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

1 Answer
Narad T.
Nov 6, 2016

The projection is #=-5/7〈-5,1,3〉#

Explanation:

The vector projection of #vecb# onto #veca# is given by
#=(veca.vecb)/(∥veca∥*∥veca∥)veca#

Here #vecb=〈6,2,1〉# and #veca=〈-5,1,3〉#

The dot product #veca.vecb=〈-5,1,3〉.〈6,2,1〉=-30+2+3=-25#

The modulus of #veca=(∥veca∥=∥〈-5,1,3〉∥=sqrt(25+1+9)=sqrt35#

#:.# the projection #=-25/(sqrt35)^2〈-5,1,3〉=-25/35〈-5,1,3〉#

#=-5/7〈-5,1,3〉#

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