Higher Outcome 4 - Official Mathematics Revision Website
Download
Report
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
41 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=9½
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 | = 22.
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)
bb 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 43
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
493
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
3k 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
ab 1
a c 1
abac 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