Vectors (5) Scaler Product Scaler Product Angle between lines Angle between lines

Angle between Vectors The angle can be measured if they are placed …. “Head-to-Head” “Tail-to-Tail”  

7 [ ] 4343 Angle between Vectors - example a = b =  |a| =  ((-1) ) |a| =  50 |b| =  ( ) = 5 |b - a| =  ( ) =  41 b - a = 5 -4 [ ]

7 [ ] 4343 Angle between Vectors - example (2) a = b =  |a| =  50 |b| = 5 |b - a| =  41 How can you find the angle now?

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 ….

a 2 = b 2 + c 2 - 2bc cos A Using the Cosine Rule... A B C angles sides  50 5  41  41 = x  50 x 5 cos A  = 61.3 o

Angle between Vectors - general case  |a| =  (a a a 3 2 ) |b| =  (b b 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 a a b b b (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 )

a 2 = b 2 + c 2 - 2bc cos A Generalizing angles sides |a||a| |b||b| |c|=|b - a|  |c| 2 = |a| 2 + |b| |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| |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 

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|

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 

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|

Parallel Vectors cos  = a.b |a||b| Occur …when cos  = 1 … so  = cos -1 (1) = 0 degrees i.e. the lines are Parallel

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

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| =  ( ) =  25 = 5 |b| =  (1 2 + (-3) 2 ) =  10 = = -8 cos  = -8 =  10  = cos -1 (0.506) = o

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| =  ( ) =  62 |b| =  (1 2 + (-2) ) =  30 = = 31 cos  = 31 =  62  30  = cos -1 (0.719) = 44.0 o