What is the projection of #<4,-6,3 ># onto #<1,5,2 >#?

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Answer 1 · 2018-06-04T08:11:13

The vector projection is #=<-2/3,-10/3,-4/3> #

The projection of #vecv# onto #vecu# is

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

#vecu = <1, 5, 2>#

#vecv= <4, -6,3>#

The dot product is

#< vecu, vecv > = <1, 5, 2> .<4, -6,3> #

#=(1xx4)+(5xx-6)+(2xx3)#

#=4-30+6#

#=-20#

The magnitude of #vecu# is

#< vecu, vecu > = ||<1, 5, 2>|| =sqrt(1^2+(5)^2+2^2)#

#=sqrt(1+25+4)#

#=sqrt30#

Therefore, the vector projection is

#proj_(vecu)(vecv)=-20/30<1, 5, 2>#

#= <-2/3,-10/3,-4/3>#