I think I have the right answer for this (Orthogonal Matrix and Distance to a plane)

http://i.imgur.com/8s7Z4gV.jpg

a) No problem, I have done this

b) I found the orthogonal basis of B to be {{1,0,1}{1/2,1,-1/2}} - not sure if correct, but I have done it several times and come up with the same answer each time

c) This one I am totally unsure about. I found the shortest distance to be sqrt(3)/3

d) The cross product of 1,0,1 and 1/2, 1, -1/2 is -1, 1, 1 so the other vector is 1, -1, -1 (perpendicular in opposite direction)

a and d I am confident about, b and c, totally unsure, but not sure how to go about checking? Is someone able to independent confirm?

Re: I think I have the right answer for this (Orthogonal Matrix and Distance to a pla

You are correct for b. I got a different answer for c. My orthogonal projection of y onto the plane was

$\displaystyle \hat{y}=c_1u_1 + c_2u_2 = u_1 + \frac{2}{3}u_2$

where

$\displaystyle c_i=\frac{y\cdot u_i}{u_i \cdot u_i}$

Re: I think I have the right answer for this (Orthogonal Matrix and Distance to a pla

for c1, I got 1, so it is 1{1,0,1} but for c2 I got 3/2 not 2/3. Are you using u1 and u2 from the set W or from the given set B?

My working is sort of as follows

$\displaystyle c2 = \frac{{[1,1,1]} \cdot {[1/2,1, -1/2]}}{{[1/2,1,-1/2]}\cdot{[1/2,1,-1/2]}}$

which I calculated to be

(1/2 + 1 - 1/2)/(1/4 + 1 + 1/4) = 1/1.5 = 2/3, which is what you got

so y^ = u1 + 2/3*u2 = 1/3[4,2,2]

and then |y - y^| = [1,1,1] - [4/3,2/3,2/3] = [-1/3,1/3,1/3] = 1/3* sqrt(1+1+1) = 1/3*sqrt(3)

Is there anything wrong with my working?

Re: I think I have the right answer for this (Orthogonal Matrix and Distance to a pla

Quote:

Originally Posted by

**lukasaurus**

and then |y - y^| = [1,1,1] - [4/3,2/3,2/3] = [-1/3,1/3,1/3] = 1/3* sqrt(1+1+1) = 1/3*sqrt(3)

I redid the calculation and got the same as you did, I missed a sign in my original calculation. Sorry, my mistake.