What is the projection of #< 2 , -6, 0># onto #< 5, -3, 8>#?

1 Answer
Truong-Son N.
May 5, 2016

Let #veca = << 2,-6,0 >># and #vecb = << 5,-3,8 >>#.

The projection of #veca# onto #vecb# is written as:

#\mathbf("proj"_(vecb)veca = (vecacdotvecb)/(||vecb||cdot||vecb||)vecb)#

So first, we have to find the dot product between #veca# and #vecb#.

#color(green)(vecacdotvecb)#

#= << 2,-6,0 >> cdot << 5,-3,8 >>#

#= 2*5 + -6*-3 + 0*8#

#= 10 + 18 = color(green)(28)#

And the norm of #vecb# is:

#color(green)(||vecb||)#

#= sqrt(vecbcdotvecb)#

#= sqrt(<< 5,-3,8 >>cdot<< 5,-3,8 >>)#

#= sqrt(5*5+ -3*-3+8*8)#

#= sqrt(25+9+64)#

#= sqrt(98) = color(green)(7sqrt2)#

Thus, the projection of #veca# onto #vecb# is:

#color(blue)("proj"_(vecb)veca)#

#= 28/(7sqrt2*7sqrt2)<< 5,-3,8 >>#

#= 28/98<< 5,-3,8 >>#

#= 2/7<< 5,-3,8 >>#

#= color(blue)(<< 10/7,-6/7,16/7 >>)# <-- let this be #vecc#.

At this point, to check that this worked, you should perform the following to determine whether the angle between the vectors is indeed #0^@#:

#vecb cdot vecc = ||vecb||cdot||vecc||costheta#

#color(blue)(theta) = arccos((vecbcdotvecc)/(||vecb||cdot||vecc||))#

#= arccos((5*10/7 + -3*-6/7 + 8*16/7)/(sqrt(5^2 + (-3)^2 + 8^2)*sqrt((10/7)^2 + (-6/7)^2 + (16/7)^2)))#

#= arccos(((50 + 18 + 128)/7)/(7sqrt2*sqrt((100+36+256)/49)))#

#= arccos(((196)/7)/(7sqrt2*sqrt8))#

#= arccos((28)/(28))#

#= color(blue)(0^@)#

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