What is the projection of #<0, 1, 3># onto #<3, 2, 1>#?

1 Answer
Apr 22, 2017

The vector projection, #mathbf p#, of #mathbfa# on #mathbfb# is:

#mathbf p = ( mathbf a cdot mathbf(hatb) )\ mathbf (hatb) = (abs(mathbf a) abs( mathbf(hatb)) cos theta) \ mathbf (hatb)#

#implies mathbf p = (abs(mathbf a) cos theta) \ mathbf (hatb#

Now:

#mathbf a cdot mathbf b = abs(mathbf a) abs( mathbf b) cos theta #

#implies cos theta = (mathbf a cdot mathbf b)/( abs(mathbf a) abs( mathbf b))#

#implies mathbf p = abs(mathbf a) (mathbf a cdot mathbf b)/( abs(mathbf a) abs( mathbf b)) mathbf (hatb#

# = (mathbf a cdot mathbf b)/( abs( mathbf b)) ( mathbf b)/( abs( mathbf b)) #

# = (mathbf a cdot mathbf b)/( abs( mathbf b)^2) mathbf b #

# = (langle 0,1,3 rangle cdot langle3,2,1 rangle)/( 3^2 + 2^2 + 1^2) langle3,2,1 rangle #

# = (5)/(14) langle3,2,1 rangle #