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Block 3
Vector Bits and Pieces
What is to be learned?
• Some wee footery bits about vectors
a.a ?
a.b = |a| |b| cosθ
a.a = |a| |a| cos00
a.a = |a| |a| (1)
a.a = a2
a(a + b) ?
= a.a + a.b
= a2
+ a.b
Need more
information to simplify
Unit Vectors
remember i j k
all have magnitude of 1
are at right angles to each other
i.i = i2 = 12
= 1 same for j.j and k.k
i.j = |i| |j| cosθ
= 1(1)cos900
= 0 same for i.k and j.k
Vector Bits and Pieces
• Breaking brackets works as normal
• a.a = a2
unit vectors
• i.i = j.j = k.k = 1
• i.j = j.k = i.k = 0
*
* i.i = i2
Evaluate a.(a + b)
a
b
a.(a + b)
= a.a + a.b
a.a
a.a
= |a||a|cosθ
= 5(5)cos00
= 25 X 1
= 25
300
2
5
a.b
a.b
= |a||b|cosθ
= 2(5)cos300
= 10 X √3
/2
= 10√3
2
= 5√3
Total = 25 + 5√3
Evaluate a.(a + b) a
b
a.(a + b)
= a.a + a.b
a.a
a.a
= |a||a|cosθ
= 3(3)cos00
= 9 X 1
= 9
300
3
6
a.b
a.b
= |a||b|cosθ
= 3(6)cos300
= 18 X √3
/2
= 18√3
2
= 9√3
Total = 9 + 9√3
Key
Question
Evaluate a.(b + c) (triangle is equilateral
side length 3 units)
a
b c
a.(b + c)
= a.b + a.c
a.b
a.b
= |a||b|cosθ
= 3(3)cos600
= 9 X ½
= 4½
600
3
3
Evaluate a.(b + c) (triangle is equilateral
side length 3 units)
a
b c
a.(b + c)
= a.b + a.c
a.c
a.c
= |a||c|cosθ
= 3(3)cos1200
= 9 X -½
= -4½
600
Must start from
same point!
3
3

600 1200
3
a.(b + c) = 4½ + -4½
= 0
b
c
Evaluate b.(a + c)
= b.a + b.c
2
a
2
a
b
c
Evaluate b.(a + c)
= b.a + b.c
2
2
b.a
Must start from
same point! 
b
c
Evaluate b.(a + c)
= b.a + b.c
2
b.a
a
2
b.a = |b| |a| cosθ
√8
450
450
= √8(2) cos450
= √8(2) (1
/√2)
= 2√4
Must start from
same point! 
= 2√8
√2
= 4
b
c
Evaluate b.(a + c)
= b.a + b.c
2
b.c
a
2
b.c = |b| |c| cosθ
√8
450
450
= √8(2) cos450
= √8(2) (1
/√2)
= 4
c2
450
b.(a + c) = 4 + 4
= 8
Must start from
same point! 

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Vector bits and pieces

  • 1. Block 3 Vector Bits and Pieces
  • 2. What is to be learned? • Some wee footery bits about vectors
  • 3. a.a ? a.b = |a| |b| cosθ a.a = |a| |a| cos00 a.a = |a| |a| (1) a.a = a2
  • 4. a(a + b) ? = a.a + a.b = a2 + a.b Need more information to simplify
  • 5. Unit Vectors remember i j k all have magnitude of 1 are at right angles to each other i.i = i2 = 12 = 1 same for j.j and k.k i.j = |i| |j| cosθ = 1(1)cos900 = 0 same for i.k and j.k
  • 6. Vector Bits and Pieces • Breaking brackets works as normal • a.a = a2 unit vectors • i.i = j.j = k.k = 1 • i.j = j.k = i.k = 0 * * i.i = i2
  • 7. Evaluate a.(a + b) a b a.(a + b) = a.a + a.b a.a a.a = |a||a|cosθ = 5(5)cos00 = 25 X 1 = 25 300 2 5 a.b a.b = |a||b|cosθ = 2(5)cos300 = 10 X √3 /2 = 10√3 2 = 5√3 Total = 25 + 5√3
  • 8. Evaluate a.(a + b) a b a.(a + b) = a.a + a.b a.a a.a = |a||a|cosθ = 3(3)cos00 = 9 X 1 = 9 300 3 6 a.b a.b = |a||b|cosθ = 3(6)cos300 = 18 X √3 /2 = 18√3 2 = 9√3 Total = 9 + 9√3 Key Question
  • 9. Evaluate a.(b + c) (triangle is equilateral side length 3 units) a b c a.(b + c) = a.b + a.c a.b a.b = |a||b|cosθ = 3(3)cos600 = 9 X ½ = 4½ 600 3 3
  • 10. Evaluate a.(b + c) (triangle is equilateral side length 3 units) a b c a.(b + c) = a.b + a.c a.c a.c = |a||c|cosθ = 3(3)cos1200 = 9 X -½ = -4½ 600 Must start from same point! 3 3  600 1200 3 a.(b + c) = 4½ + -4½ = 0
  • 11. b c Evaluate b.(a + c) = b.a + b.c 2 a 2
  • 12. a b c Evaluate b.(a + c) = b.a + b.c 2 2 b.a Must start from same point! 
  • 13. b c Evaluate b.(a + c) = b.a + b.c 2 b.a a 2 b.a = |b| |a| cosθ √8 450 450 = √8(2) cos450 = √8(2) (1 /√2) = 2√4 Must start from same point!  = 2√8 √2 = 4
  • 14. b c Evaluate b.(a + c) = b.a + b.c 2 b.c a 2 b.c = |b| |c| cosθ √8 450 450 = √8(2) cos450 = √8(2) (1 /√2) = 4 c2 450 b.(a + c) = 4 + 4 = 8 Must start from same point! 