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Vectors in a Plane and Space |
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Vectors
and a coordinate system, Cartesian vectors |
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Square of magnitude of
a vector |
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Scalar product of unit
vectors |
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Scalar or dot product
properties |
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Scalar product in the
coordinate system |
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Vectors examples |
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| Square of magnitude of
a vector |
| The scalar product of a vector with itself,
a ·
a
= a2
is the square of magnitude of a vector, that is |
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| thus, |
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in the coordinate plane, |
and |
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in 3D space. |
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| Scalar product of unit
vectors |
| The unit vectors,
i,
j and k,
along the Cartesian coordinate axes are orthogonal and their scalar products are, |
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| Scalar or dot product
properties |
| a)
k
· (a · b)
= (k · a) · b
= a · (k · b),
k
Î
R |
| b)
a
· b = b · a |
| c)
a
· (b + c)
= a · b + a
· c |
| d)
a
· a = a2
= | a |2 |
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| According to the definition of the dot product, from the
above diagram, |
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then |
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what confirms the distribution law. |
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The associative law does not hold for the dot product of more vectors, for example |
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a
· (b · c) is
not equal (a
· b) · c |
| since
a
· (b · c) is the vector
a
multiplied by the scalar b
· c,
while (a
· b) · c
is the vector c
multiplied by the scalar a
· b. |
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| Scalar product in the
coordinate system |
| The scalar product can be expressed in terms of the
components of the vectors, |
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- the scalar product in three-dimensional coordinate system, |
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- the scalar product in the coordinate plane. |
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| Vectors
examples |
| Example:
Applying the scalar product, prove
Thales’ theorem which states that an angle inscribed in a |
| semicircle is a right angle. |
| Solution:
According to diagram in
the right figure |
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| since square of a vector equal to square of its length, |
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| thus,
a
and b are orthogonal vectors as |
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| Example:
Prove the law of cosines used in the trigonometry of oblique triangles. |
| Solution:
Assuming the directions of vectors as in
the right diagram |
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Using the scalar product and substituting square of vectors
by square of their lengths, obtained is |
| a2 =
b2 + c2 -
2bc · cosa
- the law of cosines. |
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| In the same
way, |
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| Example:
Determine a parameter l so the given vectors,
a
= -2i
+ l
j
-
4k
and b
= i -
6 j + 3k
to be perpendicular. |
| Solution:
Two vectors are perpendicular if their scalar product is zero, therefore |
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| Example:
Find the scalar product of vectors, a
= -3m
+ n
and b
= 2m -
4n if
| m |
= 3 and
| n |
= 5 , and the
angle between vectors, m
and n is
60°. |
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| Example:
Given are vertices, A(-2,
0, 5), B(-3, -3,
2), C(1,
-2,
0) and
D(2,
1, 3), of a parallelogram, |
| find the
angle subtended by its diagonals as is shown in the diagram below. |
| Solution:
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| Example:
Given are points, A(-2,
-3,
1), B(3,
-1,
-4), C(0,
2, -1) and
D(-3,
0, 2), determine the scalar and
vector components of the vector AC
onto vector BD. |
| Solution:
The scalar component of the vector
AC
onto vector BD, |
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The vector component of the vector
AC
in the direction of the vector BD
equals the product of the scalar component ACBD
and the unit vector BD°
that is
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