Practice Final #7
I am not positive if I am correct, but this is what I got for #7 (I turned it into orthonormal form because I found the smaller numbers easier to work with). u1 = 1/$\sqrt{6}$(1 2 1). u2 = v2 - proj$_{u1}$v2 = (-3 -2 7) / $\sqrt{62}$ u3 = v3 - proj$_{u1}$v3 - proj$_{u2}$v3 = $\sqrt{2/35}$*(207/62, -110/62, 111/62). Thus, u3 is the closest point to that vector.
tedallen: May 30, 2015, 7:48 p.m.
my final answer was $\begin{matrix} 59/31\\126/31\\69/31\end{matrix}$.
leopham: May 31, 2015, 3:06 a.m.
I got <59/31,122/31,69/31> Here's my work
CarlHarmon: May 31, 2015, 10:25 a.m.
There is another way to approach this problem. You can calculate the orthonormal basis and then use it to find the "matrix of orthogonal projection onto V" using the equation A($A^TA)^{-1}A^T$. Then you multiply the given vector by this matrix. While this takes some time, you could then use that same matrix for multiple vectors. It's essentially a more general method.
CarlHarmon: May 31, 2015, 10:26 a.m.
Oh, and because you found an orthonormal basis, the $A^TA$ part is just the identity matrix.
tanishaharlalka: May 31, 2015, 5:42 p.m.
Got the same answer by using the projection method Professor went over in class <59/31,122/31,69/31>
James_Kercheval: May 31, 2015, 7:12 p.m.
S/o to leopham and tanishaharlalka, I got the same answer as both of you
yulduzkhonbruin: June 1, 2015, 1:47 p.m.
Yes, I have the same answer too: (59, 122, 69).
tedallen: June 1, 2015, 11:25 p.m.
Sorry. must have made an algebra mistake
brookewenig: June 2, 2015, 8:07 a.m.
Thanks for the help! I made the mistake of trying to calculate an orthonormal basis instead of finding the point closest to the vector. I now have <59/31,122/31,69/31>.
kmeneses95ND: June 5, 2015, 10:43 p.m.
Yes I have the answer <59/31, 122/31, 69/31> as well.
ShangT: June 6, 2015, 12:30 a.m.
I also have <59/31, 122/31, 69/31>