MATHEMATICS 201-NYC-05
Vectors and Matrices
Martin Huard
Fall 2015
VIII - Geometric Vectors
1. Find all vectors in the following parallelepiped that are equivalent to the given vectors.
E
F
A
B
H
D
a)
d)
g)
j)
m)
G
C
AB
AB  AE
AD  HE
AB  BC  CE
HC  HA  EC
b)
e)
h)
k)
n)
HE
AB  BC
FA  BG
AB  AD  AE
GH  BE  CE  FA
c)
f)
i)
l)
o)
HA
CD  BF
FG  FE
BC  DC  BD
AG  CB  EC  GA
2. Let ABCDEF be a regular hexagon where AB  a and FA  b .
a) Express the other sides, BC , CD , DE and EF , in terms of a and b .
b) Express FB , FC and FD in terms of a and b .
3. Let ABCDEFGH be a regular octagon where AB  a and BC  b .
a) Express all other sides in terms of a and b .
b) Express AC , AD , AE and DH in terms of a and b .
4. Consider the vectors u , v and w such that
u  4, N15E
v  3, S45W
w  6, N60W
Find the length and direction of the following vectors.
a) u  v
b) u  w
c) w  u
2v  3w
d) 12 u  13 v
e)
5
VIII – Geometric Vectors
Math NYC
5. An airplane has a maximum air speed of 500 km/h. If the plane is flying at is maximum speed
with a heading of 40 degrees west of north and the wind is blowing from north to south at 40
km/h, find the ground speed of the aircraft and the direction.
6. A jet if flying through a wind that is blowing with a sped of 40 km/h in the direction N30E .
The jet has a speed of 610 km/h in still air and the pilot head the jet in the direction N45E .
Find the speed and direction of the jet.
7. A woman launches a boat from on shore of a straight river and wants to land at the point
directly on the opposite shore. If the speed of the boat (in still water) Is 10 km/h and the river
is flowing east at the rate of 5 km/h, in what direction should she head the boat in order to
arrive at the desired landing point?
8. A boat heads in the direction N60E . The speed of the boat in still water is 24 km/h. The
water is flowing directly south. It is observed that the true direction of the boat is directly east.
Find the speed of the water and the true speed of the boat.
9. Prove that if AB  CD then AC  BD .
10. In the following parallelogram ABCD, M and N are the midpoints of AB and CD
respectively. Prove that AMCN is a parallelogram.
B
C
M
A
N
D
11. Let ABCD be any quadrilateral such that OB  OA  OC  OD where O is any point. Prove
that ABCD must be a parallelogram.
12. Prove that if the midpoints of the adjacent sides in a rectangle are joined, the resulting figure
is a rhombus.
13. Let ABCD be a parallelogram. Verify that the diagonal BD and the line AE, where E is the
midpoint of BC, intersect at a point that divides both of these segments in the ratio of 2 to 1.
14. Let ABCD be any quadrilateral on a plane. Prove that if P, Q, R and S are the midpoints of
segments AB, BC, CD and AD respectively, then PQRS is a parallelogram.
15. Prove that the line segment joining the midpoints of the diagonals in a trapezoid is parallel to
the base and is half the length of the difference between the lengths of the two bases.
Fall 2015
Martin Huard
2
VIII – Geometric Vectors
Math NYC
16. Let ABCD be a parallelogram and let E divide the segment AB in a ratio of 2 to 1 and let F
divide the segment DC in two. In what ratio does P, the point of intersection of AF and DE,
divide the segments AF and DE?
17. Determine whether the following statements are true of false. Explain
a) Two equivalent vectors have the same initial point.
b) 2 AA  5 AA
c) If AB  AC  AD then AB  AC  AD .
d) 2 AB and 3AB have the same direction.
e) 2 AB and 5BA have the same direction.
f) If u  kv then u  v  u  v .
g) There exists two nonzero vectors u and v such that u  v  u  v .
h)
i)
j)
Fall 2015
u  u  v   v
v
1
v
u and ku have the same direction.
Martin Huard
3
VIII – Geometric Vectors
Math NYC
Answers
1. a)
d)
g)
j)
m)
b) GF  DA  CB
c) GB
EF  DC  HG
e) AC  EG
f) CH  BE
AF  DG
h) FH  BD
i) EG  AC
0
k) AG
l) 0
AE  BF  DH  CG
n) BG  AH
o) EB  HC
AE  BF  DH  CG
2. a) BC  a  b
CD  b
DE  a
EF  b  a
b) FB  a  b
FC  2a
BD  2a  b
3. a) CD  2b  a
DE  b  2a
EF  a
FG  b
GH  a  2b
HA  2a  b
b) AC  a  b
DH   2b  2a
AD  1  2 b
AE  2  2 b  2a



4. a) u  v  25  12 3 , N W,   arcsin 2
3
25 12 3
d)
e)
1
2
u  13 v = 5  2 3 , N E,  =arcsin
 15
3sin 512
b) u  w =2 13  12cos 512 , N W,  =arcsin
c) w  u =2 13  12cos 512 , S W,  =arcsin

 15
1312cos 512
3sin 512
1312cos 512
1
5 2 3
 15
 15
2v  3 w
 = 65 10  6 cos 512 , S E,  =60  arcsin
5
5. 20 629  100cos 29 km/h, N W,   arcsin
6. 10 3737  488cos 12 km/h, N E,   arcsin
7. N30W
8. 12 km/h and 12 3 km/h
9.
B
D
sin 512
10  6cos 512
2sin 29
629100cos 29
61sin 29

3737  488cos 12
 40
 30
AD  AC  CD
AD  AB  BD
Thus AC  CD  AB  BD
AC  AB  AB  BD
A
Fall 2015
C
Since AB  CD
AC  BD
Martin Huard
4
VIII – Geometric Vectors
Math NYC
10. We need to show that AM  NC and AN  MC .
AB  DC
since ABCD is a parallelogram
For AM  NC ,
AM  MB  DN  NC
AM  AM  NC  NC
since M and N are the midpoints of AB and CD
2 AM  2 NC
AM  NC
For AN  MC , we have MC  MB  BC
 AM  BC since M is the midpoint ofAB
 NC  AD since AM  NC and ABCD is a parallelogram
 DN  AD since N is the midpoint of CD
 AD  DN
 AN
11. We need to show that AB  DC and BC  AD .
For AB  DC we have OB  OA  OC  OD
OB  AO  OC  DO
AO  OB  DO  OC
AB  DC
And for BC  AD ,
OB  OA  OC  OD
OD  OA  OC  OB
OD  AO  OC  BO
AO  OD  BO  OC
AD  BC
Fall 2015
Martin Huard
5
VIII – Geometric Vectors
Math NYC
12. We need to show that PQRS is a parallelogram whose sides have equal length.
B
Q
C
P
Since ABCD is a rectangle and P,Q,R and S are
midpoints, then
AP  PB  DR  RC
AS  SD  BQ  QC
R
A
D
S
PQ  PA  AS  SR  RC  CQ
QR  QP  PS  SR
 PA  BQ  SR  AP  QB
  PQ  PS  SR
  AP  QB  SR  AP  QB
  SR  PS  SR
 SR
 PS
To show that all sides are of equal length, it suffices to show that PQ  PS since we have
proved that PQRS is a parallelogram.
2
2
2
2
2
PQ  PB  BQ
 AP  AS
2
 PS
(by the Pythagorean theorem)
(by the Pythagorean theorem)
hence PQ  PS
13. We need to show that if AP  r AE then r  23 and if DP  sDB then s  23 .
E
1-r
B
1-s
r
P
C
s
Since ABCD is a parallelogram and E is the
midpoint of BC, then
BE  12 BC  12 AD
D
A
Let us express AP in terms of AB and BC in two different ways.
AP  AD  sDB
AP  r AE


 AD  s   AD  AB 


 r  AB  BC 
 r AB  BE
 AD  s DA  AB
1
2
 1  s  AD  s AB
 r AB  2r BC
 1  s  BC  s AB
By the basis theorem, we have the equations r  s,
Solving, we obtain r  23 and s  23 .
Fall 2015
Martin Huard
r
2
 1 s .
6
VIII – Geometric Vectors
Math NYC
14. We need to show that PQ  SR and QR  PS .
Q
C
B
R
P
A
D
S
For PQ  SR
PQ  PA  AS  SR  RC  CQ
 BP  AS  SR  RC  QB
Since P and Q are the midpoints of AB and BC
  PB  SD  SR  DR  BQ

 
Since S and R are the midpoints of AD and CD

  PB  BQ  SD  DR  SR
By associativity
  PQ  SR  SR
Thus 2 PQ  2SR and PQ  SR .
For QR  PS , we have QR  QP  PS  SR
  PQ  PS  SR
  SR  PS  SR
 PS

15. We need to show that MN  12 AD  BC

C
B
M
N
D
A
We have
MN  MC  CB  BN
and also MN  MA  AD  DN
 CM  AD  NB
since M and N are the midpoints of AC and DB
  MC  AD  NB
Adding, we obtain 2MN  MC  MC  CB  AD  BN  BN
 AD  BC

Thus MN  12 AD  BC
Fall 2015

Martin Huard
7
VIII – Geometric Vectors
Math NYC
16.
If AP  r AF and DP  sDE , then we need to find r and s.
Since ABCD is a parallelogram, we have
B
C
AB  DC and AD  BC .
E
Also,
by the definitions of E and F, we
P
F
have
r
s
AE  23 AB and DF  12 DC .
A
D
Let us express AP in terms of AB and
AD in two different ways.
AP  r AF

 r AD  DF
AP  AD  DP

 AD  sDE


 AD  s   AD  AB 
 AD  s DA  AE
 r AD  12 r DC
 r AD  2r AB
2
3
 1  s  AD  23s AB
By the basis theorem, we have the equations
r  1 s
r
2

2s
3
Solving, we obtain r  74 and s  73 . Hence, P divides AF in a ratio of 4 to 3 and divides
DE in a ratio of 3 to 4.
17. a) F. They may have any initial point as long as the have the same directions and
magnitude.
b) T. AA  0 so 0  0 .
c) F.
If AC , and AD have opposite direction then
B
C
AB  AC  AD .
D
A
D
A
d) F. They have opposite direction.
e) T. BA   AB .
f) F. If k  1 then u  v and we have u  v  v  v  0  0
u  v   v  v  1  v  v  2 v
g) T. If u  2v then u  v  2v  v  v and u  v  2v  v  2 v  v  v
h) T. u   u  v   u  u  v  1  v  v
i) T.
v
1

v 1
v
v
j) F. If k  1 then u and ku  u have opposite direction.
Fall 2015
Martin Huard
8