To make it easier to refer to them, let's call the first vector #vec u# and the second #vec v#. We want the project of #vec u# onto #vec v#:
#proj_vec v vec u = ((vec u*vec v)/||vec v||^2)*vec v#
That is, in words, the projection of vector #vec u# onto vector #vec v# is the dot product of the two vectors, divided by the square of the length of #vec v# times vector #vec v#. Note that the piece inside the parentheses is a scalar that tells us how far along the direction of #vec v# the projection reaches.
First, let's find the length of #vec v#:
#||vec v||=sqrt(3^2+2^2+(-3)^2) = sqrt22#
But note that in the expression what we actually want is #||vec v||^2#, so if we square both sides we just get #22#.
Now we need the dot product of #vec u# and #vec v#:
#vec u * vec v = (1xx3+(-2)xx2+3xx(-3)) = (3-4-9)= (-10)#
(to find the dot product we multiply the coefficients of #i, j and k# and add them)
Now we have everything we need:
#proj_vec v vec u = ((vec u*vec v)/||vec v||^2)*vec v = (-10/22)(3i+2j−3k)#
#=(-30/22i-20/22j+30/22k) = (-15/11i-10/11j+15/11k)#
