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2)
3)
5) Given: the coordinate system
the vector modulus
Find: OA
Solution:
OA=
OA =
Answer: OA = 13
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Let
1) From the point O we draw the vectors
2) If
3) If
The angle between two vectors
Definition:
Two vectors are called perpendicular if the angle between them is 90°.
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ABCD - is the square
AC ∩ BD = O
Find: the angles between the vectors
Calculation:
a) Since AC - is the diagonal of the square, we see that it divides the angle
b) Since ABCD - is a square, we see that the degree measure of the angle between the vectors
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ABCD - is the rhombus
BD = AB; AC ∩ BD = 0
Compute: the angle formed by the vectors
Solution:
a) By the definition of the rhombus ΔABD - is equilateral (AB = AD = BD).
Hence, all the angles in the triangle are equal to 60°. Then the angle between the vectors
b) Since the vectors
c) Since the vectors
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Definition:
A scalar product of two vectors (formula 1) is the product of the lengths of these vectors by the cosine of the angle between them.
Designation:
From the formula of the scalar product of vectors having used the cosine of the angle (1) it follows that:
1) the scalar product of vectors is greater than zero if the angle between vectors is less than 90°, i.e.
the scalar product of the vectors is less than zero if the angle between the vectors is greater than 90°, i.e.
2) If
3) If
The converse is also true, i.e. if
Conclusion:
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Given:
the vectors
the angle α = 90°
Find: scalar product of vectors
Solution:
Using the formula of the scalar multiplication of vectors by the cosine of the angle, we obtain
Answer:
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Given:
the triangle ΔABC is equilateral
Find: scalar product of vectors 1)
Solution: In an equilateral triangle, all angles are equal to 60°.
1)
2)
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Given:
the vectors
1) the angle α = 45°
2) α = 135°
Find: scalar or inner product of vectors
Solution:
1)
2)
Answer: 1) 3
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Given:
AB = a
the height BD
Find: scalar multiplication of vectors
1)
2)
Solution:
1)
2)
3)
Answer:1) -
Given:
BD ∩ AC = 0
BD = AB
1)
2)
Find: the value of the angle between the vectors
1)
Solution:
1) Consider the triangle ΔABC. This trianle is isosceles, since AB=BD.
Knowing that in the rhombus all sides are equal, we get ΔABD - is equilateral.
Then
By the property of the rhombus it follows that
Then the angle between the vectors
2) Since the sides are parallel and the vectors are co-directed:
BA || CD and
Consider a triangle ΔCBD is isosceles, since the two sides are equal: BD=BC.
By the definition of the rhombus, the triangle ΔCBD - is equilateral.
Hence, the angle
By the property of the rhombus, the angle
Then the angle between the vectors is
Answer: 1)
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Theorem:
If two vectors have coordinates
Proof:
Case 1.
Case 2.
If the vectors
We draw the vectors from an arbitrary point O.
Consider the triangle ΔOBA.
It is known that the cosine formula
c2 = a2 + b2 - 2ab • Cos α, we obtain the equality
AB2 = OB2 + OA2 - 2 • OB • OA • Cos α (3)
Using the values (*)
|
Using the formula to calculate the length of the vector having used its coordinates, we obtain
Since
|
Then it follows from the equality (4) that
(x2 - x1)2 + (y2 – y1)2 = x22 + y22 + x12 + y12 - 2
x22 -2 x2 x1 + x12 + y22 – 2 y2y1 + y12 = x22 + y22 + x12 + y12 - 2
-2 x2 x1– 2 y2y1 = - 2
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Consequences:
1) If the vectors are perpendicular, i.e.
2) By the definition of the scalar or inner product of two vectors (formula 1)
Cos α =
The formula to calculate the cosine of the angle having used the coordinates of the vectors:
To calculate the sine and the tangent of the angle between vectors by the cosine of an angle it’s used the reduction formulas and trigonometric functions.
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If
***
If
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Given:
the coordinates of the points
A(2;8), B(-1;5), C(3;1)
Solution:
Since each coordinate of the vector is equal to the difference of the corresponding coordinates of its end and the beginning
Using the formula to find the angles having used the coordinates of the vectors
Cos A =
Cos A =
Answer: Cos A =
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Given:
the angle between the vectors is equal to
magnitudes or lengths of vectors |
Find: the product of vectors (
Solution:
(
Answer: (
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Given:
the magnitude or the length of the vectors |
Find: product of vectors
Solution:
= 3
Answer:
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Given:
Find: multiplication of vectors
Solution:
Answer:
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Given:
Find: multiplication or product of vectors
Solution:
0 = 0 + (-3x)
3x = 0
x = 0
Answer: under x=0,
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Given:
the coordinates of the points
A(2;8), B(-1;5), C(3;1)
Find: the cosine of the angle of vectors
2) Cos C = ?
Solution:
1)
Since each coordinate of the vector is equal to the difference of the corresponding coordinates of its end and the beginning
Using the formula to find the angles having used the given vector coordinates
Cos B =
Cos B =
2)
Cos C =
Answer: Cos B =0, Cos C =
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Given:
Find: the length or the magnitude of the vector |
Solution:
Let us find the coordinates of the vector
Since the length of the vector is equal to the square root of the sum of the squares of its coordinates |
|
Answer: |
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Given:
ABCD is the rhombus
AB =
AC
Proof:
Since the figure ABCD is the rhombus and the parallelogram, then the vectors of the parallelogram
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Given:
the triangle ΔABC - is isosceles
AM is the median
Prove:
1) 4AM2 = AB2 + AC2 + 2AB • AC • Cos A
Proof:
1) Since the point M is the midpoint of the side BC, we see that
2
Therefore, (2
= AB2 + 2AB + 2AB • AC • Cos A + AC2 = AB2 + AC2 + 2AB • AC • Cos A
We get 4AM2 = AB2 + AC2 + 2AB • AC • Cos A
2) By the formula obtained above, it follows that
4CH2 = AC2 + BC2 + 2AC • BC • Cos C
Since the triangle ΔABC is isosceles, we see that AB = BC,
We get that 4CH2 = AC2 + BC2(=AB2) + 2AC • BC(=AB) • Cos C (= Cos A)
4CH2 = AC2 + AB2 + 2AC • AB • Cos A
4CH2 = 4AM2
2CH = 2AM | : 2
CH = AM
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Given:
ABCD is the convex quadrilateral
BD = d1 and AC = d2 are the diagonals
d1 ∩ d2 = O is the intersection point of the diagonals
Prove:
The area of the quadrilateral is equal to half the product of the diagonals by the sine of the acute angle between them
SABCD=
Proof:
The area of a quadrilateral is the sum of the areas of four triangles.
SABCD= S1 + S2 + S3 + S4 , where
S1 = SΔAOB ; S2 = SΔCOB ; S3 = SΔCOD ; S4 = SΔAOD
S1 =
S2 =
S3 =
S4 =
Adding S1 + S2 + S3 + S4, we obtain
SABCD=
+
Since OA+OC = AC, CO+OA = AC, BO + OD = BD we see that
SABCD=
The formula of the area of a convex quadrilateral:
SABCD=
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Given:
two vectors form an angle α = 150°,
the magnitudes or the lengths of vectors |
Solution:
BC2 = AB2 + AC2 - 2 AB • AC • Cos 150°
BC2 = 48 + 4 - 2 • 4
BC =
Answer: BC = |2
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Given:
The angles
a=24.6
Find:
Solution:
Using the law of sines
≈ 25.5
Answer:
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Given:
the lengths of vectors |
Find: the value of vectors
1) |
2) |
Solution: by the law of cosines
1)
AC2 = AB2 + BC2 - 2AB • BC • Cos 120°
AC2 = 25 + 64 - 80 • (- 0.5) = 129
2) BC2 = AB2 + AC2 - 2AB • AC • Cos 60°
BC2 = 89 - 80 • 0.5 = 49
BC = ±
Answer: |
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