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Micromodule 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 =

= 9 ,

b 2 =

2 = 16 ,

ab =

cos120° = 3 4 (−0, 5) = −6 are true, then

A = 3 9 + 4 (−6) − 4 16 = −61.

b) | a − b | = (a − b)2 = a2 − 2ab + b 2 = 9 − 2 (−6) + 16 = 37 .

Example 2. The given vectors are a = {1; 2; −2} and b = {3;

3; −4} . Find:

а) scalar product (4a + 3b)(a − 2b) ;

b) angle between vectors a + b and a − b .

Solution.

а)

4a + 3b = {4; 8; −8} +{9; 9; −12} = {13; 17;

−20} ,

a − 2b = {1; 2;

−2} −{6; 6; −8} = {−5; −4; 6} ,

(4a + 3b)(a − 2b) = 13 (−5) + 17 (−4) − 20 6 = −233 ;

c = a + b = {4;

−6},

d = a − b = {−2; −1;

2} ,

cos ϕ =

c d

4 (−2) + 5 (−1) − 6 2

−25

| c || d |

42 + 52 + 62 22 + 12 + 22

3 77

Therefore ϕ = π − arccos

Example 3. The given vectors are a = i − j + 4k ,

b = 6i + 5 j + 4k , c =

= −2i − 3 j + 4k . Find vector x satisfying the equalities

x a = 8 ,

x b = −3 and

x c = 13 .

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

the formula:

cos ϕ =

BA BC

3 7 + 0 0 + 4 1

| BA | | BC | =

9 + 0 + 4 49 + 0 + 1 =

25 2 =

whence ϕ = 45 . So Β = 45 .

We find a projection of the vector AB on the vector

BC according to the

formula:

PrBC

AB =

AB BC

−3 7 + 0 + (−4) 1

−25

= −

5 2

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
vector	x , if x a = 8 , x b = 10 ,		x c = 8 .
			Answers
1. а) 57; b) 2		21. 2. а)79; b) arccos		3	. 3.	x = (2; 2; 2).
1. а) 57; b) 2		21. 2. а)79; b) arccos			. 3.	x = (2; 2; 2).
			47

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 .
					4
5.1.4. а) (4a + 3b)(a − 4b) ; b) \| 2a + 3b \| , if \| a \|= 1 , \| b \|= 6 ,		ϕ =	π .
			3
5.1.5. а) (4a + 5b)(a − 2b) ; b) \| 2a − b \| , if \| a \|=	3, \| b \|= 1,	ϕ =	π	.
5.1.5. а) (4a + 5b)(a − 2b) ; b) \| 2a − b \| , if \| a \|=	3, \| b \|= 1,	ϕ =	6	.
			6

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				.
5.1.8. а) (3a + 2b)(−a + 3b) ; b) \| 2a + 3b \| , if \| a \|= 4	2, \| b \|= 3, ϕ =							4				.
				π				4
5.1.9. а) (4a + b)(3a − b) ; b) \| 2a + b \| , if \| a \|= 1, \| b \|= 4, ϕ =				π		.
			3			.	π
			3
5.1.10. а) (6a + 5b)(a + b) ; b) \| 4a − b \| , if \| a \|= 2	3, \| b \|= 1,			ϕ =					.
5.1.10. а) (6a + 5b)(a + b) ; b) \| 4a − b \| , if \| a \|= 2	3, \| b \|= 1,			ϕ =					.
		π					6
5.1.11. а) (5a + b)(a + b) ; b) \| a + b \| , if \| a \|= 3, \| b \|= 2, ϕ =		π			.
					.		2π
			3
5.1.12. а) (3a + 4b)(−3a − b) ; b) \| a + 2b \| , if \| a \|= 5, \| b \|= 2,			ϕ =							.
			ϕ =				3			.
							3
5.1.13. а) (5a + 2b)(−a + 3b) ; b) \| a + 4b \| , if \| a \|=	2, \| b \|= 3, ϕ =							π			.
5.1.13. а) (5a + 2b)(−a + 3b) ; b) \| a + 4b \| , if \| a \|=	2, \| b \|= 3, ϕ =							4			.
								4

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5.1.14. а) (4a + 3b)(3a − 2b) ; b) \| 2a + 5b \| , if \| a \|= 2, \| b \|= 4, ϕ =			π	.
			3	.
			3
5.1.15. а) (2a + 5b)(a + b) ; b) \| 2a − b \| , if \| a \|= 3, \| b \|= 1, ϕ =	π	.
	6	.
	6

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,					ϕ =								π			.
					ϕ =											.
											4
5.1.19. а) (−4a + 3b)(3a + 2b) ; b) \| 2a + b \| , if \| a \|= 2, \| b \|= 6,				ϕ =							π			.
				ϕ =							3			.
											3
5.1.20. а) (2a − 5b)(a − b) ; b) \| 2a − 3b \| , if \| a \|= 4 3, \| b \|= 1,				ϕ =							π				.
				ϕ =											.
											6
5.1.21. а) (a + b)(3a − 2b) ; b) \| a − b \| , if \| a \|= 5, \| b \|= 4,	ϕ =		π		.
	ϕ =		3		.
			3
5.1.22. а) (3a + b)(4a − b) ; b) \| a + 2b \| , if \| a \|= 3, \| b \|= 5,	ϕ =			2π				.
	ϕ =			3				.
				3
5.1.23. а) (3a + 5b)(−a − b) ; b) \| a + b \| , if \| a \|= 3 2, \| b \|= 2,			ϕ =							π		.
			ϕ =									.
				4
5.1.24. а) (−a + 3b)(3a + b) ; b) \| 2a + b \| , if \| a \|= 2, \| b \|= 3,		ϕ =				π			.
		ϕ =							.
				3
5.1.25. а) (2a − 3b)(a − b) ; b) \| 2a − b \| , if \| a \|= 3 3, \| b \|= 1,		ϕ =					π				.
		ϕ =									.
				6

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 =			π		.
5.1.28. а) (3a + 5b)(−a + b) ; b) \| a + b \| , if \| a \|= 5	2, \| b \|= 2,	ϕ =					.
		π		4
5.1.29. а) (a + 3b)(a + b) ; b) \| 2a + b \| , if \| a \|= 4, \| b \|= 3, ϕ =		π	.
		3	.
		3
5.1.30. а) (2a − b)(a − b) ; b) \| 2a − b \| , if \| a \|= 5	3, \| b \|= 2, ϕ =			π		.
5.1.30. а) (2a − b)(a − b) ; b) \| 2a − b \| , if \| a \|= 5	3, \| b \|= 2, ϕ =			6		.
				6

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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.

Fig. 1.6

Fig. 1.7

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:

	i	j	k
	i	j	k
a × b =	ax	ay	az
	bx	by	bz

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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		a, b and	c applied to the
common beginning, that is V =\| abc \| .
Note. The volume of a pyramid, built on vectors a,		b and c	is equal to 1/6 of
the parallelepiped volume.
5) If	abc > 0, then vectors a, b and c form a right-hand triple and if
abc < 0,	then they form a left-hand triple.

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) =	i	j	k	=	2	2	i +	6	2	j +	6	2	k =
	i	j	k
	6	2	2
	2	17	3		17	3		2	3		2	17
	2	17	3

= −28i − 14 j + 98k.

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:

= {3; 6; 3},

= {1; 3; − 2},

= {2; 2; 2}.

The area of ABC is determined according to the formula

S ABC = 1

Therefore

= (−12 − 9)i − (−6 − 3) j + (9 − 6)k =

3 6 3

−2

= −21i + 9 j + 3k = 3(−7i + 3 j + k);

S ABC = 1

= 1 3 (−7)2

+ 32 + 12 = 3

59.

Volume of a pyramid is equal to 1/6 of parallelepiped volume, built on vectors

AB, AC and AD, that is,

VABCD = 1 mod	3	6	3	= 1 mod(−18)
VABCD = 1 mod	1	3	−2	= 1 mod(−18)	= 3.
6	2	1	2	6
	2	1	2

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5.1.3. а) (3a + b)(2a + 3b) ; b) \| 2a − 3b \| , if \| a \|=	2 , \| b \|= 3 , ϕ =				π .
					4
5.1.4. а) (4a + 3b)(a − 4b) ; b) \| 2a + 3b \| , if \| a \|= 1 , \| b \|= 6 ,		ϕ =	π .
			3
5.1.5. а) (4a + 5b)(a − 2b) ; b) \| 2a − b \| , if \| a \|=	3, \| b \|= 1,	ϕ =	π	.
5.1.5. а) (4a + 5b)(a − 2b) ; b) \| 2a − b \| , if \| a \|=	3, \| b \|= 1,	ϕ =	6	.
			6

5.1.8. а) (3a + 2b)(−a + 3b) ; b) \| 2a + 3b \| , if \| a \|= 4	2, \| b \|= 3, ϕ =							π				.
5.1.8. а) (3a + 2b)(−a + 3b) ; b) \| 2a + 3b \| , if \| a \|= 4	2, \| b \|= 3, ϕ =							4				.
				π				4
5.1.9. а) (4a + b)(3a − b) ; b) \| 2a + b \| , if \| a \|= 1, \| b \|= 4, ϕ =				π		.
			3			.	π
			3
5.1.10. а) (6a + 5b)(a + b) ; b) \| 4a − b \| , if \| a \|= 2	3, \| b \|= 1,			ϕ =					.
5.1.10. а) (6a + 5b)(a + b) ; b) \| 4a − b \| , if \| a \|= 2	3, \| b \|= 1,			ϕ =					.
		π					6
5.1.11. а) (5a + b)(a + b) ; b) \| a + b \| , if \| a \|= 3, \| b \|= 2, ϕ =		π			.
					.		2π
			3
5.1.12. а) (3a + 4b)(−3a − b) ; b) \| a + 2b \| , if \| a \|= 5, \| b \|= 2,			ϕ =							.
			ϕ =				3			.
							3
5.1.13. а) (5a + 2b)(−a + 3b) ; b) \| a + 4b \| , if \| a \|=	2, \| b \|= 3, ϕ =							π			.
5.1.13. а) (5a + 2b)(−a + 3b) ; b) \| a + 4b \| , if \| a \|=	2, \| b \|= 3, ϕ =							4			.
								4