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

1 Answer
Feb 15, 2016

Answer:

#proj_vec v vec u = (-15/11i-10/11j+15/11k)#

Explanation:

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)#