3 y= |-51. Dec 12 1. [2 pts.) Let x = | 2 | and ompose the vector y into two com
ID: 3116245 • Letter: 3
Question
3 y= |-51. Dec 12 1. [2 pts.) Let x = | 2 | and ompose the vector y into two components: one collinear with x and another orthogonal to x. 2. Consider a plane in R3 defined by the equation (a) [3 pts.] Find two different points (any two points, let's call them A and B) in this plane and show that the vector AB that connects the points is orthogonal to the vector n = (3,-2.3)" (b) [3 pts.] Find an orthogonal projection of the vector x = (14,212)" onto the plane 3. Suppose a subspace W of R is given as a span of the following vectors 0 v2= 0 0 (a) [3 pts.] Find a basis in W (b) [3 pts.] Find a nonzero vector in W that is orthogonal to vi (c) [3 pts.] Compute a projection onto W of the vector x(3,0, 2,-4)7 using inner products and the orthogonal basis you have found above. (d) [3 pts.] Find a basis for W1Explanation / Answer
1.Let x1 = projx(y) = [(y.x)/(x.x)]x = [(21-10-12)/(9+4+1)](3,2,-1)T = -1/14(3,2,-1)T= (-3/14,-2/14,1/14)T .Also, let x2 = y- projx(y) = (7,-5,12)T-(-3/14,-2/14,1/14)T =(101/14, -68/14, 167/14)T. Then x2.x = 0. Also, y = x1+x2 where x1 is collinear with x and x2 is orthogonal to x.
3. (a) Let A be the matrix with v1,v2,v3 as columns. Then A =
1
0
2
1
-1
1
0
0
0
0
1
1
The RREF of A is
1
0
2
0
1
1
0
0
0
0
0
0
Now, apparently, v3= 2v1+v2. Thus basis for W is {v1,v2} or{(1,0,0,0)T,(0,1,0,0)T}.
(b). An arbitrary vector in W is w = x(1,0,0,0)T+y(0,1,0,0)T = (x,y,0,0)T. Now, if w is orthogonal to v1, then w.v1 = 0 or, (x,y,0,0)T.(1,1,0,0)T= 0 or, x+y = 0 or, y = -z. Then, w = (x,-x,0,0)T = x(1,-1,0,0)T. Thus, the vector (1,-1,0,0)T is in W and is orthogonal to v1.
(c ). We have projv1(x) = [(x.v1)/(v1.v1)]v1= [(3+0+0+0)/(1+1+0+0)]v1= (3/2).(1,1,0,0)T =(3/2,3/2,0,0)T and projv2(x) = [(x.v2)/(v2.v2)]v2= [(0+0+0-4)/(0+1+0+1)]v1= (-2).(0,-1,0,1)T =(0,2,0,-2)T. Then the projection of x onto W =(3/2,3/2,0,0)T+(0,2,0,-2)T = (3/2, 7/2,0,-2)T.
(d). Let (x,y,z,w)Tbe an arbitrary vector in W. Then (x,y,z,w)T.(1,1,0,0)T=0 or,x+y=0 or, x = -y and (x,y,z,w)T.(0,-1,0,1)T= 0 or, -y+w = 0 or, y = w ( then x = -y = -w) so that (x,y,z,w)T= (-w,w,z,w)T= z(0,0,1,0)T. +w(-1,1,0,1)T. Hence {(0,0,1,0)T, (-1,1,0,1)T } is a basis for W.
Please pose the 2nd question again separately.
1
0
2
1
-1
1
0
0
0
0
1
1
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