vectors. definitions scalar – magnitude only vector – magnitude and direction i am traveling at...
TRANSCRIPT
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Vectors
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Definitions
Scalar – magnitude only
Vector – magnitude and direction
I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction.
I am traveling north at 65 mph – speed is a vector. It has both magnitude and direction.
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Graphical Representation
The vector V is denoted graphically by an arrow. The length of the Arrow represents the
magnitude of the vector. The direction of the arrow represents the
direction of the vector. V
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Vector Representation
A vector may be represented by a letter with an arrow over it, e.g.
VA vector may be represented by a letter in bold
faced type, e.g. VFor ease of typing or word processing, vectors
will be represented by the bold faced type.
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Components of a Vector
In two dimensions, a vector will have an x-component (parallel to the X-axis) and a y- component (parallel to the Y-axis).
In vector terms, V = Vx + Vy
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Graphically the vector is broken into its components as follows:
Y Vy V
X Vx
V = Vx + Vy
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Vector Addition
Vectors may be added graphically by placing the tail of the second vector at the head of the first vector and then drawing a new vector from the origin to the head of the second vector.
B C C = A + B A
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The components of a vector add up to form the vector itself, i.e.
V = Vx + Vy in 2 dimensions
V Vy
Vx
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Or in three dimensions
V = Vx + Vy + Vz
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Components in 3-d
Z
V Y Vx Vz
Vy
X
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When we resolve a vector into its components, e.g.
V = Vx + Vy
the magnitude of the two component vectors is given by the relations
|Vx | = |V| cos ϴ
| Vy | = |V| sin ϴ
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The Pythagorean theorem then gives a relation between the magnitudes of the x and y components, i.e.
|V|2 = |Vx |2 + |Vy |2 in 2-dimensions
And |V|2 = |Vx |2 + |Vy |2 + |Vz |2 in 3-d
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Use of Unit Vectors i j k
It is convenient to define three unit vectors i parallel to the X axisj parallel to the Y axisk parallel to the Z axis
And to express the components of the vector in terms of a scalar times the unit vector along that axis. Vx = Vxi where Vx = | Vx |
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Z
k j Y I X
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Dot or Scalar Product
The dot or scalar product of two vectors A · BIs a scalar quantity.
A = Axi + Ayj + Azk
B = Bxi + Byj + Bzk
A · B = |A||B cos ϴ
A · B = AxBx + AyBy + AzBz
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Example of Dot Product
Consider A = 2i + j – 3k B = -i - 3j + kA·B = (2)(-1) + (1)(-3) + (-3)(1) = -2-3-3 = -8
|A| = [22 + 12 + (-3)2]1/2 = [14]1/2 = 3.74|B| = [(-1)2 + (-3)2 + (1)2]1/2 = [11]1/2 = 3.32A·B = (3.74)(3.32) cos ϴ = 12.41 cos ϴ = - 8cos ϴ = - 8/12.41 = - 0.645ϴ = cos-1 (- 0.645) = 130.2⁰
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Given a vector A = 2i + 3j – k, we can find a vector C that is normal to A by using the fact that the dot product A·C = 0 if A is normal to C.C = Cxi + Cyj + Czk
A·C = 2Cx + 3Cy – Cz = 0
You now have three unknowns and only one equation.
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• How many equations do you need to solve for three unknowns?
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• I can solve for three unknowns with only this one equation!
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A·C = 2Cx + 3Cy – Cz = 0
Let Cy = 1
2Cx + 3 – Cz = 0
Let Cx = 1
2 + 3 – Cz = 0
Cz = 5
So the vector C = i + j +5k is normal to A.
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The order of the vectors in the dot product does not affect the dot product itself, i.e.
A · B = B · A
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Cross (Vector) Product
The cross product of two vectors produces a third vector which is normal to the first two vectors, i.e.
C = A x BSo vector C is normal to both A and B.
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Calculation of C = A x B
If the vectors A and B are A = Axi + Ayj + Azk
B = Bxi + Byj + Bzk
then i j k C = Ax Ay Az
Bx By Bz
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To evaluate the determinant, it is convenient to write the i and j columns to the right and multiply along each of the diagonals 1, 2, and 3 and add them, then multiply along 4, 5, and 6 and subtract them.
1 2 3 4 5 6 i j k i j C = Ax Ay Az Ax Ay
Bx By Bz Bx By
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C = AyBzi + AzBxj + AxByk - AzByi – AxBzj - AyBxk
OrC = (AyBz – AzBy)i + (AzBx – AxBz) j + (AxBy – AyBx)k
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Example of cross product
Calculate C = A x B where A = i + 2j - 3k
B = 2i - 3j + k i j k
C = 1 2 -3 2 -3 1
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C = (2)(1)i + (-3)(2)j + (1)(-3)k - (-3)(-3)i – (1)(1)j – (2)(2)k
C = (2 – 9)i + (- 6 – 1)j + (- 3 -4)kC = -7i -7j -7k
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To check that we have not made any mistakes in calculating the cross product, we can calculate the dot product C·A which should be equal to 0 since vector C is normal to vectors A and B.
C·A = (-7)(1) + (-7)(2) + (-7)(-3) = -7 – 14 +21 = 0So we have not made any mistakes in calculating
C.
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The cross product is different if the order is reversed, i.e.
A x B = CBut
B x A = - CB x A = - A x B
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When we look at the vector C = -7i – 7j – 7k
It has the same direction as the vectorC’ = - i – j – k
But a different magnitude.|C| = 7 |C’|
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Unit Vectors
To create a unit vector g in the same direction as G, simply divide the vector by its own magnitude, e.g.
g = G/|G|If G = 2i + j – 3kThen |G| = [22 + 12 + (-3)2]1/2 = [4 + 1 + 9]1/2
= [14]1/2 = 3.74g = (2/3.74)i + (1/3.74)j – (3/3.74)k
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Useful Information
A · A = |A|2
i · i = 1 j · j = 1 k · k = 1
A x A = 0i x i = 0 j x j = 0 k x k = 0i x j = k j x k = i k x i = jj x i = -k k x j = -i i x k = -j