1. A#40 Vectors
Day 6 Paper #1
Review

2. A#40 day 6 Vectors paper1

3. N2008/SP1/TZ0

4. V e c t o r s
N2008/SP1/TZ0
(i) AO +OB = – a + b = b – a
Recall:
OB = b
OA = a
AO = -a
(ii) speed =
Recall:
Speed is the magnitude of the velocity vector.
Particle motion

5. V e c t o r s
N2008/SP1/TZ0
(b) Equation of line L: or
Particle motion

6. 2008/SP1/TZ2 Section B
Consider the points A(1, 5, 4), B(3, 1, 2) and D(3, k, 2) with (AD) perpendicular to (AB).

7. Perpendicular vectors: Scalar product is 0
2008/SP1/TZ2 Section B
(i) AB = AO +OB = – a + b = b – a
(ii) AD = AO +OD = – a + d = d – a
Perpendicular vectors: Scalar product is 0

8. Perpendicular vectors: Scalar product is 0
2008/SP1/TZ2 Section B
(b)
4 – 4(k – 5) + 4 = 0
4k + 28 = 0
k = 7
(c)
Recall:
a | b =>
Scalar
product = 0
Perpendicular vectors: Scalar product is 0

9. Method 2: The scalar product of | vectors is 0
2008/SP1/TZ2 Section B
V e c t o r s
(c) (cont) OC = OB +BC = b + BC
(d) Given AD | AB, since BC = (½)AD,
BC has the same direction as AD.
So BC | AB and ABC = 90o.
cos ABC = cos 90o = 0. ^ A D C B ^
Method 2: The scalar product of | vectors is 0

10. 2010/SP1/TZ2
2010/SP1/TZ2

11. The unit vector has magnitude of 1
2010/SP1/TZ2
V e c t o r s
BC = BA +AC = AC – AB
(b) Length (magnitude) of AB =
Unit vector is:
The unit vector has magnitude of 1

12. The scalar product of | vectors is 0
2010/SP1/TZ2
(c)
=> AB | AC.
Alternately,
cos  = 0
Vectors perpendicular
AB | AC
The scalar product of | vectors is 0

13. N09/5/MATME/SP1/ENG/TZ0/
N09/5/MATME/SP1/ENG/TZ0/

14. N09/5/MATME/SP1/ENG/TZ0/
N09/5/MATME/SP1/ENG/TZ0/

15. 2010/SP1/TZ1/Section B

16. Parallel vectors are scalar multiples of one another
2010/SP1/TZ1/Section B
(a) Equations of parallel lines:
Parallel vectors are scalar multiples of one another

17. The scalar product of | vectors is 0
2010/SP1/TZ1/Section B
(b) Equations of perpendicular lines:
(2)(-7) + (1)(-2) + (-8)(k) =
–14 – 2 – 8k = 0
– 16 – 8k = 0
k = – 2
The scalar product of | vectors is 0

18. V e c t o r s
2010/SP1/TZ1/Section B
(c) L1 and L3 intersect at A. Set L1 = L3. Solve.
(-3) + p(2) = (5) + q(-7)
(-1) + p(1) = (0) + q(-2) p + 2q =
(-25) + p(-8) = (3) + q(-2) (-8p + 2q) = -(28)
9p = -27 => p = -3
q = 2
Point A is (-9, -4, -1)
Intersecting lines

19. Position vectors OA and OB.
2010/SP1/TZ1/Section B
V e c t o r s
(d) (i) AO +OB = – a + b = b – a
Recall:
OB = b
OA = a
AO = -a
Position vectors OA and OB.

20. 2010/SP1/TZ1/Section B
V e c t o r s
(d) (ii) AC = AB +BC
(ii) magnitude =
Particle motion