Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# U 2 could be 1 2 0 5 there is a whole plane of vectors

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Unformatted text preview: U 1 D .1; 3/= 10 is w p perpendicular to u1 (and so is . 1; 3/= 10). U 2 could be .1; 2 ; 0/= 5: There is a whole plane of vectors perpendicular to u2 , and a whole circle of unit vectors in that plane. 6 All vectors w D .c; 2c/ are perpendicular to v. All vectors .x; y; z/ with x C y C z D 0 lie on a plane. All vectors perpendicular to .1; 1; 1/ and .1; 2; 3/ lie on a line. 7 (a) cos  D v  w=kvkkwk D 1=.2/.1/ so  D 60ı or =3 radians (b) cos  D 0 so  D 90ı or p radians (c) cos  D 2=.2/.2/ D 1=2 so  D 60ı or =3 =2 (d) cos  D 1= 2 so  D 135ı or 3=4. (b) True: u  .v C v/  .u v/ splits into 8 (a) False: v and w are any vectors in the plane perpendicular to u 2w/ D u  v C 2u  w D 0 (c) True, ku u  u C v  v D 2 when u  v D v  u D 0. 9 If v2 w2 =v1 w1 D v k 2 D .u 1 then v2 w2 D v1 w1 or v1 w1 Cv2 w2 D v w D 0: perpendicular! 10 Slopes 2=1 and 1=2 multiply to give 1: then v w D 0 and the vectors (the directions) are perpendicular. 11 v  w &lt; 0 means angle &gt; 90ı ; these w’s ﬁll half of 3-dimensional space. 12 .1; 1/ perpendicular to .1; 5/ c .1; 1/ if 6 2 c D 0 or c D 3; v  .w c D v  w=v  v. Subtracting c v is the key to perpendicular vectors. c v/ D 0 if 13 The plane perpendicular to .1; 0; 1/ contains all vectors .c; d; c /. In that plane, v D .1; 0; 1/ and w D .0; 1; 0/ are perpendicular. 14 One possibility among many: u D .1; 1; 0; 0/; v D .0; 0; 1; 1/; w D .1; 1; 1; 1/ and .1; 1; 1; 1/ are perpendicular to each other. “We can rotate those u; v; w in their 3D hyperplane.” p pp 15 1 .x C y/ D .2 C 8/=2 D 5; cos  D 2 16= 10 10 D 8=10. 2 16 kvk2 D 1 C 1 C    C 1 D 9 so kvk D 3I u D v=3 D . 1 ; : : : ; 1 / is a unit vector in 9D; 3 3 p w D .1; 1; 0; : : : ; 0/= 2 is a unit vector in the 8D hyperplane perpendicular to v. p p 17 cos ˛ D 1= 2, cos ˇ D 0, cos D 1= 2. For any vector v, cos2 ˛ C cos2 ˇ C cos2 2 2 2 D .v1 C v2 C v3 /=kvk2 D 1. 18 kvk2 D 42 C 22 D 20 and kwk2 D . 1/2 C 22 D 5. Pythagoras is k.3; 4/k2 D 25 D 20 C 5. 19 Start from the rules .1/; .2/; .3/ for v  w D w  v and u  .v C w/ and .c v/  w. Use rule .2/ for .v Cw/  .v Cw/ D .v Cw/  v C.v Cw/ w. By rule .1/ this is v  .v Cw/ Cw  .v Cw/. Rule .2/ again gives v  v C v  w C w  v C w  w D v  v C 2v  w C w  w. Notice v  w D w  v! The main point is to be free to open up parentheses. 20 We know that .v w/  .v w/ D v  v 2v  w C w  w. The Law of Cosines writes kvkkwk cos  for v  w. When  &lt; 90ı this v  w is positive, so in this case v  v C w  w is larger than kv wk2 . 21 2v  w  2kvkkwk leads to kv C wk2 D v  v C 2v  w C w  w  kvk2 C 2kvkkwkCkwk2 . This is .kvk C kwk/2 . Taking square roots gives kv C wk  kvk C kwk. 22 22 22 22 22 22 22 v1 w1 C 2v1 w1 v2 w2 C v2 w2  v1 w1 C v1 w2 C v2 w1 C v2 w2 is true (cancel 4 terms) 22 22 because the difference is v1 w2 C v2 w1 2v1 w1 v2 w2 which is .v1 w2 v2 w1 /2  0. Solutions to Exercises 5 23 cos ˇ D w1 =kwk and sin ˇ D w2 =kwk. Then cos.ˇ a/ D cos ˇ cos ˛ C sin ˇ sin ˛ D v1 w1 =kvkkwk...
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