Higher_Mathematics_Part_1
.pdfMicromodule 5
EXAMPLES OF PROBLEMS SOLUTION
Example 1. The angle between vectors a and b is ϕ = 120°. Knowing that a = 3 | b |= 4, calculate:
а) (3a − 2b)(a + 2b) ; b) | a − b | .
Solution.
а) A = (3a − 2b)(a + 2b) = 3a2 − 2ba + 6ab − 4b 2 = 3a2 + 4ab − 4b 2 .
If equalities |
a2 = |
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b 2 = |
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2 = 16 , |
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ab = |
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cos120° = 3 4 (−0, 5) = −6 are true, then |
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A = 3 9 + 4 (−6) − 4 16 = −61. |
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b) | a − b | = (a − b)2 = a2 − 2ab + b 2 = 9 − 2 (−6) + 16 = 37 . |
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Example 2. The given vectors are a = {1; 2; −2} and b = {3; |
3; −4} . Find: |
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а) scalar product (4a + 3b)(a − 2b) ; |
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b) angle between vectors a + b and a − b . |
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Solution. |
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а) |
4a + 3b = {4; 8; −8} +{9; 9; −12} = {13; 17; |
−20} , |
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a − 2b = {1; 2; |
−2} −{6; 6; −8} = {−5; −4; 6} , |
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(4a + 3b)(a − 2b) = 13 (−5) + 17 (−4) − 20 6 = −233 ; |
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b) |
c = a + b = {4; |
5; |
−6}, |
d = a − b = {−2; −1; |
2} , |
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cos ϕ = |
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c d |
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4 (−2) + 5 (−1) − 6 2 |
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−25 |
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| c || d | |
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42 + 52 + 62 22 + 12 + 22 |
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3 77 |
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Therefore ϕ = π − arccos |
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25 |
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Example 3. The given vectors are a = i − j + 4k , |
b = 6i + 5 j + 4k , c = |
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= −2i − 3 j + 4k . Find vector x satisfying the equalities |
x a = 8 , |
x b = −3 and |
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x c = 13 . |
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Solution. Let x = {x1, x2 , x3} , then the condition |
x a = 8 |
equals to the |
equation x1 − x2 + 4x3 = 8. Similarly we receive two other equations 6x1 + 5x2 +
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+ 4x3 = −3 and −2x1 − 3x2 + 4x3 = 13. By solving the system of linear equations
x1 − x2 + 4x3 = 8,
6x1 + 5x2 + 4x3 = −3,−2x1 − 3x2 + 4x3 = 13,
we get values: x1 = −1, x2 = −1, x3 = 2.
Answer: x = {−1; −1; 2).
Example 4. Points Α(−1; − 2; 4) , Β(−4; −2; 0) , C(3; − 2;1) are vertices of a triangle АBС. Find Β and a projection of a vector AB on a vector BC .
Solution. We find coordinates of the vectors BA and BC that coincide with the corresponding sides of a triangle:
BA = {3; 0; 4} , BC = {7; 0;1} .
We find cosines of the angle ϕ between vectors |
BA and BC according to |
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the formula: |
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cos ϕ = |
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BA BC |
3 7 + 0 0 + 4 1 |
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25 |
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1 |
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| BA | | BC | = |
9 + 0 + 4 49 + 0 + 1 = |
25 2 = |
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whence ϕ = 45 . So Β = 45 . |
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We find a projection of the vector AB on the vector |
BC according to the |
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formula: |
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PrBC |
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AB = |
AB BC |
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−3 7 + 0 + (−4) 1 |
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−25 |
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5 2 |
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BC |
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Micromodule 5
CLASS AND HOME ASSIGNMENT
1. Vectors a and b form an angle ϕ = 60°. Knowing that | a |= 2 | b |= 5 , calculate: а) (4a − b)(2a + 3b) ; b) | a − 2b | .
2. The given vectors are a = {1; 4; −1} and b = {0; 3; −2} . Find: а) a scalar product (a + 3b)(3a − 2b) ;
b) an angle between vectors 2a + b and a − b ;
c) a projection of a vector −3a + 2b on a vector a − b .
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3. |
Vectors |
a = 2i − j + 3k , |
b = i + 4k , |
c = i − j + 4k are given. Find a |
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x , if x a = 8 , x b = 10 , |
x c = 8 . |
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Answers |
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1. а) 57; b) 2 |
21. 2. а)79; b) arccos |
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x = (2; 2; 2). |
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Micromodule 5
SELF-TEST ASSIGNMENTS
5.1. Evaluate:
5.1.1.а) (4a + 7b)(a − 2b) ; b) | 2a − 3b | , if | a |= 2 , | b |= 5 , ϕ = 60°.
5.1.2.а) (2a + 5b)(3a − 2b) ; b) | a − 3b | , if | a |= 3 , | b |= 4 , ϕ = 23π .
5.1.3. а) (3a + b)(2a + 3b) ; b) | 2a − 3b | , if | a |= |
2 , | b |= 3 , ϕ = |
π . |
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5.1.4. а) (4a + 3b)(a − 4b) ; b) | 2a + 3b | , if | a |= 1 , | b |= 6 , |
ϕ = |
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5.1.5. а) (4a + 5b)(a − 2b) ; b) | 2a − b | , if | a |= |
3, | b |= 1, |
ϕ = |
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5.1.6.а) (5a + 3b)(a + 2b) ; b) | a − b | , if | a |= 3, | b |= 4, ϕ = π3 .
5.1.7.а) (2a + 4b)(−3a − b) ; b) | a + 2b | , if | a |= 3, | b |= 2, ϕ = 23π .
5.1.8. а) (3a + 2b)(−a + 3b) ; b) | 2a + 3b | , if | a |= 4 |
2, | b |= 3, ϕ = |
π |
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5.1.9. а) (4a + b)(3a − b) ; b) | 2a + b | , if | a |= 1, | b |= 4, ϕ = |
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5.1.10. а) (6a + 5b)(a + b) ; b) | 4a − b | , if | a |= 2 |
3, | b |= 1, |
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5.1.11. а) (5a + b)(a + b) ; b) | a + b | , if | a |= 3, | b |= 2, ϕ = |
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5.1.12. а) (3a + 4b)(−3a − b) ; b) | a + 2b | , if | a |= 5, | b |= 2, |
ϕ = |
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5.1.13. а) (5a + 2b)(−a + 3b) ; b) | a + 4b | , if | a |= |
2, | b |= 3, ϕ = |
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5.1.14. а) (4a + 3b)(3a − 2b) ; b) | 2a + 5b | , if | a |= 2, | b |= 4, ϕ = |
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5.1.15. а) (2a + 5b)(a + b) ; b) | 2a − b | , if | a |= 3, | b |= 1, ϕ = |
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5.1.16.а) (a + b)(a − 2b) ; b) | a − b | , if | a |= 3, | b |= 4, ϕ = π3 .
5.1.17.а) (3a + 5b)(3a − b) ; b) | a + 2b | , if | a |= 3, | b |= 2, ϕ = 23π .
5.1.18. а) (3a + 2b)(−a + 3b) ; b) | a + 3b | , if | a |= 6 2, | b |= 3, |
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5.1.19. а) (−4a + 3b)(3a + 2b) ; b) | 2a + b | , if | a |= 2, | b |= 6, |
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5.1.20. а) (2a − 5b)(a − b) ; b) | 2a − 3b | , if | a |= 4 3, | b |= 1, |
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5.1.21. а) (a + b)(3a − 2b) ; b) | a − b | , if | a |= 5, | b |= 4, |
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5.1.22. а) (3a + b)(4a − b) ; b) | a + 2b | , if | a |= 3, | b |= 5, |
ϕ = |
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5.1.23. а) (3a + 5b)(−a − b) ; b) | a + b | , if | a |= 3 2, | b |= 2, |
ϕ = |
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5.1.24. а) (−a + 3b)(3a + b) ; b) | 2a + b | , if | a |= 2, | b |= 3, |
ϕ = |
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5.1.25. а) (2a − 3b)(a − b) ; b) | 2a − b | , if | a |= 3 3, | b |= 1, |
ϕ = |
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5.1.26.а) (a + b)(a − 2b) ; b) | 3a + b | , if | a |= 3, | b |= 4, ϕ = π3 .
5.1.27.а) (2a + b)(4a − b) ; b) | 5a + 2b | , if | a |= 3, | b |= 2, ϕ = 23π .
5.1.28. а) (3a + 5b)(−a + b) ; b) | a + b | , if | a |= 5 |
2, | b |= 2, |
ϕ = |
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5.1.29. а) (a + 3b)(a + b) ; b) | 2a + b | , if | a |= 4, | b |= 3, ϕ = |
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5.1.30. а) (2a − b)(a − b) ; b) | 2a − b | , if | a |= 5 |
3, | b |= 2, ϕ = |
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5.2. Find a dot product pq, an angle between vectors p and q and a projection of the vector p onto the vector q if:
5.2.1.p = 2a + 4b , q = 3a − b , a = {−1; 3; 4}, b = {−5;1; 2} .
5.2.2.p = 5a + 2b , q = 2a − 3b , a = {−2; 1; 2} , b = {−2; 4; 3} .
5.2.3.p = −2a + 7b , q = 3a − 2b , a = {2; − 3; 4} , b = {−1; − 1; 3} .
5.2.4.p = 3a + 4b , q = 2a − b , a = {−4; 3; 1}, b = {2; 2; 1} .
5.2.5.p = a − 4b , q = 2a − 3b , a = {−4; 3; 2} , b = {−2; 4; 5}.
5.2.6.p = −a + 3b , q = 2a + b , a = {3; 3;1} , b = {−2; − 3; − 2} .
5.2.7.p = 3a + 2b , q = 2a − 6b , a = {4; 3; 2} , b = {2; 1; 4} .
5.2.8.p = −3a + 7b , q = 3a − b , a = {−1; 0; 4} , b = {−3; 1; 2} .
5.2.9.p = −3a − 2b , q = 2a + b , a = {3; 2; 1} , b = {−1; − 2; − 2} .
5.2.10.p = 5a − 4b , q = 2a − b , a = {0; − 4; 4} , b = {−2; − 3; 3} .
5.2.11.p = −a + 4b , q = a + b , a = {−1; 4; 4} , b = {3;1; − 2} .
5.2.12.p = 3a − 2b , q = 2a − 3b , a = {−5; 1; 2} , b = {−3; 4; 3} .
5.2.13.p = −3a − 4b , q = a − b , a = {0; − 2; 2} , b = {−2; − 3; 0}.
5.2.14.p = −3a + 2b , q = 2a + 9b , a = {2; 3; 2} , b = {2; − 1; − 4} .
5.2.15.p = 5a − 3b , q = a − 3b , a = {−2; 4; 2} , b = {−3; 0; 3} .
5.2.16.p = −a + 3b , q = a + 2b , a = {3; − 2;1} , b = {−1; − 2; 2}.
5.2.17.p = 3a − 2b , q = 3a + 2b , a = {2; 1; 2} , b = {2; − 1; − 2} .
5.2.18.p = 3a − b , q = 4a − b , a = {−1; 2; 3}, b = {4; 1; − 3} .
5.2.19.p = −3a + b , q = 2a + b , a = {3; − 2; 0} , b = {1; − 2; 2} .
5.2.20.p = 2a − 3b , q = a − 2b , a = {0; − 2;1} , b = {−2; − 4; 0} .
5.2.21.p = −a − 2b , q = −a + 2b , a = {−1; 2;1} , b = {4; 1; − 2} .
5.2.22.p = 3a − 2b , q = a + 2b , a = {−2; 4; 3} , b = {−3; 0; 2}.
5.2.23.p = 2a − b , q = 3a + b , a = {0; − 2; 2} , b = {−2; − 3; 0}.
5.2.24.p = −3a + 2b , q = 4a + b , a = {2; 1; 4} , b = {2; − 1; − 3}.
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5.2.25.p = 2a − b , q = 5a − 2b , a = {−2; 4; 0} , b = {−4; 0; 2} .
5.2.26.p = −a − 2b , q = −a − 3b , a = {4; − 1; 0} , b = {1; − 2; 3} .
5.2.27.p = 3a − b , q = a + b , a = {2; 0; 4} , b = {3; − 1; − 2}.
5.2.28.p = a − b , q = 3a + 4b , a = {−1; 4;1} , b = {3; 1; − 2} .
5.2.29.p = 3a + 2b , q = 4a + b , a = {3; − 2; 0} , b = {1; − 4; 3} .
5.2.30.p = 2a − 3b , q = 2a − b , a = {0; − 1; 1} , b = {−2; − 1; 0} .
5.3. Find a vector x provided
5.3.1.x (i − j + k ) = 1 , x (i + 3k ) = 5 , x (i − 3 j + 6k ) = 2 .
5.3.2.x (i − j + 2k ) = 2 , x (i − j + 5k ) = 8 , x (i − 4 j + 8k ) = 2 .
5.3.3.x (i − j + 3k ) = 5 , x (i − 2 j + 7k ) = 11, x (i − 5 j + 10k ) = 2.
5.3.4.x (i − j + 4k ) = 10 , x (i − 3 j + 9k ) = 14 , x (i − 6 j + 12k ) = 2.
5.3.5.x (i − j + 5k ) = 17 , x (i − 4 j + 11k ) = 17 , x (i − 7 j + 14k ) = 2 .
5.3.6.x (i − j + 6k ) = 26 , x (i − 5 j + 13k ) = 20 , x (i − 8 j + 16k ) = 2 .
5.3.7.x (i − j + 7k ) = 37 , x (i − 6 j + 15k ) = 23 , x (i − 9 j + 18k ) = 2 .
5.3.8.x (i − j − k ) = 5 , x (i + 2 j − k ) = −1 , x (i − j + 2k ) = 2 .
5.3.9.x (i − j − 3k ) = 17 , x (i + 4 j − 5k ) = −7 , x (i + j − 2k ) = 2 .
5.3.10.x (i − j − 4k ) = 26 , x (i + 5 j − 7k ) = −10 , x (i + 2 j − 4k ) = 2 .
5.3.11.x (i − j − 5k ) = 37 , x (i + 6 j − 9k ) = −13 , x (i + 3 j − 6k ) = 2 .
5.3.12.x (i − j + k ) = 3 , x (i − j + 2k ) = 1 , x (3i + j + 10k ) = −1.
5.3.13.x (3i − 2 j + 3k ) = 10 , x (2i − j + 3k ) = 4 , x (5i + j + 12k ) = 0 .
5.3.14.x (2i − j + 2k ) = 7 , x (3i − j + 4k ) = 7 , x (7i + j + 14k ) = 1.
5.3.15.x (5i − 2 j + 5k ) = 18 , x (4i − j + 5k ) = 10 , x (9i + j + 16k ) = 2.
5.3.16.x (3i − 2 j + 3k ) = 11 , x (5i − j + 6k ) = 13 , x (11i + j + 18k ) = 3.
5.3.17.x (i − j ) = 10 , x (6i − j + 7k ) = 16 , x (13i + j + 20k ) = 4.
5.3.18.x (i + 2 j + k ) = 6 , x (2i + j + 2k ) = 9 , x (−3i + j + 4k ) = −4.
5.3.19.x (i + j + k ) = 5 , x (3i + j + 2k ) = 11 , x (−5i + j + 2k ) = −5 .
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5.3.20.x (3i + 2 j + 3k ) = 14 , x (4i + j + 3k ) = 14 , x (−7i + j) = −6 .
5.3.21.x (2i + j + 2k ) = 9 , x (5i + j + 4k ) = 17 , x (9i − j + 2k ) = 7 .
5.3.22.x (5i + 2 j + 5k ) = 22 , x (6i + j + 5k ) = 20 , x (11i − j + 4k ) = 8 .
5.3.23.x (3i + j + 3k ) = 13 , x (7i + j + 6k ) = 23 , x (13i − j + 6k ) = 9 .
5.3.24.x (5i + 2 j − 3k ) = −2 , x (i + j + k ) = 3 , x (i − j + 4k ) = 0 .
5.3.25.x (5i + j − 2k ) = 1 , x (2i + j) = 3 , x (i − 2 j + 5k ) = −1.
5.3.26.x (5i − k ) = 4 , x (3i + j − k ) = 5 , x (i − 3 j + 6k ) = −2 .
5.3.27.x (5i − j) = 7 , x (4i + j − 2k ) = 9 , x (i − 4 j + 7k ) = −3 .
5.3.28.x (5i − 2 j + k ) = 10 , x (5i + j − 3k ) = 15 , x (i − 5 j + 8k ) = −4.
5.3.29.x (5i + 4 j − 5k ) = −8 , x (−i + j + 3k ) = 9 , x (i + j + 2k ) = 2 .
5.3.30. x (5i + 5 j − 6k ) = −11, x (−2i + j + 4k ) = 15, x (i + 2 j + k ) = 3.
Micromodule 6
BASIC THEORETICAL INFORMATION
CROSS AND TRIPLE PRODUCTS
Cross product of two vectors, its algebraic and geometrical properties. The coordinate form. Triple product of three vectors, its algebraic and geometrical properties. The coordinate form. The condition of complanarity of three vectors.
Literature: [1, chapter 4], [4, part 3, p. 3.2], [6, chapter 2, §§ 5,6], [7, chapter1, § 4], [10, chapter1, § 2], [11, chapter1, § 2].
6.1. Cross product
Definition 1.39. The cross product of two vectors a and b is said to be a vector c satisfying the following conditions:
1) a module of the vector c is determined according to the formula: | c |=| a || b | sin ϕ, where φ is an angle between vectors a and b ;
2) the vector c is perpendicular to each of vectors a and b ;
3) the vectors a , b and c form a right-hand triple (that is, if we look from the end of the resulting vector c then the shortest turn from the first vector a to
the second vector b can be seen counterclockwise (Fig. 1.6)).
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A cross product designation is a × b.
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6.2. Cross product properties
Geometrical interpretation of a cross product. The module of cross product is
equal to the area of a parallelogram built on vectors a and b applied to the common beginning (Fig.1.7).
1) anticommutativity of multiplication:
a × b = −b × a ;
2)(λa)× b = λ(a × b) ; a × (λb) = λ (a × b) ;
3)a × (b + c) = a × b + a × c .
Note. If the coordinates of a triangle ABC are known, then its area can be determined according to the formula:
S ABC = 12 AB × AC .
6.3. Cross product of two vectors in coordinate form
Let vectors a = {ax , ay , az }, b = {bx , by , bz } be given by their coordinates in
Cartesian coordinate system. Then their cross product is determined according to the formula:
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or
a × b = (aybz − azby )i − (axbz − azbz ) j + (axby − aybx )k.
6.4. Triple product
Definition 1.40. A number abc equal to the dot product of the vector a × b by the vector c is called the mixed (triple) product of three vectors a , b and c.
abc = (a × b) c.
Properties:
1) If any two multipliers are interchanged in a triple product, then the triple product changes its sign. For example,
abc = −cba.
2)In case of a cycle interchange of multipliers a triple product is not changed.
3)Signs of cross and dot products can be interchanged in a mixed product:
(a × b) c = a (b × c) = (b × c) a.
4) Geometrical content of a mixed product. A module of a mixed product abc is
equal to the volume of parallelepiped, built on vectors |
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6) Condition of three vectors complanarity.
Vectors a, b and c are complanar if and only if abc = 0.
6.5. Triple product of three vectors in coordinate form
Let vectors a = {ax , ay , az }, b = {bx , by , bz }, c = {cx , cy , cz } be given by
their coordinates in Cartesian coordinate system.Then their triple product is determined according to the formula
ax ay az abc = bx by bz .
cx cy cz
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Micromodule 6
EXAMPLES OF PROBLEMS SOLUTION
Example 1. Vectors a = {2; 3; 1} and b = {−2; 4; 0} are given. Find the vector
(2a − b) × (3a + 2b). Solution. Firstly we find
2a − b = {4; 6; 2}− {−2; 4; 0} = {6; 2; 2},
3a + 2b = {6; 9; 3}+ {−4; 8; 0} = {2; 17; 3}.
Then
(2a − b)× (3a + 2b) = |
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Answer: −28i − 14 j + 98k.
Example 2. Find the area of ABC and volume of a pyramid, whose vertices are at the points A(2; –1; 1). B(5; 5; 4), C(3; 2; –1), D(4; 1; 3).
Solution. Let us find the coordinates of vectors AB, AC and AD, on which the pyramid is built:
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Volume of a pyramid is equal to 1/6 of parallelepiped volume, built on vectors
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