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 | |

√

vector theory (15/3/2014)media.freeola.com/other/28841/vectortheory230514.pdf · vector theory...

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