CHAPTER 2 THE DOT CROSS PRODUCTS AND To determine the angle between the vectors ; we have  A and B sin   A B AB  162 142  52  477 Now A B   32  22  42  29 A A   22  32  22  17 B B  477  129 17 sin 477 or 2917 790 Example 2.12 If , and 3iˆ 2kˆ 2iˆ 2 2kˆ A   B   kˆ C  iˆ  ˆj ˆj ˆj A , Find (a) B C B . (b)  A C   7  5kˆ A B ˆj (a) A iˆ 7 5kˆ 2 2kˆ 24iˆ 7 5kˆ      B C iˆ ˆj ˆj ˆj    5 5k B  C  0ˆ ˆ (b) B 3ˆ 0ˆ  2k 5 5k A  C  15 ˆ ˆ 15ˆ ˆ A B , Thus showing the need for parentheses in to avoid B C A  C A B  C ambiguity. Example 2.13 The area of a parallelogram with sides and is A B A B i.e. area of parallelogram A  h B h  25 sin  where  h A
CHAPTER 2 THE DOT AND CROSS PRODUCTS To determine the angle  between the vectors A and B ; we have A B  AB sin  Now A B  162 142  52  477 A  A  32  22  42  29 B  B  22  32  22  17 477  129 17 sin  or 477 2917 790 Example 2.12 If A  3iˆ ˆj  2kˆ , B  2iˆ  ˆj kˆ and C  iˆ 2 ˆj  2kˆ Find (a) (b) A B C , A B  C . (a) (b) A B   7 ˆj  5kˆ A B C  iˆ  7 ˆj  5kˆ iˆ 2 ˆj  2kˆ 24iˆ  7 ˆj 5kˆ B  C     0ˆ 5 ˆ 5k A B  C 3ˆ ˆ  2k 0ˆ 5 ˆ 5k  15ˆ 15 ˆ Thus A B C A B  C , showing the need for parentheses in A B  C to avoid ambiguity. Example 2.13 The area of a parallelogram with sides A and B is A B i.e. area of parallelogram  h B A h where h  A sin  25 