Transcript Vectors - Alford Academy

Vectors
ab
b
a
(i) a  b  b  a
Addition is commutative
(ii ) a  b  a  b
(iii) a  (b  c)  (a  b)  c Addition is associative
(iv) The negative of a vector u has the same magnitude and direction
as u but the opposite sense.
(v) Subtraction of vectors can be defined as a  b  a  (b)
(vi) If vector u is multiplied by a scalar k, then the product ku is a vector in the
same direction as u but k times the magnitude.
If k  1, then the magnitude is increased
If k  1, then the magnitude is decreased
If k  0, then ku has the same sense as u
If k  0, then ku has the opposite sense to u
u
(k  l )u  ku  lu
2u
k (u  v)  ku  kv
Vector Directions
Hold your right hand out in front of you as if to shake hands
x direction
y direction
z direction
-
Fingers
From Palm
From Thumb
Given A(a1 , a2 , a3 )
 a1 
then a   a2  This is the position vector of A
 
a 
 3
If A and B have position vectors a and b respectively, then AB  b  a
If P divides AB in the ration m : n, then
AP
m

PB n
mb  na
P
mn
 u1 
If u   u2  , then u  u12  u22  u32 (magnitude)
 
u 
 3
A unit vector is a vector of magnitude 1.
ua 
1
a
a
Unit vectors in the x direction, y direction and z direction are denoted i, j and k.
1
i  0
 
0
 
0
j  1
 
0
 
0
k  0
 
1
 
Any vector can be expressed in terms of i, j and k for example;
2
 3   2i  3 j  5k
 
 5 
 
P
The projection of a point P on a
plane is the point P’, at the foot
of the perpendicular to the plane
passing through P.
P’
The projection of a line PQ on a
plane is the line P’Q’, which joins
the projections of P ands Q on
the plane
P
P’
Q

Q’
The angle between a line and a plane , is the angle between
the line and its projection on the plane. (degrees or radians)
The scalar product or dot product of the vectors a and b is defined by;
a . b  a b cos
Where  is the angle between the positive directions of a and b.
Note that b cos is the magnitude of the projection of b on a
b

b cos 
a.b is scalar
a.a  a a cos0  a
2
a.b  b.a
a.(b  c)  a.b  a.c
a.b  0 and a  0, b  0  a and b are perpendicular
 a1 
 b1 
a   a2  and b   b2 
 
 
a 
b 
 3
 3
a.b  a1b1  a2b2  a3b3
a1b1  a2b2  a3b3
cos  
ab
Direction Ratios
 a1 
Given a   a2  these components uniquely determine the magnitude and direction of a.
 
a 
 3
Any scalar multiple of a has the same direction but not necessarily the same sense as a.
It follows that the ratio a1:a2:a3 can be used to describe uniquely the direction ratio of
the vector. If two vectors have equal direction ratios then they are parallel.
Direction Cosines
If  ,  and  are the angles the vector a makes with ox, oy and oz
 u1 
respectively and u   u2  is a unit vector in the direction of a
 
u 
 3
then we know;
u1  u . i cos   cos 
u2  u . j cos   cos 
u3  u . k cos   cos 
 cos  
giving u   cos  


 cos  


These components are called the direction cosines of a
Note:
u  cos 2   cos 2   cos 2   1
 cos 2   cos 2   cos 2   1
This means that ,  and  are not independent.
Find
(a) the direction ratio
(b) the direction cosines
of the vector a = 6i + 8j +24k
(a) Direction ratio = 6 : 8 : 24
3 : 4 :12
(b) a  6  8  24  26
2
6
26
8
cos  
26
24
cos  
26
cos  
2
2
 ua 
6
8
24
i
j k
26 26
26
Page 44 Exercise 1 Questions 6 – 8
Page 46 Exercise 2 Questions 1, 2, 9
TJ Exercise 1
The Vector Product
Two vectors can be combined by the binary operation, the scalar product, to
produce a number (or scalar).
It is also possible to define an operation where two vectors, a and b, multiply to
produce a vector. This is referred to as the vector product, or cross product, as
denoted a × b.
Any two non parallel vectors, a and b, define a plane. Let n be a unit vector
perpendicular to this plane (a normal) so that a, b and n form a right handed
system.
Any two non parallel vectors, a and b, define a plane. Let n be a unit vector
perpendicular to this plane (a normal) so that a, b and n form a right handed
system.
a along the fingers
u
a
b from the palm
n along the thumb.
b
The angle between a and b need NOT be 900
The vector product is defined as
a  b  n a b sin 
where  is the angle between the positive directions of a and b.
If either a = 0, or b = 0, then n is not defined and a × b is defined as 0.
The following properties follow directly from the definition.
a  b is a vector with the same sense and direction as n
a  b = a b sin  is the area of the parallelogram defined by a and b
a  a =n a a sin   0
a  ka =n a k a sin   0 : parallel vectors have a vector product of zero
a  0, and b  0 and a  b  0  a and b are parallel vectors
a  b =  (b  a) : The vector product is not commutative.
i  j =k i j sin 90o  k
i  j =k i j sin 90o  k
Similarly:
i
i j  k
jk  i
k i  j
ik   j
k  j  i
j  i  k
k
j
The vector product is distributive over addition.
a  b  c   a  b  a  c
and
 a  b  c  a  c  b  c
b
3
600
3
a
a  b  3  3sin 600
9

2
b  a  3  3sin 600
9

2
Same magnitude but different sense. (see page 48)
Component form of the Vector Product
 a1 
 b1 
Suppose a   a2  and b   b2  then
 
 
a 
b 
 3
 3
a  b  (a1 i  a2 j  a3 k )  (b1 i  b2 j  b3 k )
Using the distributive law:
 a1b1 (i  i )  a1b2 (i  j )  a1b3 (i  k ) 
a2b1 ( j  i )  a2b2 ( j  j )  a2b3 ( j  k ) 
a3b1 (k  i )  a3b2 (k  j )  a3b3 (k  k )
 0  a1b2 (k )  a1b3 ( j )  a2b1 (k )  o  a2b3 (i )  a3b1 ( j )  a3b2 (i )  0
 (a2b3  a3b2 )i  (a3b1  a1b3 ) j  (a1b2  a2b1 )k
a  b  (a2b3  a3b2 )i  (a3b1  a1b3 ) j  (a1b2  a2b1 )k
This is normally written as:
a  b  (a2b3  a3b2 )i  (a1b3  a3b1 ) j  (a1b2  a2b1 )k
We do this so as to use the following shortcut notation
i
j
k
a  b  a1 a2
b1 b2
a3
b3
Calculate the area of a triangle whose vertices are
A(2, -1, 3), B(5, -1, 2), C(2, 3, 4)
1
Area = AB AC sin 
2
1
 AB  AC
2
B

C
A
i
j
k
a  b  3 0 1
0 4 1
 3  0
1
  0  4
2    
 1   1 
1
 (0  4)i  (3  0) j  (12  0)k
2
1
 4i  3 j  12k
2
1
13

16  9  144  units 2
2
2
Calculate the shortest distance from the point P(1, 2, 3) to the straight line
passing through the points A(1, 3, -2) and B(2, 2, -1)
PP '  AP sin 
P
AP

P’
B

AB AP sin 

AB  AP
AB
AB
AB
A
1
0
AB  b  a   1 , AP  p  a   1
 
 
1
5
 
 

(5  1)i  (5  0) j  ( 1  0) k
3

i
j
k
a  b  1 1 1
0 1 5
4i  5 j  1k
3
16  25  1

3
 14 units
The Scalar Triple Product
A parallelopiped is a region of 3D space
bounded by three pairs of parallel lines.
a  h
b
Its volume can be calculated by multiplying
the area of one plane in a pair and the
perpendicular distance between the planes.
V = Ah
c
Area of Base = b  c
Suppose that the parallelopiped is defined
by three vectors a, b and c which form a
right handed system!
Distance between the planes = h  a cos
V  b  c a cos   a.(b  c)
Now any of the 3 pairs of parallel planes could have been used.
V  b  c a cos   a.(b  c) (Dot Product)
Now any of the 3 pairs of parallel planes could have been used.
V  a.(b  c)  b.(c  a)  c.(a  b)
NOTE:
a.(b  c) is a number NOT a vector
a.(b  c)  (a  b).c In this expression the dot and cross are interchangeable.
a.(b  c) is called the scalar triple product denoted [a, b, c]
If a, b or c  0, then a.(b  c) is defined as 0.
Component form of the scalar product
 a1 
a   a2 
 
a 
 3
 b1 
b   b2 
 
b 
 3
 c1 
c   c2 
 
c 
 3
 b2 c3  b3c2 
b  c   b3c1  b1c3 


 bc b c 
 1 2 2 1
 a1  b2c3  b3c2 
a.(b  c)   a2  b3c1  b1c3   a1 (b2 c3  b3c2 )  a2 (b3c1  b1c3 )  a3 (b1c2  b2c1 )
 

 a  b c  b c 
 3  1 2 2 1 
 a1 (b2 c3  b3c2 )  a2 (b1c3  b3c1 )  a3 (b1c2  b2 c1 )
a1 a2
a.(b  c)  b1
c1
b2
c2
a3
b3
c3
Page 52 Exercise 4
TJ Exercise 2
Equation of a line in 3 dimensions
A line in space is completely determined when we know the direction in which it
runs and we know a point on the line. Its direction is unambiguously described by
stating a vector parallel to the line. Such a vector is known as a position vector.
P
Consider the line L which passes
through the point A(x1, y1, z1) with
direction vector u = ai + bj + ck. Let
P(x, y, z) be any point on the line L.
A
In Component Form this becomes
p
a
u
AP  tu for some scalar t. p  a  tu
L
p  a  tu
1
O
P
Consider the line L which passes
through the point A(x1, y1, z1) with
direction vector u = ai + bj + ck. Let
P(x, y, z) be any point on the line L.
A
p  a  tu
AP  tu for some scalar t. p  a  tu
In Component Form this becomes
 x1   a   x1  at 
 y   t  b    y  bt 
 1    1

 z   c   z  ct 
 1    1

Giving
x  x1  at , y  y1  bt , z  z1  ct
2
x  x1 y  y1 z  z1


t
a
b
c
3
Also:
p
a
u
 x
 y 
 
z
 
L
1
O
We have three form of the equation of a line in space.
1.
p  a  tu
2. x  x1  at , y  y1  bt , z  z1  ct
x  x1 y  y1 z  z1
3.


t
a
b
c
Equation 1 is known as the vector equation.
The system of equations (2) is the parametric form. (t being the parameter).
The system of equations (3) is the symmetric form.
(also referred to as the standard or canonical form)
•‘t’ is often omitted in the symmetric form but has to be inserted to convert to other forms.
•If any of the components of the direction vector is zero then some parts of the symmetric form
will be undefined in which case the parametric form is better.
•Each point on L is uniquely associated with a value of the parameter t.
•The equation of a particular line is NOT unique.
x 1 y  2 z  4
x  2 y z 1


and
 
2
4
6
1
2
3
Both equations represent a line that
passes through (3, 2, 2) and is
parallel to i + 2j + 3k.
(a) Write down the symmetric form of the equations of the line which passes
through (1, -2, 8) and is parallel to 3i + 5j + 11k.
(b) Does the point (-2, -7, -3) lie on the line?
(a) Direct substitution gives:
x 1 y  2 z  8


3
5
11
(b) Substituting (-2, -7, -3) we get:
2  1
 1
3
3  8
7  2
 1
 1
11
5
Since the results are consistent (-1), the point lies on the line.
Find the equations of the line passing through A(2, 1, 3) and B(3, 4, 5)
The line MUST be parallel to AB
Since the line passes through A we get:
 3  2  1
AB  b  a   4    1    3 
     
 5  3  2
     
x  2 y 1 z  3


t
1
3
2
x 3 y 4 z 5
NOTE: we could have chosen point B.


t
1
3
2
Page 66 Exercise 9A Questions 1(a), (b), 2(a), 3(a), (c), (e) and 5
Page 67 Exercise 9B Question 2
TJ Exercise 3
The Equation of a Plane
A plane in space can be uniquely identified if:
•3 points on the plane are known
•2 Lines on the plane are known
•1 point on the plane and a normal to the plane are known
Suppose that relative to a right handed set of axes, we have a
plane called  . Let P( x, y, z ) be a typical point on the plane and
a
let a   b  be a normal to the plane passing through the plane at Q.
 
c
 
Since QP is perpendicular to a,
a
P

Q
a. QP  0
a.( p  q)  0
a. p  a.q  0
Since both a and Q are fixed a.q is a constant on the plane.
Let k = a . q then
a. p  k  0
 a. p  k
a
  b .
 
c
 
 x
 y  k
 
z
 
If we know the normal a and any point P we can easily compute k
 ax  by  cz  k
This is the equation of the plane 
To find the equation of a plane we must be able to reduce given data to
(i) The components of a suitable normal vector
(ii) A point on the plane
P, Q, R and S are the points (1, 2, 3), (2, 1, -4), (1, 1, 1) and (7, -6, 5) respectively.
(a) Find the equation of the plane perpendicular to PQ which contains the point P
(b) which of the other points lie on the plane?
 2 1   1 
(a) PQ  q  p   1  2    1 

  
 4  3   7 

  
 1  1
 k   1  . 2   1  2  21  22
  
 7   3 
  
 Equation of the plane = x  y  7 z  22
(b) R(1,1,1) does not satisfy the equation
S(7,-6,5) does satisfy the equation so lies on the plane.
Find the equation of the plane which passes through the points A(-2, 1, 2)
B(0, 2, 5) ans C(2, -1, 3).
Strategy: Find a Normal to the plane.
Two vectors that lie on the plane are AB and AC .
AB  AC is a normal vector to the plane.
 2
 4
AB   1  AC   2 
 
 
 3
1
 
 
Using Point A to find k
i
j
k
AB  AC  2 1 3
4 2 1
 1 6   7 
  (2  12)    10 

  
 4  4   8 

  
k  14  10  16  20
Required Equation of the plane = 7 x  10 y  8z  20
Vector Equation of a plane
A plane can be defined by any two non zero non parallel vectors which lie upon it.
Let AB and AC be such a pair defining the plane  . Let R be any point on the plane.
Then AB and AC and AR are coplanar.
R
AR
AR  t AB  u AC
r  a  t (b  a)  u(c  a)
AC
AB
u AC
t AB
r  a  t (b  a)  u(c  a)  r  (1  t  u)a  tb  uc
This is known as the vector equation of a plane.
r  a  t (b  a)  u(c  a) can be written as r  a  tb  uc
Where A is a point on the plane, and b and c are vectors parallel to the plane.
(a) Find the equation of the plane, in vector form, which contains the points
A(1, 2, -1), B(-2, 3, 2) and C(4, 5, 2).
(b) Find the point on the plane corresponding to the parameter values t=2, u=3
1
 2 
 4
(a) a   2  , b   3  , c   5 
 
 
 
 1
 2
 2
 
 
 
 1   2   4 
r  (1  t  u )a  tb  uc  (1  t  u )  2   t  3   u  5 
     
 1  2   2 
     
 1  t  u  2t  4u   1  3t  3u 
  2  2t  2u  3t  5u    2  t  3u 


 


 1  t  u  2t  2u 

1

3
t

3
u


 
4
(b) When t  2, u  3 r  13  . So the required point is R(4, 13, 14)
 
14 
 
Find the cartesian equation of the plane whose vector equation is
r  (2  t  3u )i  (1  t  3u ) j  (3  t  7u )k
r  2i  ti  3ui  j  t j  3u j  3k  tk  7uk
 2i  j  3k  t (i  j  k )  u (3i  3 j  7k )
Expressed in component form
 2  1   3 
r   1   t  1   u  3 
     
 3   1  7 
     
comparing with r  a  tb  uc
1
3
 
 
We see that the point (2, 1, 3) lies in the plane and 1 and -3
 
 
 1
 -7 
 
 
are vectors lying in the plane.
Hence a vector normal to the plane is
Using the point (2, 1, 3) to find k,
10 x  4 y  6 z  34
1  3
 10 
i j k
 1    3 
 4 

     1 1 1


 1  7 
 6 
    3 3 7


 10   2 
 4  . 1   34

 
 6   3 

 
or 5x  2 y  3z  17
Page 57 Exercise 6 Questions 1, a, b, 2a, 3, 4a, c, 5a, 9, 10
Page 63 Exercise 8 As many as you can manage (A/B)
TJ Exercise 4
Intersection of 2 lines
Two lines in space can be
•Parallel
•Intersect at a point
•Skew. (not parallel but never intersect)
Strategy for finding the point of Intersection of two lines
•Express the equations of the lines in parametric form using parameters t1 and t2
•Equate the corresponding expressions for x, y and z producing 3 equations in
two unknowns
•Use two of the equations to find the values of t1 and t2
•Substitute these values into the third equation. If they satisfy then the point of
intersection has been found. If not, then the lines do not intersect.
Show that the lines with equations x  5  ( y  2)  z and
x  12 y  3 z  5


intersect and find the point of intersection.
5
2
4
Equating:
Using parameters t1 and t2;
x  t1  5
x  5t2  12
t1  5t2  7
1
y  t1  2
y  2t2  3
t1  2t2  1
2
z  t1
z  4t2  5
t1  4t2  5
3
1  2  3t2  6  0  t2  2  t1  3
Substituting these values into equation 3 shows that they satisfy the equation.
Hence the lines intersect.
x  2, y  1, z  3
(2, 1, -3)
Find the acute angle between the lines with equations
x  2 y  1 z  11


and x  t  3, y  t  4, z  8
1
2
1
The second equation can be rewritten as
1
 -1
This gives the vectors  -2  and  1 
 
 
 -1 
0
 
 
cos  
1  2  0  3

 1500
2
6 2
Hence the acute angle is 300
x 3 y  4 z 8


1
1
0
Angle between two planes
π2 A
P
The angle between two planes can be constructed by
picking any point B on the line of intersection of the planes
and drawing perpendiculars to the line BA and BC on both
planes. The angle ABC is the required angle.
B
π1
C
Viewing from P along the line of intersection we can see
that the angle ABC is the same as the angle between
the normal.
Find the angle between the planes with equations, x + 2y + z = 5 and x + y = 0.
1
1
By in spection the normals are: a   2  and b   1 
 
 
1
 0
 
 
cos 
a.b 1  1  0
3


ab
2
6 2
  300
Page 70 Exercise 11 Questions 1 and 2
Page 59 Exercise 7A Questions 1, 2 and 3
TJ Exercise 5 questions 1 and 2
Angle between a line and a plane
90-
u
a

 Is the compliment of the angle between the line
and the normal to the plane
sin   cos(90   )
Use modulus since <90
a.u

au
Strategy
•Change the equation of the line to parametric form if necessary.
•Substitute x y and z into the equation of the plane.
•Solve this subsequent equation to obtain a value of the parameter, t, and hence
the coordinates of the point of intersection.
x  7 y  11 z  24


and the plane
3
4
13
6 x  4 y  5 z  28, find
Given the line
(a) The point of intersection
(b) The angle the line makes with the plane.
(a) x  3t  7 y  4t  11 z  13t  24  (3t  7,4t  11,13t  24)
Substitute this into the equation of the plane.
6(3t  7)  4(4t  11)  5(13t  24)  28
31t  62  0
t  2
 point of intersection (1, 3, -2)
(b) a  (6i  4 j  5k )
u  (3i  4 j  13k )
(b) a  (6i  4 j  5k )
u  (3i  4 j  13k )
a.u
sin   cos(90   ) 
au


6  3  4  4  5 13
36  16  25  9  16  169
31
77  194
0.254
 14.70 to 3 significant figures
Page 68 Exercise 10 Questions 1, 2, 3, 4
TJ Exercise 5 Question 3
Line of intersection of two planes
Two planes must either be parallel or intersect in a line.
To determine the equations of the line of intersection we need to know its direction
vector and a point on the line.
To find a point on the line:
•The line must either cross the (x,y) plane (which has equation z = 0) or be
parallel to it.
•If it crosses it, set z = 0 in the equations of both planes to obtain a pair of
simultaneous equations in x and y. Solving them will provide the required point
on the line (x1, y1, 0)
•If the line is parallel to the (x,y) plane then a similar point can be found on the
(x,z) plane by a similar strategy. (set y = 0)
To find the direction vector:
•The line of intersection lies in both planes.
•Its direction vector is therefore perpendicular to the normal vector in each plane.
•Thus the direction vector is parallel to the vector product of these normal vectors.
Find the equations of the line of intersection of the planes with equations
x  2 y  3 z  1 and 2 x  y  z  3
Let z  0
x  2 y  1 and 2 x  y  3
2(2 y  1)  y  3  5 y  2  3
y  1  x  1
 (1, 1,0) lies on the line of intersection.
Normal Vectors are:
u  i  2 j  3k
 5 
u  v  1 2 3   5 
 
2 1 1  5 
i
j
k
v  2i  j  k
x 1 y 1 z
 Equation of the line is


5
5 5
x 1 y 1 z
also


1
1 1
For the intersection of 3 planes look at page 76 and 77.
Page 72 Exercise 12 Questions 1, 2
Page 78 Exercise 15 Questions 1a, c, 2a, c
TJ Exercise 5 Questions 5 to 12
END OF TOPIC