vector theory (15/3/2014)media.freeola.com/other/28841/vectortheory230514.pdf · vector theory...
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Vector Theory (15/3/2014)
Contents
(1) Equation of a line
(i) parametric form
(ii) relation to Cartesian form
(iii) vector product form
(2) Equation of a plane
(i) scalar product form
(ii) parametric form
(iii) converting between scalar product and parametric forms
(3) Angle between two direction vectors
(4) Perpendicular vectors
(i) vector perpendicular to given (2D) vector
(ii) vector perpendicular to two given (3D) vectors
(5) Intersections (lines are 3D)
(i) point of intersection of two lines
(ii) point of intersection of a line and a plane
(iii) line of intersection of two planes
(6) Shortest distances
(i) from a point to a plane
(ii) between two parallel planes
(iii) between parallel lines / from a point to a line
(iv) between two skew lines
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(1) Equation of a line
(i) parametric form (2D example; but can be extended to 3D)
The vector equation of the line through the points A & B can be
written in various forms:
(a) r = a + d
(b) r = a + (
(c) r = a +
(a weighted average of ; when ; when
; when
, is the average of ; the diagram
shows
)
(d) ( ) = (
) + (
) or (
)
where a = (
) and d = (
) is any vector in the direction from A
to B
(normally d1 & d2 are chosen to be integers with no common
factor)
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Note the difference between (a) the vector equation of the line
through the points A & B and (b) the vector : The vector
has magnitude |AB| (the distance between A & B) and is in the
direction from A to B.
Whereas the vector equation of the line through A & B is the
position vector of a general point P on the line, with completely
different magnitude and direction to that of the vector
(ii) relation to Cartesian form
( ) = (
) + (
)
the straight line through with gradient
(iii) vector product form (3D lines only)
r = a + d can be written as ( )
(since and are parallel)
or
eg line through (1, 0, 1) and (0, 1, 0):
( ) (
) (
)
|
| (
)
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Thus equation is (
) (
)
Note: Textbooks tend to write the determinant with the elements
transposed (it gives the same result though).
(2) Equation of a plane
(i) scalar product form
Let be the position vector of a point in the plane,
and ( ) be a general point in the plane.
Let be a vector perpendicular to the plane.
As and are perpendicular,
(a constant)
(Cartesian form)
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Example
If ( ) and (
), then
(Another way of thinking of this is that, since is a point on the
plane, it is a solution of , so that , or )
(ii) parametric form
This is an extension of the parametric form of the vector equation
of a line.
Let and be non-zero vectors in the plane (that are not parallel
to each other).
Then
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Note that and are direction vectors, whilst is a position
vector. and can of course be determined from 2 points and
in the plane, as and (or )
(iii) converting between scalar product and parametric
forms
(a) to convert from scalar product to parametric form
Example
Suppose that the equation of the plane is
Let and , so that and a general point
is
( ) (
) (
) (
) (
)
(b) to convert from parametric to scalar product form
Method 1
Example: ( ) (
) (
) ( )
Then eliminate s and t to obtain an equation in .
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Method 2
For the above example, create normal vector: (
) ( )
|
|
giving
(3) Angle between two direction vectors
Example 1
To find the acute angle between the line with equation
( ) (
) and the plane with equation (
)
The two direction vectors in this case are (
) and (
)
Then (
) (
) |
| |
| (*),
so that
√ √
√
This gives
Whether is acute or obtuse depends on the relative direction of
the normal vector to the plane and the direction vector of the
line - see the diagram below.
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In this case, the angle we want (between the plane and the line) is
in the diagram.
Thus
Example 2
If we need to find the angle between two planes, then the angle in
question will be in the diagram below. This will be acute, so
that we expect to be obtuse (as ). If one of the
normals to the planes has its direction reversed, then we obtain
an acute angle from the scalar product result (*), and this has to
be converted to the required angle by subtracting from .
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(4) Perpendicular vectors
(i) vector perpendicular to given (2D) vector
Example
Given direction vector ( ): gradient is
; hence perpendicular
gradient
and perpendicular direction vector is (
) or
(
)
(ii) vector perpendicular to two given (3D) vectors
Let given vectors be and
Method 1
Method 2
Let ( ) be required vector.
Then eliminate two of from and (*)
to give a direction vector in terms of parameter .
eg (
)
(note: form of eq'ns (*) ensures that will be multiples of )
which is equivalent to the direction vector ( )
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(5) Intersections (lines are 3D)
(i) point of intersection of two lines
Note: Lines may not have a point of intersection, if the equations
are not consistent; in which case they are termed 'skew'.
Example: intersection of
where has equation ( ) (
) ( )
and has equation ( ) (
) (
)
Eliminate , to give ( ) =(
)
(ii) point of intersection of a line and a plane
Example: has equation ( ) (
) ;
plane has equation (
)
Then (( ) (
)) (
) creates a linear equation in .
(It is possible that the line is either parallel to the plane or lies in
the plane; in which case the term corresponding to
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(
) (
) above will vanish, since the scalar product will be
zero; then the remaining numbers will only be consistent if the
vector corresponding to ( ) lies in the plane; ie if the scalar
product corresponding to ( ) (
) equals the right-hand side.)
(iii) line of intersection of two planes
Method 1
Starting with the equations of the planes in the form , let
(eg) , to obtain ( ) (
); ie will be expressible
as linear functions of .
Note that we are effectively choosing a point on the line which has
coordinate 0 (and coordinates ).
Example 2 covers an unusual case.
Example 1
Planes
Let , so that and
So that the equation of the line of intersection of the planes is:
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( ) (
)
Example 2
Planes
This implies that and ,
so that the equation of the line of intersection of the planes is:
( ) (
) , where any value can be chosen for p
Method 2
The required line will be perpendicular to the normal vectors of
both planes. Therefore the vector product of the normal vectors
to the two planes has the direction vector of the required line.
Using Example 1 above, with planes ,
( ) (
) |
|
In order to find the equation of the line, we just need a point on it;
ie a point on both planes, so that
eg let ; then
and the equation of the line is (
) (
)
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Note: This can be seen to be equivalent to ( ) (
) in
Example 1, as follows:
Let , so that
Then (
) (
) (
)
(6) Shortest distances
(i) shortest distance from a point to a plane
(See "Equation of a plane" to convert between the scalar product
and parametric forms of the equation of a plane, if necessary.)
Method 1
Example 1
Point, P is ( ) ; plane has equation (*)
The position vector of the point in the plane at the shortest
distance from P is:
( ) (
) for some (to be determined), as (
) is the
direction vector normal to the plane.
Since this point lies in the plane, it satisfies (*);
hence (**)
giving
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The shortest distance is the distance travelled from P to the plane,
along the direction vector (
); ie
|(
)|
Example 2
In the special case where P is the origin O (and the plane has
equation as before), (**) becomes
ie , where is the normal to the plane, (
)
As before, the shortest distance from O to the plane is
| |
| |
Alternatively, if the equation of the plane is given in 'normalised'
form (ie the direction vector has unit magnitude; the word
'normal' being used here in a different sense to that of the normal
to a plane);
ie
, then the distance required is simply the
right-hand side of the equation.
Method 2
Using the above example, we can find the equation of the plane
parallel to and passing through ( ).
The equation of the parallel plane will be
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ie
From the special case of the Origin in Method 1, the distance
between the two planes (and hence between the point and the
plane) is
√
Note: This method gives rise to the standard formula:
√
, as the shortest distance from the point (
)
to the plane
Method 3
Using the same example, where P is ( ) and the plane has
equation (*),
we first of all find a point Q in the plane (as in the diagram above)
and create the vector
The required distance will then be the projection of onto
(the normal to the plane); namely
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In this case, putting (say) in (*) gives
, so that
(
) , and (
)
Then (
)
and the shortest distance
√
(ii) distance between two parallel planes
Using the method for finding the shortest distance from the origin
to a plane (method 1, example 2 of "shortest distance from a point
to a plane"), the two planes need first of all to be put into
normalised form; the constant term of each equation then gives
the distance of the plane from the origin, so that the distance
between the planes is then the difference between the constant
terms.
Example: Find the distance between the planes
and
As √ , the normalised equations are
and
so that the distance between the planes is
(iii) distance between parallel lines / shortest distance from
a point to a line
Assuming that A and B are given points on the two lines, and that
is the common direction vector:
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Method 1
Let C be the point on with parameter , so that (*)
Then we require ( )
Solving this equation for and substituting for in (*) gives ,
and the distance between the two lines is then .
Example
Let lines be ( ) (
) and (
) ( )
If ( ) , (
) and ( ) (
),
then (
) (
)
Hence (
) and the distance between the lines is
√ √
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Method 2
Having obtained the general point, (
) on in
Method 1, we can minimise the distance BC by finding the
stationary point of either or :
Then
and
, as before
Method 3
As ,
In the above example, =(
) and
|
| (
)
Then √
√
√
√ √
(iv) shortest distance between two skew lines
Method 1
eg & ; A has position vector
etc
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XY is shortest distance, as it is perpendicular to both and
unit vector in direction of XY is
AE = XY ; (as CE is in the plane of the 'back wall' of
the cuboid - because C lies on )
So AE is the projection of onto the direction of XY; ie onto
So XY = AE = |( )
| (the modulus sign ensuring that the
distance is +ve)
Note: This method can't be used to find the distance between two
parallel lines, as | | , since
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Example 1a: To find the shortest distance between the lines
(
) (
) and (
) (
)
The direction normal to the two lines is
|
| (
) ; and we can take (
) instead
As √ the unit vector in this direction is
(
)
We then require ((
) (
))
(
)
(
) (
)
so that the required distance is
or 9.2
Method 2
Find the vector perpendicular to both and , as in Method 1:
Then the equation of the plane with normal , containing line
(ie the front face of the cuboid in Method 1) will be
Similarly the equation of the plane with normal , containing line
(ie the back face of the cuboid) will be
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The distance between these two planes (ie XY) is obtained by first
adjusting the equations of the planes, so that they are based on a
normal vector of unit magnitude.
Thus
and
Then
| |
| | [See "Distance between two parallel
planes"]
[Note that this method is algebraically equivalent to method 1.]
Example 1b (Lines as in 1a)
From Example 1a,
(
)
Then the equations of the planes in which the front and back faces
of the cuboid in method 1 lie are
[ ]
and
[ ]
So the distance between the two planes, and hence between the
two lines is
Method 3
Referring to the earlier diagram, suppose that X and Y have
position vectors & respectively.
Then, if is the vector normal to both and ,
(*)
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(ie Y is reached by travelling first to X and then along XY) and XY
will then
(*) gives 3 simultaneous equations in
(
) (
) , from which can be found
Example 1c: (Lines as in 1a)
From Example 1a, (
)
We need to find such that
(
) (
) (
) (
) (
)
So
or (
)(
) (
)
(
)
(
)(
)
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and
√
Method 4
As in method 3, suppose that X and Y have position vectors
& respectively.
Then
and (*)
Solving (*) enables to be determined,
from which | | can be found
Example 1d: (Lines as in 1a)
(
) (
) (
) (
)
(
)
Then (
) (
)
and (
) (
)
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and
ie or
and or
or (
) (
) (
)
(
)
(
) (
)
(
)
(
)
Then (
)
(
)
and | |
√