Transcript Higher Outcome 4 - Official Mathematics Revision Website

3D Coordinates
In the real world points in space can be located
using a 3D coordinate system.
For example, air traffic controllers find the location
a plane by its height and grid reference.
z
y
O
(x, y, z)
x
Write down the coordinates for the vertices
z
E (0, 1, 2)
A (6, 1, 2)
B (6, 0, 2)
2
(0, 0, 2) F
H (0, 1, 0 )
G
(0,0, 0) O
y
6
D (6, 1, 0)
1 x
C
(6, 0, 0)
3D vectors are defined by 3 components.
For example, the velocity of an aircraft taking off
can be illustrated by the vector v.
z
(7, 3, 2)
2
y
v
2
3
O
7
x
Any vector can be represented in terms of the
i , j and k
Where i, j and k are unit vectors (one unit long)
in the x, y and z directions.
z
y
k
O
j
i
x
 1
 
i=  0 
0
 
0
 
j =  1
0
 
0
 
k=0
 1
 
Any vector can be represented in terms of the
i , j and k
Where i, j and k are unit vectors
in the x, y and z directions.
z
(7, 3, 2)
v
y
2
v = ( 7i+ 3j + 2k )
x
7
 
v =  3
 2
 
3
O
7
Magnitude of a Vector
A vector’s magnitude (length) is represented by |v|
A 3D vector’s magnitude is calculated using
Pythagoras Theorem in 3D
v  x y z
2
2
2
(3 , 2 , 1)
z
v
y
v  32  22  12
v  14
1
2
O
3
x
Find the unit vector in the direction of u
 3 
 
u 0 
  4
 
u  32  02  (4)2
1
Unit vector  u
5
u  25
 3 
1 
Unit vecto r   0 
5 
  4
u 5
 3 


5


Unit vecto r   0 
  4 
 5
Addition of vectors
3
 2
 
 
Let u   4  and v   5 
1 
 2
 
 
Then u + v
 3   2  5
     
4
+

5


1
     
1   2   3 
     
Addition of Vectors
For vectors u and v
a
d 
 
If u   b  and v   e  then
f
c 
 
 
a
d 
a d 
 
 


u + v  b  + e  = b  e 
c 
f
c  f 
 
 


Negative vector
BA is the negative of AB
For any vector u
u   u   u   u   0
a
 a 
 
 
If u   b  then  u   b 
c 
 c 
 
 
Subtraction of vectors
6
 2
 


Let a   5  and b   4 
 2
3
 
 
Then a  b
6
 
 5
3
 
 2  4




- 4   1
 2 1
   
Subtraction of Vectors
For vectors u and v
a
d 
 
 
If u   b  and v   e  then
c 
f
 
 
a
d 
a d 
 
 


u  v  b   e  = b  e 
c 
f
c  f 
 
 


Multiplication by a scalar ( a number)
x
 kx 
 
If a vector v   y  then kv   ky 
 kz 
z 
 
 
The vector kv is parallel to vector v ( different size )
Hence if u = kv then u is parallel to v
Conversely if u is parallel to v then u = kv
CombiningVectors
 2 
 1 
  4
 
 
 
p   5 , q   0 , r   2  Express 2 p  q  1 r
2
  7
  1
 0 
 
 
 
 4   1    2

    
1
2 p  q  r   10    0    1 
2
  14    1   0 

    
 41 2 


1
2 p  q  r   10  0  1 
2
  14  1  0 


 5 


1
2p q  r   9 
2
  13 


in component form
Show that the two vectors are parallel.
6 
 12 
 
w   9  then z   18 
 24 
 12 
 
 
If z = kw then z is parallel to w
12 
6 
 
 
z  18  = 2  9 
 24 
 12 
 
 
z  2w
Position Vectors
A (3,2,1)
 3
 
a   2
 1
 
z
a
y
1
2
O
3
The position vector of the point A(3 , 2 , 1)
is OA written as a
x
Position Vectors
If R is ( 2 , -5 , 1) and S is (4 , 1 , -3)
4
Then RS is s – r =
1
-3
2
2
–
-5
1
=
6
-4
Position Vectors as a journey
If R is ( 2 , -5 , 1) and S is (4 , 1 , -3)
x component of RS is 2 to 4 = 2
y component of RS is -5 to 1 = 6
z component of RS is 1 to -3 = -4
2
RS = 6
-4
Vectors as a journey
Express VT in terms
of f, g and h
–h
VT = VR + RS + ST
VT = – h – f + g
–f
+g
Collinear Points
A is (0 , -3 , 5), B is (7 , -6 , 9) and C is (21 , -12 , 17). Show
that A , B and C are collinear stating the ratio AB:BC.
 7 
 14 
 7 
 
 
 
AB    3  BC    6  BC  2  3 
 4 
 8 
 4 
 
 
 
BC = 2AB
AB and BC are parallel (multiples of each other)
through the common point B, and so must be collinear
A
1
B
2
AB : BC = 1 : 2
C
Collinear Points
Given that the points S(–4, 5, 1), T(–16, –4, 16) and
U(–24, –10, 26) are collinear, calculate the ratio in which T
divides SU.
  12 
  4




ST    9   3  3 
 15 
 5 




S
3
T
  8
  4
 


TU    6   2  3 
 10 
 5 
 


2
ST : TU = 3 : 2
U
Using Vectors as journeys
PQRS is a parallelogram with P(3 , 4 , 0), Q(7 , 6 , -3) and
R(8 , 5 , 2). Find the coordinates of S.
Draw a sketch
P(3 , 4 , 0)
The journey from
Q to R is the same
S
as the journey
from P to S
 1
 
QR    1 From P to S
 5
 
Q(7 , 6 , -3)
R(8 , 5 , 2)
S(3 + 1 , 4 – 1 , 0 + 5)
S(4 , 3 , 5)
The point Q divides the line joining P(–1, –1, 0) to
R(5, 2 –3) in the ratio 2:1.
Find the co-ordinates of Q.
R(5 , 2 , -3)
PQ = 2/3 PR
6
PQ = 2/3
1
4
3
= 2
-3
-2
Q
The journey
from P to Q
2
Q(-1 + 4, -1 + 2, 0 – 2)
Q(3 , 1 , – 2)
P(-1 , -1 , 0)
P divides the line joining S(1,0,2) and T(5,4,10) in
the ratio 1:3. Find the coordinates of P.
SP = 1/4 ST
4
SP = 1/4
T(5 , 4 , 10)
1
4
= 1
8
2
The journey
from S to P
3
P
P(1 + 1, 0 + 1, 2 + 2)
1
P(2 , 1 , 4)
S(1 , 0 , 2)
The scalar product
The scalar product is defined as being:
a . b = |a| |b| cos θ
0 ≤ θ ≤ 180
a
θ
a and b must be divergent,
ie joined tail to tail
b
The Scalar Product
Find the scalar product for a and b when
|a|= 4 , |b|= 5 when (a) θ = 45o (b) θ = 90o
a . b = |a| |b| cos θ
 a . b = 4 × 5 cos 45o
 a . b = 20 × 1/√2 × √2/√2
 a . b = 10√2
a . b = |a| |b| cos θ
 a . b = 4 × 5 cos 90o
 a . b = 20 × 0
a.b=0
When θ = 90o
The Scalar Product
This equilateral triangle has
sides of 3 units. p . q
p . q = |p| |q| cos θ
 p . q = 3 × 3 cos 60o
 p . q = 9 × 1/2
 p . q = 41/2
The Scalar Product
This equilateral triangle has
sides of 3 units. p . (q + r) r
p . (q + r)= p . q + p . r
r
60o
 p . q = 3 × 3 cos 60o
p.q=3×3×½
 p . q = 41/2
p and r are not
divergent so move r
 p . r = 3 × 3 cos 60o
p.r=9½
p . (q + r) = p . q + p . r = 9
 p . r = 4½
If a and b are perpendicular
then
a.b=0
Component Form Scalar Product
If a =
a1
a2
a3
and
b=
b1
b2
b3
a . b = a1b1 + a2b2 + a3b3
Angle between Vectors
To find the angle between two vectors we simply use
the scalar product formulae rearranged
a . b = |a| |b| cos θ
cos θ =
a.b
|a| |b|
a . b = a1b1 + a2b2 + a3b3
cos θ =
a1b1 + a2b2 + a3b3
|a| |b|
Find the angle between the two vectors below.
p = 3i + 2j + 5k and q = 4i + j + 3k
3 |p| = √(32 + 22 + 52)
4
|q| = √(42 + 12 + 32)
p= 2
q= 1
|q| = √26
5 |p| = √38
3
a . b = a1b1 + a2b2 + a3b3
cos θ =
= 3×4 + 2×1 + 5×3
= 29
a1b1 + a2b2 + a3b3
=
|a| |b|
29
= 0∙923
√38 × √26
θ = cos-1 0∙923 = 22∙7o
Perpendicular Vectors
a . b = | a | | b | cosƟ
If a and b are perpendicular then
a.b=0
cos 90o = 0
If a and b are perpendicular then
+ a2b2 + a3b3 = 0
a1b1
If a and b are perpendicular then
Show that if a =
3
2
-1
and
b=
a.b=0
1
2
7
a and b are perpendicular
a . b = a1b1 + a2b2 + a3b3
 a . b = 3×1 + 2×2 + (-1)×7
a.b=3+4 –7
a.b=0
 a and b are perpendicular
Properties of a Scalar Product
Two properties that you need to be aware of
a.b=b.a
a .( b + c)= a . b + a . c
If | p | = 5 and | q | = 4, find p . (p + q)
p
60o
p . (p + q) = p . p + p . q
q
= | p | × | p |cos 0o + | p | × | q |cos60o
=5×5×1+5×4×½
= 25 + 10
= 35
T divides PR in the ratio 5:4
P(-2,-1,-4)
Show that Q, T and S are
collinear, and find the ratio
in which T divides QS Q(1,5,-7)
Find the acute angle
between the diagonals of
PQRS
 9  5
5   
5
PT  PR   9    5 
9   
9
 9  5
T(-2+5,-1+5,-4+5)
T(3 , 4 , 1)
 5
 
 5
 5
 
S(7,2,17)
R(7,8,5)
 4 
 2
 2
 
 
 
QT    1 TS    2   2  1
 16 
 8
 8
 
 
 
TS = 2QT, vectors are parallel
through the common point T, so
Q , T , S are collinear
QT : TS = 1 : 2
P(-2,-1,-4)
Find the acute angle between
the diagonals of PQRS
S(7,2,17)
Q(1,5,-7)
 9
 6 
 


PR   9  QS    3 
 9
 24 
 


 243
62  32  24 2  621
PR  9  9  9
2
QS 
2
2
PR.QS  9  6  9  (3)  9  24
 243
R(7,8,5)
PR.QS
cos 
PR QS
243
 0  626
cos 
243  621
  51.2o
Vectors u and v are defined by u = 3i + 2j and v = 2i – 3j + 4k
Determine whether or not u and v are perpendicular.
3
2
u . v = 2 . -3
0
4
 u . v = 3×2 + 2×(-3) + 0×4
u.v=6–6 +0
u.v=0
Hence vectors are perpendicular
For what value of t are the vectors u and v perpendicular ?
 t 
 
u   2 
 3
 
v
2
 
 10 
t 
 
t
2
u . v = -2 . 10
3
t
u . v = 0 if vectors perpendicular
 u . v = t ×2 + (-2)×10 + 3×t
 u . v = 5t – 20
u . v = 0  5t – 20 = 0
t=4
VABCD is a pyramid with rectangular base ABCD.
The vectors AB, AD and AV are given by
AB = 8i + 2j + 2k, AD = -2i + 10j – 2k and
AV = i + 7j + 7k Express CV in component form.
Look for an alternative route from C to V
along vectors you know.
Note BC = AD and AB = DC
CV = CB + BA + AV
8
AB = 2
2
CV = -AD – AB + AV
=
=
2
-10 2
-5
-5
7
8
2
2
+
-2
AD = 10
-2
1
7
7
CV = –5i – 5j + 7k
1
AV = 7
7
The diagram shows two vectors a and b, with | a | = 3 and | b | = 22.
These vectors are inclined at an angle of 45° to each other.
a) Evaluate
i) a.a
ii) b.b
iii) a.b
b) Another vector p is defined by p = 2a + 3b
Evaluate p.p and hence write down | p |.
i)
a  a  a a cos0  3  3 1  9
iii) a  b  a b cos 45  3  2 2 
b)
p  p   2a  3b   2a  3b 
ii)
bb  2 2 2 2
1
6
2
 4a.a  12a.b  9b.b
 36  72  72
 180
p  180  6 5
p2
8
Vectors p, q and r are defined by p = i + j – k, q = i + 4k, r = 4i – 3j
Express p – q + 2r in component form
a)
b) Calculate p.r
c)
Find |r|
a)
p  q  2r   i  j - k    i  4k   2  4i  3 j   8i  5 j - 5k
b)
p.r   i  j - k  .  4i  3 j   p.r  1 4  1 (3)  (1)  0
 p.r  1
c)
r 
42  (3) 2
 r  16  9 
r 5
The diagram shows a point P with co-ordinates
(4, 2, 6) and two points S and T which lie on the x-axis.
If P is 7 units from S and 7 units from T,
find the co-ordinates of S and T.
S (a, 0, 0)
T (b, 0, 0)
PS 2  49  (4  a)2  22  62
 a  43
 49  (4  a)2  40
 9  (4  a)2
 a  7 or a  1
hence there are 2 points on the x axis that are 7 units from P
S (1, 0, 0)
and
T (7, 0, 0)
i.e. S and T
The position vectors of the points P and Q are
p = –i + 3j + 4k and q = 7i – j + 5k respectively.
a) Express PQ in component form.
b) Find the length of PQ.
PQ  q - p


 7   1
   
PQ   1 -  3 
5 4
   
8
 
PQ   4   8 i  4 j  k
1
 
PQ  82  (4)2  12  64  16  1  81  9
P
PQR is an equilateral triangle of side 2 units.
PQ  a, PR  b, and QR  c
a
60°
b
Evaluate a.(b + c) and hence identify
two vectors which are perpendicular.
60°
60°
Q
a.(b  c)  a.b  a.c
120o
R
c
1
a.b  a b cos60  a.b  2  2  2  a.b  2
NB for a.c vectors must diverge ( so angle is 120° )
 1
a.c  a c cos120  a.c  2  2    2 
Hence
a.(b  c)  0
so,
a
 a.c   2
is perpendicular to
b+c
Calculate the length of the vector 2i – 3j + 3k
Length
 2  (3) 
2
2
 3
2
 493
 16
4
 4 
1 
Find the value of k for which the vectors   and 
 are
3


 2
perpendicular
 k 1
 1 


 
Put Scalar product = 0
 1   4 
  

 2  3   0
  1  k  1
  

 4  6  ( k  1 )  0
 4  6  k  1  0
 4  6  k  1  0
 3k  0
k 3
A is the point (2, –1, 4), B is (7, 1, 3) and C is (–6, 4, 2).
If ABCD is a parallelogram, find the co-ordinates of D.
The journey B to A is the same as
the journey from C to D
  5


BA    2 
 1 


That journey from C gives:
D(–6 + [-5], 4 + [-2], 2 + 1)
D(–11, 2 , 3)
The vectors a, b and c are defined as follows:
a = 2i – k,
b = i + 2j + k,
c = –j + k
a) Evaluate a.b + a.c
b) From your answer to (a), make a deduction about the vector b + c
 2   1
   
a) a  b   0    2 
  1  1 
   
 2   0 
   
a  c   0     1
  1  1 
   
ab 1
a  c  1
abac  0
b)
a  b  a  c  a  (b  c )
b + c is perpendicular to a
In the square based pyramid,
all the eight edges are of length 3 units.
AV  p, AD  q, AB  r ,
Evaluate
p.(q + r)
Triangular faces are all equilateral
p.(q  r )  p.q  p.r
p.q  p q cos60
p.r  p r cos60
1
2
p.(q  r )  4  4
1
2
p.q
1
 3 3
2
p.r
1
 3 3
2
p.(q  r )  9
p.q  4
p.q
1
2
1
4
2
A and B are the points (-1, -3, 2)
and (2, -1, 1) respectively.
(-1,-3,2)
B and C are the points of trisection of AD.
That is, AB = BC = CD.
Find the coordinates of D
 3 
 
AB   2 
  1
 
 9 


AD   6 
  3


AD  3  AB
 D is (-1 + 9, -3 + 6, 2 – 3)
 D is (8, 3, –1)
(2,-1,1)
The point Q divides the line joining P(–1, –1, 0) to
R(5, 2 –3) in the ratio 2:1.
R
1
Q
Find the co-ordinates of Q.
2
(5,2,-3)
P
(-1,-1,0)
 6 


PR   3 
  3


2
PQ  PR
3
 Q is (-1 + 4, -1 + 2, 0 – 2)
 Q is (3, 1, –2)
 6 

2 
PQ    3 
3 


3


 4 


PQ   2 
  2


VABCD is a pyramid with rectangular base ABCD.
VA = – 7i – 13j – 11k, AB = 6i + 6j – 6k, AD = 8i – 4j – 4k ;
K divides BC in the ratio 1:3.
Find VK in component form.
VK
= VA + AB
+ ¼BC
= VA + AB
+ ¼AD
= – 7i – 13j – 11k
6i + 6j – 6k
2i – j – k
= i – 8j – 18k
 1 


8
  18 


A is the point (2, –5, 6), B is (6, –3, 4) and C is (12, 0, 1).
Show that A, B and C are collinear
and determine the ratio in which B divides AC
 2 
 
AB  2   1 
  1
 
 4 


AB   2 
  2


 6 


BC   3 
  3


 2 
 
BC  3   1 
  1
 
AB and BC are scalar multiples, so are parallel.
B is common.
A, B, C are collinear
A
2
B
3
C
B divides AB in ratio 2 : 3