What is the projection of #< 5 ,- 3, 9 ># onto #< 5, -2 , -4 >#?

1 Answer
Oct 10, 2016

#"proj"_v u = color(green)(< -5/9,2/9,4/9>)#

Explanation:

In general for vectors #vecu# and #vecv#
the projection of #vecu# unto #vecv# is given by
#color(white)("XXX")"proj_v u = ( (vecu * vecv)/(abs(abs(vecv))^2)) * vecv#

#color(white)("XXXXXXXXXXXX")#[If you need a derivation of this equation,
#color(white)("XXXXXXXXXXXXX")#ask as a separate question.]

For the given case
#color(white)("XXX")vecu= < 5,-3,9># and
#color(white)("XXX")vecv = <5,-2,-4>#

#vecu * vecv = (5 * 5) + ((-3) * (-2)) + (9 * (-4))#
#color(white)("XXXX") = 25 +6 -36#
#color(white)("XXXX") = -5#

#abs(abs(vecv))^2 = 5^2+(-2)^2+(-4)^2#
#color(white)("XXX")=25+4+16#
#color(white)("XXX")=45#

#((vecu * vecv)/(abs(abs(vecv))^2)) = (-5)/(45)#
#color(white)("XXXXXX")=-1/9#

#"proj"_v u = (-1/9)vecv#
#color(white)("XXX")=(-1/9) * < 5, -2, -4>#
#color(white)("XXX")=<-5/9, 2/9, 4/9>#