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