What is the projection of #(3i + 2j - 6k)# onto # (3i - j - 2k)#?

1 Answer
Narad T.
Nov 2, 2016

The answer is #=19/(7sqrt14)(3i-j-2k)#

Explanation:

Let #veca=〈3,-1,-2〉# and #vecb=〈3,2,-6〉#
Then the vector projection of #vecb# upon #veca# is
#(veca.vecb)/(∥veca∥∥vecb∥)veca#
The dot product #veca.vecb=〈3,-1,-2〉.〈3,2,-6〉=9-2+12=19#
The modulus #∥veca∥=sqrt(9+1+4)=sqrt14#
The modulus #∥vecb∥=sqrt(9+4+36)=sqrt49=7#
the projection is #=19/(7sqrt14)〈3,-1,-2〉#

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