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Published byMiranda Holt Modified over 9 years ago
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Vectors (5) Scaler Product Scaler Product Angle between lines Angle between lines
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Angle between Vectors The angle can be measured if they are placed …. “Head-to-Head” “Tail-to-Tail”
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7 [ ] 4343 Angle between Vectors - example a = b = |a| = ((-1) 2 + 7 2 ) |a| = 50 |b| = (4 2 + 3 2 ) = 5 |b - a| = (5 2 + -4 2 ) = 41 b - a = 5 -4 [ ]
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7 [ ] 4343 Angle between Vectors - example (2) a = b = |a| = 50 |b| = 5 |b - a| = 41 How can you find the angle now?
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a 2 = b 2 + c 2 - 2bc cos A The Cosine Rule A B C a b c angles sides … is used for working out angles and sides in non-right angled triangles It is ….
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a 2 = b 2 + c 2 - 2bc cos A Using the Cosine Rule... A B C angles sides 50 5 41 41 = 50 + 25 - 2 x 50 x 5 cos A = 61.3 o
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Angle between Vectors - general case |a| = (a 1 2 + a 2 2 + a 3 2 ) |b| = (b 1 2 + b 2 2 + b 3 2 ) |c| = ((b 1 -a 1 ) 2 + (b 2 -a 2 ) 2 +(b 3 -a 3 ) 2 ) |c| 2 = (b 1 -a 1 ) 2 + (b 2 -a 2 ) 2 +(b 3 -a 3 ) 2 |c| 2 = a 1 2 + a 2 2 + a 3 2 + b 1 2 + b 2 2 + b 3 2 -2(a 1 b 1 + a 2 b 2 + a 3 b 3 ) Expand and rearrange |c| 2 = |a| 2 + |b| 2 -2(a 1 b 1 + a 2 b 2 + a 3 b 3 )
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a 2 = b 2 + c 2 - 2bc cos A Generalizing angles sides |a||a| |b||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(a 1 b 1 + a 2 b 2 + a 3 b 3 ) |a| 2 + |b| 2 -2(a 1 b 1 + a 2 b 2 + a 3 b 3 ) = |a| 2 + |b| 2 - 2 |a||b| cos -2(a 1 b 1 + a 2 b 2 + a 3 b 3 ) = - 2 |a||b| cos a 1 b 1 + a 2 b 2 + a 3 b 3 = |a||b| cos
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Generalizing (cont.) a 1 b 1 + a 2 b 2 + a 3 b 3 = |a||b| cos cos = a 1 b 1 + a 2 b 2 + a 3 b 3 |a||b|
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The Scaler Product a 1 b 1 + a 2 b 2 + a 3 b 3 = |a||b| cos The scaler product is defined as... Previously, … was proved the value of … a 1 b 1 + a 2 b 2 + a 3 b 3 or |a||b| cos The scaler product is written as... a.ba.b … it’s also known as the dot product a.b = a 1 b 1 + a 2 b 2 + a 3 b 3 a.b = |a||b| cos
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Scaler Product (cont.) cos = a 1 b 1 + a 2 b 2 + a 3 b 3 |a||b| becomes a.b = a 1 b 1 + a 2 b 2 + a 3 b 3 cos = a.b |a||b|
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Parallel Vectors cos = a.b |a||b| Occur …when cos = 1 … so = cos -1 (1) = 0 degrees i.e. the lines are Parallel
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Perpendicular Vectors cos = a.b |a||b| a.ba.b If = 0, …then cos = 0 … so = cos -1 (0) = 90 degrees i.e. the lines are Perpendicular So, if a.b = 0 then the lines are perpendicular
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Example (2D) - angle between vectors Given: a = 3i + 4j and b = i - 3j The scaler product is written as... a.ba.b a.b = (3 x 1) + (4 x -3) The j components The i components cos = a.b |a||b| |a| = (3 2 + 4 2 ) = 25 = 5 |b| = (1 2 + (-3) 2 ) = 10 = 4 - 12 = -8 cos = -8 = 0.506 5 10 = cos -1 (0.506) = 120.4 o
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Angle between 3D Vectors The scaler product is written as... a.ba.b a.b = (2 x 1) + (3 x -2) + (7 x 5) cos = a.b |a||b| |a| = (2 2 + 3 2 + 7 2 ) = 62 |b| = (1 2 + (-2) 2 + 5 2 ) = 30 = 2 - 6 + 35 = 31 cos = 31 = 0.719 62 30 = cos -1 (0.719) = 44.0 o
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