In R3 let V be the subspace spanned by a(6,-3,0,-3,6) (1,0,-6,0,-1) c = Then Vha
ID: 3137650 • Letter: I
Question
In R3 let V be the subspace spanned by a(6,-3,0,-3,6) (1,0,-6,0,-1) c = Then Vhas dimension dim(V)because .A each vector has 5 coordinates B-these vectors have differing numbers of nonzero coordinates C-these vectors form an orthogonal basis of V . D-being orthogonal makes these vectors linearly dependent Enter the letter which marks your response: Given an arbitrary vector x-(2,2,5,7,-5) in IR3, we find the point in V closest to x by projecting x orthogonally onto V. It has the position vector: xy proja(x) + projb(x) + proje(x) In the case at hand, we find Once we have Xv. finding the point closest to x in the orthogonal complement W-V position vector: of Vin R is easy. It has the In the case at hand, we find:Explanation / Answer
6
3
1
-3
6
0
0
1
-6
-3
-6
0
6
-3
-1
The RREF of A is
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
It implies that a,b,c are linearly independent so that dim(V) = 3.Further, these vectors form an orthogonal basis for V. Option C.
2. We have Xv = proja V+ projb V+ projc V = [(x.a)/(a.a)]a+ [(x.b)/(b.b)]b+ [(x.c)/(c.c)]c = [(12-6+0-21-30) / (36+9+0+9+36)]a +[(6+12+5-42+15)/ (9+36+1+36+9)]b +[(2+0-30+0+5)/(1+0+36+0+1)]c = (-1/2)(6,-3,0,-3,6)+(-4/91)(3,6,1,-6,-3) +(-23/38)(1,0,-6,0,-1)= (-12923/3458,225/182, 12406/3458,321/182,-7825/3458).
3. XW = X –XV = (2,2,5,7,-5)- (-12923/3458,225/182, 12406/3458,321/182,-7825/3458) = (19839/3458,139/182,4884/3458,11800/3458,-16381/3458).
6
3
1
-3
6
0
0
1
-6
-3
-6
0
6
-3
-1
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