Vector Theory (24 pages; 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 lthrough the points A & B can be written in various forms:

(a) r = a + λd

(b) r = a + λ(b−a¿

(c) r = (1− λ)a + λb

(a weighted average of a∧b; when λ=0 , r=a; when λ=1 , r=b; when

λ=12 , r is the average of a∧b; the diagram shows λ=13)

(d) ( xy ) = (a1a2) + λ (d1d2) or (a1+ λd1a2+ λd2)

where a = (a1a2) and d = (d1d2) 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 A⃗B: The vector A⃗B 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 r of a general point P on the line, with completely different magnitude and direction to that of the vector A⃗B .

(ii) relation to Cartesian form

( xy ) = (a1a2) + λ (d1d2) ⇒ λ=x−a1d1

=y−a2d2

⇒ y=a2+d2d1. (x−a1 )

the straight line through (a1 , a2) with gradient d2d1

(iii) vector product form (3D lines only)

r = a + λd can be written as (r−a )×d=0

(since r−a and d are parallel)

or r ×d=a×d

eg line through (1, 0, 1) and (0, 1, 0):

d=(010)−(101)=(−11−1) a×d=|i 1 −1

j 0 1k 1 −1|=−i+k=(−101 )

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Thus equation is r ×(−11−1)=¿ (−101 )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 a be the position vector of a point in the plane,

and r=( xyz ) be a general point in the plane.

Let n be a vector perpendicular to the plane.

As r−a and n are perpendicular, (r−a) .n=0

⇒r . n=a .n=p (a constant)

⇒ nx x+n y y+nz z=p (Cartesian form)

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Example

If a=(124) and n=(−1211−9 ), then

−12 x+11 y−9 z=−12 (1 )+11 (2 )−9 (4 )=−26

(Another way of thinking of this is that, since a is a point on the plane, it is a solution of r . n=p, so that p=a .n, or n .a)

(ii) parametric form

This is an extension of the parametric form of the vector equation of a line.

Let b and c be non-zero vectors in the plane (that are not parallel to each other).

Then r=a+(λb+μc )

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Note that b and c are direction vectors, whilst a is a position vector. b and c can of course be determined from 2 points p and q in the plane, as p−a and q−a (or p−q)

(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 2 x+3 y+z=4

Let x=s and y=t , so that z=4−2 s−3 t and a general point is

( xyz )=( st

4−2 s−3t )=(004)+s (10

−2)+ t(01

−3)

(b) to convert from parametric to scalar product form

Method 1

Example: r=( xyz )=(124)+s (−135 )+t (231)

→x=1−s+2 t

y=2+3 s+3 t

z=4+5 s+t

Then eliminate s and t to obtain an equation in x , y∧z.

Method 2

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For the above example, create normal vector: (−135 )×(231) ¿|i −1 2j 3 3k 5 1|=−12 i+11 j−9k

giving −12 x+11 y−9 z=−12 (1 )+11 (2 )−9 (4 )=−26

(3) Angle between two direction vectors

Example 1

To find the acute angle between the line with equation

r=(234)+λ (−1−21 ) and the plane with equation r .( 51−1)=1

The two direction vectors in this case are (−1−21 ) and ( 51−1)

Then (−1−21 ) .( 51−1)=| 12−1||

51

−1|cosθ (*),

so that cosθ=(−1 ) (5 )+ (−2 ) (1 )+(1)(−1)

√1+4+1√25+1+1=

−8√162

=−0.62854

This gives θ=128.9°

Whether θ is acute or obtuse depends on the relative direction of the normal vector to the plane (n) and the direction vector of the line (d ) - see the diagram below.

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In this case, the angle we want (between the plane and the line) is

α in the diagram.

Thus α=90−(180−128.9 )=38.9 °

Example 2

If we need to find the angle between two planes, then the angle in question will be α in the diagram below. This will be acute, so that we expect θ to be obtuse (as θ=180−α). 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 180 °.

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(4) Perpendicular vectors

(i) vector perpendicular to given (2D) vector

Example

Given direction vector (23): gradient is 32 ; hence perpendicular

gradient ¿−23 and perpendicular direction vector is (−32 ) or ( 3−2)

(ii) vector perpendicular to two given (3D) vectors

Let given vectors be a and b

Method 1

a×b

Method 2

Let r=( xyz ) be required vector.

Then eliminate two of x , y∧z from r . a=0 and r . b=0 (*)

to give a direction vector in terms of parameter x , y∨z.

eg ( x2 x3 x) (note: form of eq'ns (*) ensures that y∧z will be multiples of x)

which is equivalent to the direction vector (123)

(5) Intersections (lines are 3D)

(i) point of intersection of two lines

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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 l1∧l2 ,

where l1 has equation r=( xyz )=( 1−6−1)+λ (123)

and l2 has equation r=( xyz )=(972)+μ(23

−1)Eliminate λ∧μ, to give ( xyz ) =( 3−25 )

(ii) point of intersection of a line and a plane

Example: l1has equation r=(234)+λ (12

−1) ; plane has equation r .( 51−1)=1 Then ((234 )+ λ(

12

−1)).(51

−1)=1 creates a linear equation in λ.

(It is possible that the line is either parallel to the plane or lies in the plane; in which case the term corresponding to

λ ( 12−1).(51

−1) above will vanish, since the scalar product will be zero;

then the remaining numbers will only be consistent if the vector

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corresponding to (234) lies in the plane; ie if the scalar product

corresponding to (234) .(51

−1) equals the right-hand side.)

(iii) line of intersection of two planes

Method 1

Starting with the equations of the planes in the form r . n=d, let

(eg) x=λ, to obtain ( xyz )=( λaλ+bcλ+d ); ie y∧z will be expressible as linear

functions of λ.

Note that we are effectively choosing a point on the line which has x coordinate 0 (and y∧z coordinates b∧d).

Example 2 covers an unusual case.

Example 1

Planes 2 x+z=3∧x+ y−z=2

Let x=λ, so that z=3−2 λ and y=2+ (3−2 λ )−λ=5−3 λ

So that the equation of the line of intersection of the planes is:

r=(053)+ λ(1

−3−2)

Example 2

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Planes 2 x+z=4∧x−z=−1

This implies that x=1 , z=2 and y= λ,

so that the equation of the line of intersection of the planes is:

r=(1p2 )+λ (010) , 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 2 x+z=3∧x+ y−z=2,

(201)×( 11−1)=|i 2 1j 0 1k 1 −1|=−i+3 j+2 k

In order to find the equation of the line, we just need a point on it; ie a point on both planes, so that 2 x+z=3∧x+ y−z=2

eg let z=0; then x=32∧ y=1

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and the equation of the line is r=(32120)+μ(−132 )

Note: This can be seen to be equivalent to r=(053)+ λ(1

−3−2) in Example

1, as follows:

Let 0+2 μ=3−2 λ, so that μ=32− λ

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Then (32120)+μ(−132 )=( λ

5−3 λ3−2 λ)

(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 (123) ; plane has equation 4 x+3 y−12 z=26 (*)

The position vector of the point in the plane at the shortest distance from P is:

(123)+λ (43

−12) for some λ (to be determined), as ( 43−12) is the direction

vector normal to the plane.

Since this point lies in the plane, it satisfies (*);

hence 4 (1+4 λ )+3 (2+3 λ )−12 (3−12λ )=26 (**)

giving λ=413

The shortest distance is the distance travelled from P to the plane,

along the direction vector ( 43−12); ie 413|( 43−12)|= 4

13(13 )=4

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Example 2

In the special case where P is the origin O (and the plane has equation 4 x+3 y−12 z=26 as before), (**) becomes

4 (4 λ )+3 (3 λ )−12 (−12 λ )=26

ie λ ¿n∨¿2=26¿, where n is the normal to the plane, ( 43−12)As before, the shortest distance from O to the plane is

λ|n|=26|n|

=2613

=2

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 413x+ 313

y−1213z=2, 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 4 x+3 y−12 z=26 and passing through (123).The equation of the parallel plane will be

4 x+3 y−12 z=4 (1 )+3 (2 )−12 (3 )

ie 4 x+3 y−12 z=−26

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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 26−(−26)

√42+32+(−12)2=5213

=4

Note: This method gives rise to the standard formula:

¿n1 p1+n2 p2+n3 p3−d∨ ¿√n12+n22+n32

¿ , as the shortest distance from the

point ( p1p2p3) to the plane n1 x+n2 y+n3 z=d

Method 3

Using the same example, where P is (123) and the plane has equation

4 x+3 y−12 z=26 (*),

we first of all find a point Q in the plane (as in the diagram above) and create the vector P⃗Q

The required distance will then be the projection of P⃗Q onto n (the

normal to the plane); namely ¿ P⃗Q .n∨ ¿¿n∨¿¿

¿

In this case, putting y=z=0 (say) in (*) gives x=132 , so that

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Q=(13200

) , and P⃗Q=(112

−2−3

)Then P⃗Q .n=(112 ) (4 )+(−2 ) (3 )+(−3 ) (−12 )=52

and the shortest distance ¿52

√42+32+(−12)2=5213

=4

(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

3 x+4 y+12 z=13 and 3 x+4 y+12 z=39

As √32+42+122=13, the normalised equations are

113

(3 x+4 y+12 z )=1 and 113 (3 x+4 y+12 z )=3

so that the distance between the planes is 3−1=2

(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 d is the common direction vector:

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Method 1

Let C be the point on l1 with parameter k , so that c=a+k d (*)

Then we require d . (c−b )=0

Solving this equation for k and substituting for k in (*) gives c, and the distance between the two lines is then ¿c−b∨¿.

Example

Let lines be r=(102)+ λ(3

−11 ) and r=(−2−1

−1)+λ (−311)

If a=(102) , b=(−2−1−1) and c=(102)+k (−

311),

then ( 3−11 ) .(1+3k+2−k+12+k+1 )=0 → 9+9k+k−1+k+3=0

→11k+11=0→k=−1

Hence c=(−211 ) and the distance between the lines is

√(−2+2)2+(1+1)2+(1+1)2=√8

Method 2

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Having obtained the general point, C=(1+3k−k2+k ) on l1 in Method 1, we

can minimise the distance BC by finding the stationary point of either BC or BC2:

BC2=(1+3k+2)2+(−k+1)2+(2+k+1)2

¿11k2+22k+19

Then ddk

(BC ¿¿2)=22k+22¿

and ddk

(BC ¿¿2)=0⇒k=−1¿, as before

Method 3

As BC=ABsinθ, BC=¿ A⃗B×d∨ ¿¿d∨¿¿

¿

In the above example, A⃗B=(−3−1−3) and A⃗B×d=¿

|i −3 3j −1 −1k −3 1 |=−4 i−6 j+6k=−2( 23−3)

Then BC=2√4+9+9√9+1+1

=2√22√11

=√8

(iv) shortest distance between two skew lines

Method 1

eg l1:r=a+ λb & l2: r=c+μd ; A has position vector a etc

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XY is shortest distance, as it is perpendicular to both l1 and l2

unit vector in direction of XY is b×d

¿b×d∨¿¿

AE = XY ; ∠CEA=90° (as CE is in the plane of the 'back wall' of the cuboid - because C lies on l2)

So AE is the projection of c−a onto the direction of XY; ie onto b×d

¿b×d∨¿¿

So XY = AE = |(c−a ) . (b×d )¿b×d∨¿¿| (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 |b×d|=0, since sinθ=0

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Example 1a: To find the shortest distance between the lines

l1:r=( 0−20 )+ λ( 22−1) and l2: r=(−1125 )+ λ( 7−144 )

The direction normal to the two lines is

b×d=|i 2 7j 2 −14k −1 4 |=( −6

−15−42) ; and we can take ( 2514) instead

As √4+25+196=15 , the unit vector in this direction is

115 ( 2514) We then require (( 0−20 )−(−1125 )) . 115 ( 2514)= 1

15 ( 1−14−5 ) .( 2514 )

¿ 115

(2−70−70 )=−13815

=−465

so that the required distance is 465 or 9.2

Method 2

Find the vector perpendicular to both b and d , as in Method 1: n=b×d

Then the equation of the plane with normal n , containing line l1 (ie the front face of the cuboid in Method 1) will be r . n=a .n

Similarly the equation of the plane with normal n , containing line l2 (ie the back face of the cuboid) will be r . n=c .n

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.

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Thus r . n

¿n∨¿= a .n¿n∨¿¿

¿ and r . n

¿n∨¿= c .n¿n∨¿¿

¿

Then XY=¿ a .n|n|

− c .n|n|

∨¿ [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, n|n|=

115 ( 2514)

Then the equations of the planes in which the front and back faces of the cuboid in method 1 lie are

115

(2 x+5 y+14 z )= 115 [2 (0 )+5 (−2 )+14 (0 ) ]=−10

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and 115

(2 x+5 y+14 z )= 115 [2 (−1 )+5 (12 )+14 (5 ) ]=12815

So the distance between the two planes, and hence between the

two lines is ¿128−(−10)

15=13815

=465

=9.2

Method 3

Referring to the earlier diagram, suppose that X and Y have position vectors r=a+λ Xb & r=c+μY d respectively.

Then, if n is the vector normal to both b and d ,

c+μY d=a+ λXb+k n (*)

(ie Y is reached by travelling first to X and then along XY) and XY will then ¿k∨n∨¿

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(*) gives 3 simultaneous equations in λ , μ∧k :

(c1+μY d1c2+μY d2c3+μY d3)=(a1+λXb1+k n1a2+ λXb2+k n2

a3+ λXb3+k n3) , from which k can be found

Example 1c: (Lines as in 1a)

From Example 1a, n=( 2514 )We need to find k such that

(−1125 )+μY ( 7−144 )=( 0−20 )+λX ( 22−1)+k (

2514)

So 7 μY−2 λX−2k=1

−14 μY−2 λX−5 k=−14

4 μY+λX−14 k=−5

or ( 7 −2 −2−14 −2 −54 1 −14)(

μYλ Xk )=( 1−14

−5 )→(μYλ Xk )= 1

7 (33 )+14 (30 )+4(6) ( . . .. . .

−6 −15 −42)(1

−14−5 )

→k=414675

and XY=414675

×√4+25+196=9.2

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Method 4

As in method 3, suppose that X and Y have position vectors r=a+λ Xb & r=c+μY d respectively.

Then X⃗Y=c+μY d−(a+λX b)

and X⃗Y .b=X⃗Y .d=0 (*)

Solving (*) enables λX∧μY to be determined,

from which |⃗XY| can be found

Example 1d: (Lines as in 1a)

X⃗Y=(−1125 )+μY ( 7−144 )−( 0−20 )−λX ( 22−1)

¿( −1+7 μY−2 λX14−14 μY−2 λX5+4 μY+ λX )

Then ( −1+7 μY−2 λX14−14 μY−2λX5+4 μY+λX ). ( 22−1)=0

and ( −1+7 μY−2 λX14−14 μY−2λX5+4 μY+λX ). ( 7−14

4 )=0→−2+14 μY−4 λX+28−28 μY−4 λX−5−4 μY− λX=0

and

−7+49μY−14 λX−196+196μY+28 λX+20+16 μY+4 λX

¿0

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ie →21−18 μY−9 λX=0 or 7−6 μY−3 λX=0

and −183+261 μY+18 λ X=0 or −61+87 μY+6 λX=0

or (−6 −387 6 )(μYλX)=(−761 )

→(μYλ X)=1

−36+261 ( 6 3−87 −6)(−761 )

¿ 1225 (141243)= 1

75 (4781)

Then X⃗Y=¿ ( −1+7 μY−2 λX14−14 μY−2λX5+4 μY+λX )= 1

75 ( 92230644)and |⃗XY|=

175 √476100=9.2

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