Transcript C4.8 Vectors 2 - Heathcote School

A2-Level Maths:
Core 4
for Edexcel
C4.8 Vectors 2
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Contents
The scalar product
The scalar product
The vector equation of a line
Intersecting lines
Examination-style questions
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The scalar product
Two vectors can be multiplied together to give the scalar
product, also known as the dot product.
The result of this multiplication is a scalar quantity, hence the
name.
The scalar product of two vectors a and b is defined as:
a.b = a b cos
where θ is the angle between a and b when they are placed tail
to tail.
Note that θ is always taken to be between 0° and 180°.
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The scalar product
Whether the scalar product of two vectors is positive, negative
or zero depends on the angle between them.
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Commutative and distributive properties
The scalar product is said to be commutative.
In other words, the order of a and b is not important:
a.b = b.a
We can show this as follows:
a.b = a b cos
= b a cos
= b.a
The scalar product is also distributive over addition:
a.(b + c) = a.b + a.c
We can prove this fact using a diagram.
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Commutative and distributive properties
Suppose we arrange the vectors a, b and c as follows:
c
B
b
O
β
Q
This is the vector b + c.
S
We now introduce the points
P, Q, R and S as shown.
Let θ be the angle AOB …
b+c
… α the angle AOQ …
θ α
P
a
R
A … and β the angle SBQ.
Now, using trigonometry:
OP = |b| cos θ
OR = |b + c| cos α
PR = |c| cos β
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Commutative and distributive properties
Since OR = OP + PR we can write:
|b + c| cos α = |b| cos θ +|c| cos β
Multiplying through by |a| gives:
|a||b + c| cos α = |a||b| cos θ + |a||c| cos β
Therefore,
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a.(b + c) = a.b + a.c
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Perpendicular and parallel vectors
We have seen that, since cos 90° = 0:
If a.b = 0 then either a = 0, b = 0 or a and b are perpendicular.
In particular, for the unit base vectors i, j and k:
i.j = i.k = j.i = j.k = k.i = k.j = 0
Also, if two vectors are parallel the angle between them is
taken as 0°.
Since cos 0° = 1 we can conclude from the scalar product that:
If two vectors a and b are parallel, a.b = |a||b|.
In particular, for the unit base vectors i, j and k:
i.i = i.j = k.k = 1
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Scalar products of vectors in component form
These two results make it easy to find the scalar product of two
vectors given in component form. For example,
Find the scalar product of a = i + 2j + k and b = –i + 4j + 3k.
a.b = (i + 2j + k ).(–i + 4j + 3k)
= i.(–i + 4j + 3k) + 2j.(–i + 4j + 3k) + k.(–i + 4j + 3k)
= –i.i + 4i.j + 3i.k – 2j.i + 8j.j + 6j.k – k.i + 4k.j + 3k.k
Since i.j = i.k = j.i = j.k = k.i = k.j = 0, this reduces to:
a.b = –i.i + 8j.j + 3k.k
And since i.i = j.j = k.k = 1,
a.b = –1 + 8 + 3 = 10
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Scalar products of vectors in component form
In general, if a = a1i + a2j + a3k and b = b1i + b2j + b3k then
a.b = a1b1+ a2b2 + a3b3.
Prove that the vectors a = –3i + j + 2k and b = 2i + 8j – k
are perpendicular.
a.b = (–3i + j + 2k).(2i + 8j – k )
= (–3 × 2) + (1 × 8) + (2 × –1)
= –6 + 8 – 2
=0
Since a and b are both non-zero vectors, and their scalar
product is 0, they must be perpendicular.
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Finding the angle between two vectors
A useful application of the scalar product is in finding the angle
between two vectors.
To do this we can write the scalar product in the form:
cos =
a.b
ab
For example,
Find the angle between the vectors a and b where
5
a =  1
2
 
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 2 
0
b
=
and
 
3
 
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Finding the angle between two vectors
 5   2 
a.b =  1 .  0 
23
  
= (5× 2) + ( 1×0) + (2×3)
= 10 + 0 + 6
= 4
Also,
a = 52 +12 + 22 = 30
and
b = 22 + 02 + 32 = 13
cos  =
a.b
4
=
a b
30 13
 = 101.69 (to 2 d.p.)
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Contents
The vector equation of a line
The scalar product
The vector equation of a line
Intersecting lines
Examination-style questions
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The vector equation of a line
The vector equation of a line is given in terms of two vectors.
Then first vector tells you how to get from the origin to the line
and the second vector tells you the direction of the line.
For example, suppose we have the line given by the Cartesian
equation
2x
y
3
5
2
This line passes through the point (0, 3) and has gradient 5 .
Let A be the point (0, 3) so:
0
OA = a =  
3
The gradient tells us that for every 5 units moved along the
line in the x-direction we move 2 units in the y-direction.
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The vector equation of a line
5
The direction of the line is therefore given by the vector   .
 2
Let’s call this vector b.
Let R(x, y) be a general point on the
b
line.
R
If r is the position of vector of R then
A
we can write:
r
OR = r = OA  AR
a
AR is given by a scalar multiple of b so:
O
r = OA + t AB
We can therefore write a vector equation of this line as:
0 5
r =    t   where t is a scalar.
3  2
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The vector equation of a line
This vector equation is not unique since we can use the
position vector of any point of the line and the direction can be
given by any multiple of the direction vector.
For example, the equations
5 5
r =    t  ,
5  2
 5   10 
r =    t   , and
5  4 
 5   15 
r = t

1

6
  

are also vector equations of this line.
In general, a vector equation of a line that passes through the
point A with position vector a and in the direction of the vector
b is
r = a  tb
where t is a scalar parameter.
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A line through two given points
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A line through two given points
In general if two points on a line A and B have position vectors
a and b, then the equation of the line is given by
r = a + t(b – a)
For example,
Find the vector equation of the line passing through the points
A(3, 7, –7) and B(–9, 1, –4).
 12 
 4 
 6  = 3   2 


 
 3 
 1


 
 3   4 
   
A vector equation of this line is therefore r =  7   t  2  .
 7   1 
This can also be written as
   
3 
OA =  7 
 7 
 
 9  3 
AB =  1  7  =
 4   7 


r = 3i + 7j – 7j + t(–4i –2j + k)
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Contents
Intersecting lines
The scalar product
The vector equation of a line
Intersecting lines
Examination-style questions
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Intersecting lines
In two dimensions, any two given lines must either be parallel
or they must intersect.
For example, the two lines given by the vector equations:
r = 2i – j + s(4i – 2j) and r = –i + 5j + t(–6i + 3j)
must be parallel because the direction vectors 4i – 2j and
–6i + 3j are multiples of each other.
The lines given by the vector equations:
r = 3i – 8j + s(–i + 2j) and r = 4i – j + t(i + j)
are not parallel and so they must intersect.
This intersection occurs when the i components and the j
components of both lines are equal. Equating these gives:
1
3–s=4+t
–8 + 2s = –1 + t
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Intersecting lines
11 – 3s = 5
–3s = –6
s=2
Substituting this into 1 gives:
3–2=4+t
t = –3
We can now find the point of intersection by substituting s = 2
into the equation of the first line or by substituting t = –3 into
the equation of the second line.
When s = 2 the equation of the first line becomes:
1 – 2 :
This is the position vector
r = 3i – 8j + 2(–i + 2j) = i – 4j of the intersection point.
The point of intersection is therefore (1, –4).
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The angle between two intersecting lines
Suppose we are given the vector equations of two lines and we
want to find the angle between them.
We can do this using the direction vectors of each line.
In our last example we had the two lines given by the vector
equations r = 3i – 8j + s(–i + 2j) and r = 4i – j + t(i + j).
The directions of these two lines is given by the vectors –i + 2j
and i + j.
The angle between them, θ, is therefore given by:
( i + 2 j).(i + j)
cos =
1+ 4 1+1
1
=
5 2
  = 71.56 (to 2 d.p.)
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Intersecting lines
In three dimensions, any two given lines can either be parallel,
they can intersect or they can be skew.
Two lines are said to be skew
if they are neither parallel nor
do they intersect.
As in two dimensions, we can
conclude that two lines are
parallel if the vectors giving
their directions are multiples
of each other.
To work out whether two lines intersect we form three
simultaneous equations by equating the i, j and k components
to give three equations in two unknowns.
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Intersecting lines
If the values given by the first and second equation satisfy the
third equation then the lines intersect.
If they don’t satisfy the third equation then they are skew.
For example,
Determine whether the lines given by the vector equations
r = 2i + j – 3k + s(i – 5j + 2k) and r = 8i – 9j + 4k + t(4i + 3k)
intersect. If they do intersect give the coordinates of their point
of intersection.
Equating the coefficients of i, j and k:
2 + s = 8 + 4t
1 – 5s = –9
–3 + 2s = 4 + 3t
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2
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Intersecting lines
–5s = –10
s=2
Substituting this into 1 gives:
2 + 2 = 8 + 4t
4t = –4
t = –1
If these values of s and t satisfy equation 3 then the lines
intersect.
–3 + 2s = –3 + 4 = 1
4 + 3t = 4 – 3 = 1
LHS = RHS so the lines intersect.
When s = 2 the equation of the first line becomes:
r = 2i + j – 3k + 2(i – 5j + 2k) = 4i – 9j + 7k
The point of intersection is therefore (4, –9, 7).
Rearranging 2 :
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Converting from vector to Cartesian form
If we are given the vector equation of a line and we want to
write it in Cartesian form we can do this by writing the vector
equation as a set of parametric equations. For example,
Find the Cartesian equation of the line r = 5i – j + t(–2i + 3j).
 x
Since r is a general point on the line we can write it as   .
 y
The vector equation of the line can therefore be written as:
 x   5   2 
 y  =  1 + t  3 
     
The equation of the line is therefore given by the parametric
equations:
x = 5 – 2t
y = –1 + 3t
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Converting from vector to Cartesian form
Rearranging to make t the subject of these equations gives:
5x
t=
2
y +1
t=
3
Equating these gives us the Cartesian form of the equation of
the line.
y +1 5  x
=
3
2
2 y + 2 =15  3 x
2 y + 3 x  13 = 0
The same method can be used for a line given in three
dimensions.
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Converting from vector to Cartesian form
Find the Cartesian equation of the line
r = 6i + 3j – 4k + t(5i – 4j + 7k).
 x
 y
Since r is a general point on the line we can write it as   .
z
 
The vector equation of the line can therefore be written as
 x  6   5 
 y  =  3  + t  4 
     
 z   4   7 
     
The equation of the line is therefore given by the parametric
equations:
x = 6 + 5t
y = 3 – 4t
z = –4 + 7t
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Converting from vector to Cartesian form
Rearranging to make t the subject of these equations gives:
x6
t=
5
3 y
t=
4
z+4
t=
7
Equating these gives us the Cartesian form of the equation of
the line.
x6 3 y z+4
=
=
5
4
7
In general the Cartesian form of a line given by the vector
equation r = a1i + a2j + a3k + t(b1i + b2j + b3k ) is
x  a1 y  a2 z  a3
=
=
b1
b2
b3
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Contents
Examination-style questions
The scalar product
The vector equation of a line
Intersecting lines
Examination-style questions
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Examination-style question 1
Relative to a fixed origin O, the points A and B have position
vectors –2i – 2j – k and mi + 4j – 7k respectively, where m is
a constant.
C is a point such that OABC is a rectangle.
a) Find the value of the constant m.
b) Write down the coordinates of the point C.
c) Show the ratio of the rectangle’s height to its width is 1 : 3.
a) Since OABC is a rectangle OA must be perpendicular to AB .
AB = OB  OA
= (mi + 4j – 7k) – (–2i – 2j – k)
= (m + 2)i + 6j – 6k
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Examination-style question 1
OA. AB = (–2i – 2j – k) .(m + 2)i + 6j – 6k)
= –2(m + 2) + (–2 × 6) + (–1 × –6)
= –2m – 4 – 12 + 6
= –2m – 10
Since OA and AB are perpendicular OA. AB = 0, so
2m = –10
m = –5
b) OC is the side opposite AB and so the position vector of the
point C is the same the vector AB .
Using the value m = –5,
OC = AB = –5i + 6j – 6k
 The coordinates of C are (–5, 6, –6).
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Examination-style question 1
c) The height of the rectangle is given by the magnitude of the
vector OA and the width of the rectangle is given by the
magnitude of the vector AB .
OA = ( 2)2 + ( 2)2 + ( 1)2
= 9
=3
OB = ( 3)2 + (6)2 + ( 6)2
= 81
=9

height : width
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=
3:9
=
1 : 3 as required.
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Examination-style question 2
The points A, B and C have coordinates (2, –4, 6), (2, –3, –1)
and (1, –2, 3).
a) Write down the position vectors of the points A, B and C
relative to a fixed origin O.
b) Calculate the cosine of angle ABC in its simplest form.
c) Find the exact area of the triangle ABC.
d) Write down the equation of the line AB in the form
r = a + tb.
The point D has coordinates (a, b, c) and lies on the line AB
such that the line CD is perpendicular to the line AB.
e) Find the values of a, b and c and hence give the
coordinates of the point D.
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Examination-style question 2
a) OA = 2i  4 j + 6k
OB = 2i  3 j  k
OC = i  2 j + 3k
b) Let θ be the angle ABC:
BA = OA  OB
A
= 2i – 4j + 6k – 2i + 3j + k
θ
C
B
= – j + 7k
BC = OC  OB
= i – 2j + 3k – 2i + 3j + k
= – i + j + 4k
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Examination-style question 2
The cosine of the angle θ is given by
cos =
BA.BC
BA BC
=
=
(0 × 1) + ( 1×1) + (7 × 4)
12 + 72 12 +12 + 42
27
50 18
27
=
900
27
=
30
9
=
10
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Examination-style question 2
c) The area of triangle ABC is given by
Area =
1
2
AB BC sin
From part a), |AB||BC| = 30. Also, sin θ = 1  cos2  so
Area = 21 × 30 × 1  cos2 
= 15 × 1 
81
100
19
= 15 ×
100
15 19
=
10
3 19
=
units squared
2
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Examination-style question 2
d) A vector equation of the line AB is given by
 
r = OA + t AB
Since BA = –j + 7k, an equation of the line AB is
r = 2i  4 j + 6k + t  j  7k 
e) The coordinates of the point D are (a, b, c).
Since D lies on the line AB we can write
ai + b j + cz = 2i  4 j + 6k + t  j  7k 
Equating the i, j and k components gives
a=2
b = –4 + t
c = 6 – 7t
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Examination-style question 2
The position vector of the point D can therefore be written in
terms of the parameter t as:
OD = 2i + (t  4) j + (6  7t )k
We are also told that CD is perpendicular to AB.
CD = OD  OC
= 2i +(t  4) j +(6  7t )k  i + 2 j  3k
= i +(t  2) j +(3  7t )k
The scalar product of AB and CD is 0 so
 j  7k  . i + (t  2) j + (3  7t )k  = 0
(1×0) +(t  2)  7(3  7t ) = 0
t  2  21+ 49t = 0
50t = 23
t = 0.46
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Examination-style question 2
So for the point D (a, b, c):
a=2
b = –4 + t
= –4 + 0.46
= –3.54
c = 6 – 7t
= 6 – (7 × 0.46)
= 2.78
 The coordinates of the point D are:
(2, –3.54, 2.78)
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