#

#=sqrt122#

Therefore,

#proj_(vecu)(vecv)=7/sqrt122<4, -5, 9>#