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Vectors (5)Vectors (5)•Scaler ProductScaler Product
•Angle between linesAngle between lines
Angle between VectorsThe angle can be measured if they are placed ….
“Head-to-Head” “Tail-to-Tail”
-1 7[ ] 4
3[ ]Angle between Vectors - example
a = b =
|a| = ((-1)2 + 72)
|a| = 50
|b| = (42 + 32)= 5
|b - a| = (52 + -42)= 41
b - a = 5 -4[ ]
-1 7[ ] 4
3[ ]Angle between Vectors - example (2)
a = b =
|a| = 50|b| = 5
|b - a| = 41
How can you find the angle now?
a2 = b2 + c2 - 2bc cos A
The Cosine Rule
A B
C
ab
c
anglessides
… is used for working out angles and sides in non-right angled triangles
It is ….
a2 = b2 + c2 - 2bc cos AUsing the Cosine Rule ...
A
B C
anglessides 505
41
41 = 50 + 25 - 2 x 50 x 5 cos A
= 61.3o
3
2
1
a
a
a
a
3
2
1
b
b
b
b
Angle between Vectors - general case
33
22
11
3
2
1
3
2
1
ab
ab
ab
a
a
a
b
b
b
abc
|a| = (a12 + a2
2 + a32)
|b| = (b12 + b2
2 + b32)
|c| = ((b1-a1)2 + (b2-a2)2 +(b3-a3)2)
|c|2 = (b1-a1)2 + (b2-a2)2 +(b3-a3)2
|c|2 = a12 + a2
2 + a32 + b1
2 + b22 + b3
2 -2(a1 b1+ a2 b2+
a3 b3)
Expand and rearrange
|c|2 = |a|2 + |b|2 -2(a1 b1+ a2 b2+ a3 b3)
a2 = b2 + c2 - 2bc cos A
Generalizing
anglessides |a||b|
|c|=|b - a|
|c|2 = |a|2 + |b|2 - 2|a||b| cos
c=b - a
Cosine Rule
|c|2 = |a|2 + |b|2 -2(a1 b1+ a2 b2+ a3 b3)|a|2 + |b|2 -2(a1 b1+ a2 b2+ a3 b3) = |a|2 + |b|2 - 2|a||b| cos -2(a1 b1+ a2 b2+ a3 b3) = - 2|a||b| cos
a1 b1+ a2 b2+ a3 b3 = |a||b| cos
Generalizing (cont.)
a1 b1+ a2 b2+ a3 b3 = |a||b| cos cos = a1 b1+ a2 b2+ a3 b3
|a||b|
The Scaler Product
a1 b1+ a2 b2+ a3 b3 = |a||b| cos
The scaler product is defined as ...
Previously,
… was proved
the value of … a1 b1+ a2 b2+ a3 b3
or |a||b| cos
The scaler product is written as ...a.b… it’s also known as the dot product
a.b = a1 b1+ a2 b2+ a3 b3
a.b = |a||b| cos
Scaler Product (cont.)
cos = a1 b1+ a2 b2+ a3 b3
|a||b|
becomes
a.b = a1 b1+ a2 b2+ a3 b3
cos = a.b|a||b|
Parallel Vectors
cos = a.b|a||b|
Occur
…when cos = 1… so = cos-1(1) = 0
degreesi.e. the lines are Parallel
Perpendicular Vectors
cos = a.b|a||b|
a.bIf = 0,
…then cos = 0… so = cos-1(0) = 90
degreesi.e. the lines are Perpendicular
So, if a.b = 0 then the lines are perpendicular
Example (2D) - angle between vectors
Given: a = 3i + 4j and b = i - 3j
The scaler product is written as ...a.b
a.b = (3 x 1) + (4 x -3)
The j componentsThe i components
cos = a.b|a||b|
|a| = (32 + 42) = 25 = 5
|b| = (12 + (-3)2) = 10
= 4 - 12 = -8
cos = -8 = 0.506 510
= cos-1(0.506) = 120.4o
Angle between 3D VectorsAngle between 3D Vectors
7
3
2
a
5
2
1
b
The scaler product is written as ...a.b
a.b = (2 x 1) + (3 x -2) + (7 x 5)
cos = a.b|a||b|
|a| = (22 + 32 + 72) = 62
|b| = (12 + (-2)2 + 52) = 30
= 2 - 6 + 35 = 31
cos = 31 = 0.719 6230 = cos-1(0.719) = 44.0o
