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Mathematics (Математика). Учебное пособие

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101

=--

ì ï í

ï î

zyx

=++

1432

zyx

=++

16234

zyx

115 --

D =

321

= 5(4 – 9) + (2 – 12) – (3 – 8) = –25 – 10 + 5 = –30;

234

110 --

= D1/D = 1;

3214

= (28 – 48) – (42 – 32) = –20 – 10 = –30.

2316

105 -

3141

= 5(28 – 48) – (16 – 56) = –100 + 40 = –60.

2164

= D2/D = 2;

015 -

= D3/D = 3.

1421

= 5( 32 – 42) + (16 – 56) = –50 – 40 = –90.

1634

Solve and check yourself using the answer:

=+-+

xxxx

4321

=-+-

1232

xxxx

4321

Answer: {1, 2, 3, 4}.

=+-

xxx

431

=-+-

xxxx

4321

ì ï

ï í

ï ï

2. Solve by the method of Kramer the following system of equations:

x1 + x2 + x3 + x4 = 5,

x1 + 2x2 – x3 + 4x4 = –2,

2x1 – 3x2 – x3 – 5x4 = –2,

3x1 + x2 +2x3 + 11 x4 = 0.

Answer: {1, 2, 3, –1}

102

Topic 8. А analytical geometry

с ab

uuurrr

(0,1,1);(1,1,0)

(2,2,0)

(2,3,1)

с ab

uuurrr

(0,1,1);(1,1,0)

с ab

uuurrr

(0,1,1);(1,1,0)

с ab

uuuruurr

(0,1,1);(1,1,0)

с ab

uuurrr

(0,1,1);(1,0,0)

с ab

uuurrr

(0,1,3);(1,1,0)

Introduction. This workshop is devoted to the discussion of the following concepts: different equations of straight lines on the plane and in space, conditions parallelism and perpendicularity direct in space, the cross product and its geometrical meaning.

Linear operations on vectors in the coordinates.

Given the set of vectors in a rectangular coordinate system

),,,();,,(

zyxbzyxa linear operations on them in the coordi-

BBBAAA

nates have the form:

);;();;;(

zyxazzyyxxcba

, if

aaaa

AAABABABA

=×+++=+

Example 1 : Find the coordinates of the vector

ab.

Solution: Find the coordinates of the vector

dinate of a vector b by a number 2. Get the vector

corresponding coordinates of the vectors, we obtain the vector

Solve and check yourself using the answer:

1. Find the coordinates of the vector

ab--

2. Find the coordinates of the vector

ab--

3. Find the coordinates of the vector

ab--

4. Find the coordinates of the vector

ab--

5. Find the coordinates of the vector

ab--

Answer (-2, 1, 1)

. Answer (2, -3, 1)

. Answer (-1, -1, 2)

. Answer (-2,- 1, 1)

. Answer (-2, 1, 3)

. Multiply each coor-

. Adding the

, if

103

Solve:

с ab

uuurrr

(0,1,1);(1,1,0)

с ab

uuuruurr

(0,1,1);(1,1,0)

с ab

uuurrr

(0,1,15);(1,7,0)

с ab

uuurrr

(0,1,8);(1,1,4)

с ab

uuurrr

(0,1,1);(1,1,0)

136

с ab

uuurrr

(0,1,1);(1,1,0)

с ab

uuuruurr

(0,1,1);(1,1,0)

с ab

uuuruurr

(0,1,1);(1,1,0)

с ab

uuurrr

(0,1,7);(3,1,0)

с ab

uuuruurr

(0,1,1);(1,1,0)

126

с ab

uuuruuurr

(0,1,1);(1,1,0)

с ab

uuurrr

(0,1,5);(7,1,0)

с ab

uuurrr

(0,1,1);(1,1,0)

с ab

uuurrr

(0,6,1);(1,6,0)

1. Find the coordinates of the vector

ab--

2. Find the coordinates of the vector

ab--

, if

3. Find the coordinates of the vector

ab--

4. Find the coordinates of the vector

ab--

5. Find the coordinates of the vector

ab--

6. Find the coordinates of the vector

ab--

7. Find the coordinates of the vector

ab--

8. Find the coordinates of the vector

ab--

9. Find the coordinates of the vector

ab--

10. Find the coordinates of the vector

ab--

11. Find the coordinates of the vector

ab--

12. Find the coordinates of the vector

ab-

, if

13. Find the coordinates of the vector

ab--

14. Find the coordinates of the vector

ab--

, if

104

15. Find the coordinates of the vector

136

с ab

uuuruuurr

(0,1,1);(1,1,0)

rrr

rrrrr

rrr

ab--

The scalar product of vectors.

Definition. Scalar product of vectors equal to the product of the lengths of these vectors by the cosine of the angle between them.

If we consider the vectors gular coordinate system, then

Using the resulting equality, we obtain the formula to calculate the angle between the vectors:

and

= ï

ïï

= xa xb + ya yb + za zb;

=jcos

ïcosj

, if

is called a number

),,();,,(

zyxbzyxa

in the rectan-

bbbaaa

zzyyxx

bababa

Example 2. To calculate (5

.,3,2 baba

^==

- 5

+ 6

Because

Example 3. To find the angle between vectors

,32 kjia

++=

that is

= (1, 2, 3),

= 6 + 8 – 6 = 8:

cosj =

Example 4. Find the scalar product (3

5614

rrr

= (6, 4, -2)

=== bаba

+ 3

)(2

–

- 3

= 10

kjib

246 -+=

14142

.3/^,6,4

– 2

), if

)×(5

1327403

=-=- ba

0,9,4

=×==×==× babbbaaa

5641636;14941 =++==++= ba

– 6

), if

arccos;

====

, if

105

rrr

rrrrrrr

rrr

rrrrrrrrrrr

rrrrr

rrrrrrrrrrr

- 18

- 10

+ 12

+ 12×36 = 240 – 336 + 432 = 672 – 336 = 336.

Example 5. To find the angle between vectors

++= kjib

that is

= (3, 4, 5),

= 12 + 20 – 15 =17 :

cos28

,543 kjia

bbaa

= (4, 5, -3)

354 -+=

6428161512

and

5092516;5025169 =++==++= ba

+×××-×=+-

, if

cosj =

Example 6. For what m vectors jima are perpendicular?

Example 7. Find the scalar product

5050

= (m, 1, 0);

(

)(

cba

432 ++

) =

cba

765 ++

arccos;

= (3, -3, -4)

+= and

^^^,3,2,1

211815141210

432 ++

====== cbcabacba

cba

+×+×+×+×+×+× cbbbbacabaaa

1;033 =Þ=-=× mmba

and

kjib

433 --=

, if

cba

765 ++

10282420 =×+×+×+ cccbac

= 10 +

ccbbcbcabaaa

×+×+×+×+×+× 2818453427

+ 27 + 51 + 135 + 72 + 252 = 547.

Example 8. To find the angle between vectors a = 2m+4n и b = m-n, where m and n – there are unit vectors and the angle between m and n is equal to 120 degrees (120о).

Solution: let ab = (2m+4n) (m-n) = 2 m

= 2 – 4+2cos120o = – 2 + 2(-0.5) = -3; a =

2 – 4n2

+2mn =

; a2 = (2m+4n) (2m+4n) =

= 4 m2 +16mn+16 n2 = 4+16(-0.5)+16=12, so a =

. b = b2;

106

b2 = (m-n)(m-n) = m2 -2mn+ n2 = 1-2(-0.5)+1 = 3, so b = 3.

®®®

Finally we have: cos j =

= -1/2, Þ j = 120

123

Solve and check yourself using the answer:

1. The cosine of the angle between vectors

Answer:

2. The cosine of the angle between vectors

Answer:

3. If

ç è

Answer:

4. If

æ ç

ç è

Answer:

– unit vectors, и

®®®®

+×

baba 43

– unit vectors, и

+×÷÷

ç è

is…

÷ ø

®®®®

baba 43

is…

÷ ø

®®

{2;1;3} и

{-2;1;3} и

then the scalar product

,3=-

,3=+

then the scalar product

{0;1;1} is…

5. The work force

++= kjiF 3

for the rectilinear movement of a

material point from position A (-1; 2; 0 ) in position (2; 1; 3 ) will be equal…

Answer: 3

6. The work force

++= kjiF 3 for the rectilinear movement of a

material point from position А (-1; 3; -3 ) in position В (2; 1; 3 ) will be equal…Answer: 3

7. сos угла между векторами

Answer:

{2;1;0} и

{0;-1;1} равен

107

8. The cosine of the angle between vectors

rrr

pjj

££³

rrr

Answer:

9. If

æ ç è

æ ç

ç è

the following conditions:

÷ ø

Answer: -1

10. If 3и,3,2 =+==

Answer : -3.

Vector product of two vectors.

Definition. The vector product of vectors is called a vector satisfying

®®®®

baba

®®®®

æ ç è

is…

-×

baba 2

÷ ø

®®®®

baba then the scalar product

®®®®

ç è

ö ÷

+×÷÷

baba 2

is…

÷ ø

, where j s the angle between vectors

sinbac

×=

{2;-1;0} и

then the scalar product

3и,3,2 =+==

{0;1;1} is…

and

2) the vector

Denoted:

Fig.1 – graphic representation of vector product of vectors

0;0sin

is orthogonal vectors

form the right set of vectors.

bac

´=

],[ bac

Fig.1

and

108

If the vectors

rrr

(xa, ya, za) и

(xb, yb, zb) are in the Cartesian rectan-

gular coordinate system with unit vectors

kji

, then

zyx

aaa

zyx

bbb

Geometric meaning of vector product of vectors is the area of the

parallelogram constructed on the vectors

Example 9. Find the vector product of two vectors kjia

and kjib

So,

= (2, 5, 1);

=´ 717

32 -+= .

kji

152

321

= (1, 2, -3)

++= 52

kjikji

-+-=+

Example 10. Knowing the vectors AB(-3,-2,6) and BC(-2,4,4), calculate the length of the altitude AD of a triangle ABC.

Solution. Denoting the area of the triangle ABC over S, we get:

S = 1/2 BC AD. Then AD=2S/BC, BC =

2 2 2

( )- + +2 4 4

= 6,

S = 1/2 çAB ´ACç. AC=AB+BC, so, the vector AC has coordinates

i j k

AC(-5,2,10). AB´AC =

-3 -2 6

= i (-20 -12) – j (30 -30) + k (- 6 – 10) =

-5 2 10

2 2

16 2 1

= -16(2`i +`k ). çAB´ACç =

AD =

165

( )+

= 165; S = 85,

Solve and check yourself using the answer:

1. To find the base area BCD pyramid АBCD if the vertices have co-

ordinates A(0; 0; 1), B(2; 3; 5), C(6; 2; 3), D(3; 7; 2).

109

Answer:

rrr

2. To calculate the area of a triangle with vertices А(2, 2, 2), В(4, 0, 3),

С(0, 1, 0).

Answer:

3. To find the area of a parallelogram constructed on the vectors

Answer: 4.

The mixed product of vectors. (Compositional product of vectors)

Definition. Mixed product of vectors

equal to the scalar product of a vector

product of the vectors

Denoted:

2/510

(units 2)

baba

++ 3;3

, if

××

or (

cba

(units 2).

and

.30^;1

=== baba

by a vector equal to the vector

and

, is the number

The mixed product

parallelepiped constructed on the vectors

Fig.2 – graphic representation of mixed product of vectors

),,(

cba =

Volume of a triangular pyramid formed by the vectors equal

is numericaly equal to the volume of the

cba

××

and

Fig.2

zyx

111

zyx

222

zyx

333

and

110

(

)

143(2(83)3(212)4(116))

---

==-=---+-+---=

cba

Example 11. To prove that the points A(5; 7; 2), B(3; 1; -1), C(9; 4; -

4), D(1; 5; 0) lie in the same plane.

)1;6;2(

--=

)2;3;4(

--=

)2;2;4(

--=

Find the coordinates of the vectors:

Find the mixed product of vectors obtained:

162

234

=×× ADACAB

224

162

0150

0100

160

0150

0100

Thus, the vectors are coplanar, hence the points A, B, C and D lie in the same plane.

Example 12. To find the volume of pyramids and the length of the altitude, placed on the face BCD, if the vertices have coordinates A(0; 0;

1), B(2; 3; 5), C(6; 2; 3), D(3; 7; 2).

)4;3;2(

---=

Find the coordinates of the vectors:

)3;4;1(

)2;1;4(

--=

The volume of the pyramid

234

412

(223068)20(units)

=++=

To find the length of the altitude of the pyramid will first find the base area BCD.

kji

BCBD

-=´

341 kjikji

214

222

.171011)161()122()38(

---=--++----=

510289100121171011

=++=++=´ BCBD

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