12 cm0268 geometric computing2

8/10/2019 12 CM0268 Geometric Computing2

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Lines, Curves and Surfaces in 3D

Lines in 3D

In 3D theimplicit equationof a line is defined as theintersection oftwo planes. (More on this shortly)

The parametric equation is a simple extension to 3D of the 2D form:

x = x0+f t

y = y0+gt

z = z0+ht
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Parametric Lines in 3D

v

w

tw

v +tw

v0= (x0, y0, z0)

p= (x,y,z)

x = x0+f t

y = y0+gt

z = z0+ht

This is simply anextensionof the vector form in 3D

The line is normalised whenf2 +g2 +h2 = 1
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Perpendicular Distance from a Point to a Line in 3D

For the parametric form,

x = x0+f ty = y0+gt

z = z0+ht

This builds on the 2D example we met earlier, line par point dist 2d.

The 3D form isline par point dist 3d.dx = g * ( f * ( p ( 2 ) - y 0 ) - g * ( p(1) - x0 ) ) ...

+ h * ( f * ( p ( 3 ) - z 0 ) - h * ( p(1) - x0 ) );

dy = h * ( g * ( p ( 3 ) - z 0 ) - h * ( p(2) - y0 ) ) ...

- f * ( f * ( p ( 2 ) - y 0 ) - g * ( p(1) - x0 ) );

d z = - f * ( f * ( p ( 3 ) - z 0 ) - h * ( p(1) - x0 ) ) ...

- g * ( g * ( p ( 3 ) - z 0 ) - h * ( p(2) - y0 ) );

dist = sqrt ( dx * dx + dy * dy + dz * dz ) ...

/ ( f * f + g * g + h * h );

The value of parameter, t, where the point intersects the line isgiven by:t = (f*( p(1) - x0) + g*(p(2) - y0) + h*(p(3) - z0)/( f * f + g * g + h*h);
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Line Through Two Points in 3D (parametric form)

P

Q

The parametric form of a line through two points, P(xp, yp, zp)andQ(xq, yq, zq) comes readily from the vector form of line (again a simpleextension from 2D):

Set base to pointPVector along line is(xqxp, yqyp, zqzp)The equation of the line is:

x = xp+ (xqxp)ty = yp+ (yqyp)tz = zp+ (zqzp)t

As in 2D,t = 0givesP andt = 1givesQNormalise if necessary.
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Implicit SurfacesAn implicit surface (just like implicit curves in 2D) of the form

f(x,y,z ) = 0

We simply add theextrazdimension.

For example:

A plane can be represented

ax+by+cz+d= 0

A sphere can be represented as

(xxc)2 + (yyc)2 + (zzc)2 r2 = 0which is just the extension of the circle in 2D to 3D where thecentre is now(xc, yc, zc)and the radius isr.


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Implicit Equation of a Plane

(a,b,c)

O

d

The plane equation:

ax+by+cz+d = 0

Is normalised ifa2 +b2 +c2 =1.

Like 2D thenormal vectorthe surface normal is given by avector n= (a,b,c)

a, bandc are the cosine angles which the normal makes withthex-,y- andz-axes respectively.


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Parametric Equation of a Plane

(f2, g2, h2)

(f1, g1, h1)

O

(x0, y0, z0)

x = x0+f1u+f2v

y = y0+g1u+g2v

z = z0+h1u+h2v

This is an extension of parametric line into 3D where we nowhave two variable parametersuandv that vary.

(f1, g1, h1) and (f2, g2, h2) are two different vectors parallel to theplane.


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Parametric Equation of a Plane (Cont.)

(f2, g2, h2)

(f1, g1, h1)

O

(x0, y0, z0)

x = x0+f1u+f2v

y = y0+g1u+g2vz = z0+h1u+h2v

A point in the plane is found by adding proportion u of onevector to a proportionv of the other vector

If the two vectors are have unit length and are perpendicular,then:

f21+g2

1+h2

1 = 1

f22+g2

2+h2

2 = 1

f1f2+g1g2+h1h2 = 0 (scalar product)


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Distance from a 3D point and a Plane

J(xj , yj, zj)

d

The distance,d, between a point,J(xj, yj, zj), and animplicitplane,ax+by+cz+d= 0is:

d=axj+byj+czj+d

a2 +b2 +c2

This is very similar the 2D distance of a point to a line.

The MATALB code to achive this is,plane imp point dist 3d.m:

norm = sqrt ( a * a + b * b + c * c );

if ( norm == 0.0

error ( PLANE Normal = 0! );

end

dist = abs ( a * p(1) + b * p(2) + c * p(3) + d ) / norm;
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Angle Between a Line and a Plane

If the plane is in implicit formax+ by+cz+d = 0and line is inparametric form:

x = x0+f t

y = y0+gt

z = z0+ht

then the angle, between the line and the normal the plane (a,b,c)

is:= cos1(af+bg+ch)

The angle,, between the line and the plane is then:

=

2


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Angle Between a Line and a Plane (cont)

If either line or plane equations are not normalised the we mustnormalise:

= cos1 (af+bg+ch)

(a2 +b2 +c2), = cos1

(af+bg+ch)(f2 +g2 +h2)

, = cos1 (af+bg+ch)

(a2 +b2 +c2)(f2 +g2 +h2)

The angle,, is as before:

=

2The MATLAB code to so this isplanes imp angle line 3d.m:

norm1 = sqrt ( a1 * a1 + b1 * b1 + c1 * c1 );

if ( norm1 == 0.0 )

angle = Inf;

return

end

norm2 = sqrt ( f * f + g *g + h * h );

if ( norm2 == 0.0 )

angle = Inf;

return

end

cosine = ( a1 * f + b 1 * g + c 1 * h) / ( norm1 * norm2 );

angle = pi/2 - acos( cosine );
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Angle Between Two Planes

Given twonormalisedimplicitplanes

a1x+b1y+c1z+d1= 0

and

a2x+b2y+c2z+d2= 0

The angle between them,, is the angle between the normals:

= cos1(a1a2+b1b2+c1c2)


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Angle Between Two Planes (MATLAB Code)

The MATLAB code to do this isplanes imp angle 3d.m:

norm1 = sqrt ( a1 * a1 + b1 * b 1 + c 1 * c1 );if ( norm1 == 0.0 )

angle = Inf;

return

end

norm2 = sqrt ( a2 * a2 + b2 * b 2 + c 2 * c2 );

if ( norm2 == 0.0 )angle = Inf;

return

end

cosine = ( a1 * a2 + b1 * b2 + c1 * c2 ) / ( norm1 * norm2 );

angle = acos ( cosine );
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Intersection of a Line and Plane

P(x, y)

If the plane is in implicit formax+ by+cz+d = 0and line is inparametric form:

x = x0+f t

y = y0+gt

z = z0+htThe the point,P(x, y), where the is given by parameter,t:

t=(ax0+by0+cz0+d)

af+bg+ch


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Intersection of a Line and Plane (MATLAB Code)

The MATLAB code to do this isplane imp line par int 3d.m

tol = eps;

norm1 = sqrt ( a * a + b * b + c * c );if ( norm1 == 0.0 )

error ( Norm1 = 0 - Fatal error! )

end

norm2 = sqrt ( f * f + g * g + h * h );

if ( norm2 == 0.0 )

error ( Norm2 = 0 - Fatal error! )

end

denom = a * f + b * g + c * h;

if ( abs ( denom ) < tol * norm1 * norm2 ) % The line and the plane may be parallel.

i f ( a * x 0 + b * y 0 + c * z 0 + d = = 0 . 0 )

intersect = 1;

p(1) = x0;

p(2) = y0;

p(3) = z0;

else

intersect = 0;

p(1:dim_num) = 0.0;end

else

intersect = 1;

t = - ( a * x 0 + b * y 0 + c * z0 + d ) / denom; % they must intersect.

p(1) = x0 + t * f;

p(2) = y0 + t * g;

p(3) = z0 + t * h;

end
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Intersection of Three Planes

Three planes intersect at point.Two planes intersect in a linetwo lines intersect at a pointSimilar problem to solving in 2D for two line intersecting:

Solvethreesimultaneous linear equations:

a1x+b1y+c1z+d1 = 0

a2x+b2y+c2z+d2 = 0

a3x+b3y+c3z+d3 = 0


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Intersection of Three Planes (MATLAB Code)

The MATLAB code to do this is,planes 3d 3 intersect.m:

tol = eps;

b c = b 2*c 3 - b 3*c2;

a c = a 2*c 3 - a 3*c2;

a b = a 2*b 3 - a 3*b2;

det = a1*b c - b 1*a c + c 1*ab

if (abs(det) < tol)

error(planes_3d_3_intersct: At least to planes are parallel);end;

else

d c = d 2*c 3 - d 3*c2;

d b = d 2*b 3 - d 3*b2;

a d = a 2*d 3 - a 3*d2;

detinv = 1/det;

p(1) = (b1*dc - d1*bc - c1*db)*detinv;p(2) = (d1*ac - a1*dc - c1*ad)*detinv;

p(3) = (b1*ad + a1*db - d1*ab)*detinv;

return;

end;
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Intersection of Two Planes

Two planes

a1x+b1y+c1z+d1 = 0

a2x+b2y+c2z+d2 = 0

intersect to form a straight line:

x = x0+f t

y = y0+gt

z = z0+ht


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Intersection of Two Planes (Cont.)

(f , g , h)may be found by finding a vector along the line. This isgiven by the vector cross product of(a1, b1, c1)and(a2, b2, c2):

f g ha1 b1 c1a2 b2 c2

(x0, y0, z0)then readily follows.


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Intersection of Two Planes (Matlab Code)

The MATLAB code to do this is,planes 3d 3 intersect.m:

tol = eps;

f = b 1*c 2 - b 2*c1;g = c 2*a 2 - c 2*a1;

h = a 1*b 2 - a 2*b1;

d e t = f*f + g*g + h*h;

if (abs(det) < tol)

error(planes_3d_2intersect_line: Planes are parallel);

end;

else

d c = d 1*c 2 - c 1*d2;

d b = d 1*b 2 - b 1*d2;

a d = a 1*d 1 - a 2*d1;

detinv = 1/det;

x 0 = ( g*d c - h*db)*detinv;

y0 = -(f*dc +h*ad)*detinv;

z 0 = ( f*db +g*ab)*detinv;

return;

end;

l h h h i
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Plane Through Three Points

Just as two points define a line,three points define a plane (this is theexplicitform of a plane):

Pk

Pl

Pj

A plane may be found as follows:

LetPjbe the to point(x0, y0, z0)in the planeForm two vectors in the planePlPjandPkPjForm another vector from a general pointP(x, y)in the plane to

Pj

The the equation of the plane is given by:

xxj yyj zzjxkxj ykyj zkzjxlxj ylyj zlzj

a, b, canddcan be found by expanding the determinant above

Pl Th h Th P i (MATLAB C d )


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Plane Through Three Points (MATLAB Code)

The MATLAB code to do this,plane exp2imp 3d.m

a = ( p2(2) - p1(2) ) * ( p3(3) - p1(3) ) ...

- ( p2(3) - p1(3) ) * ( p3(2) - p1(2) );

b = ( p2(3) - p1(3) ) * ( p3(1) - p1(1) ) ...

- ( p2(1) - p1(1) ) * ( p3(3) - p1(3) );

c = ( p2(1) - p1(1) ) * ( p3(2) - p1(2) ) ...

- ( p2(2) - p1(2) ) * ( p3(1) - p1(1) );

d = - p2(1) * a - p2(2) * b - p2(3) * c;
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Parametric Surfaces

The general form of a parametric surface is

r= (x(u, v), y(u, v), z(u, v)).

This is just like a parametric curve except we now have two

parametersuandv that vary.

u

v

P t i S f C li d


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Parametric Surface: Cylinder

For example, a cylindrical may be represented in parametric form as

x= x0+r cos u y=y0+r sin u z=z0+v.

!1 0

1 2

3 4

5

!3

!2

!1

0

1

2

3

0

2

4

6

8

10

Parametric Surface: Cylinder (MATLAB Code


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Parametric Surface: Cylinder (MATLAB Code

The MATLAB code to plot the cylinder figure iscyl plot.m

p0 = [2,0,0] % x_0, y_0, z_0

r = 3; %radius

n = 360;

hold on;

for v = 1:10

for u = 1:360

theta = ( 2.0 * pi * ( u - 1 ) ) / n;

x = p0(1) + r * cos(theta);y = p0(2) + r * sin(theta);

z = p0(3) + v;

plot3(x,y,z);

end

end

Parametric Surface: Sphere
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Parametric Surface: Sphere

A sphere is represented in parametric form as

x= xc +sin(u) sin(v) y=yc +cos(u) sin(v) z=zc +cos(v)

MATLAB code to produce a parametric sphere is at HyperSphere.m
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Piecewise Shape Models

Polygons

A polygon is a 2D shape that is bounded by a closed path or circuit,composed of a finite sequence of straight line segments.

We can represent a polygon by a series of connected lines:

P1(x1, y1)

P2(x2, y2)

P3(x3, y3)

P4(x4, y4)

P5(x5, y5)

P6(x6, y6)

P7(x7, y7)

e1

e2

e3

e4e5e6

e7

Polygons (Cont )


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Polygons (Cont.)

P1(x1, y1)

P2(x2, y2)

P3(x3, y3)

P4(x4, y4)

P5(x5, y5)

P6(x6, y6)

P7(x7, y7)

e1

e2

e3

e4e5e6

e7

Each line is defined by two vertices the start and end points.

We can define a data structure which stores a list of points(coordinate positions) and the edges define by indexes to twopoints:

Points:{P1(x1, y1), P2(x2, y2), P3(x3, y3), . . .}Points define thegeometryof the polygon.

Edges:Edges :{e1= (P1, P2), e2= (P2, P3), . . .Edges define thetopologyof the polygon.If you traverse the polygon points in an ordered direction

(clockwise) then the lines define a closed shape with the insideon the right of each line.

3D Objects: Boundary Representation


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In 3D we need to represent the objects faces, edges and vertices and

how they are joined together:



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Topology The topology of an object records the connectivity of

the faces, edges and vertices. Thus,Each edge has a vertex at either end of it (e1is terminated by

v1andv2), and,

each edge has two faces, one on either side of it (edgee3 liesbetween facesf1andf2).

A face is either represented as an implicit surface or as a parametricsurface

Edges may be straight lines (like e1), circular arcs (like e4), andso on.

Geometry This described the exact shape and position of each ofthe edges, faces and vertices. The geometry of a vertex is just itsposition in space as given by its(x,y,z )coordinates.


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Geometric TransformationsSome Common Geometric Transformations:

2D Scaling:

xk 00 yk

2D Rotation:

cos sin sin cos 2D Shear (xaxis):

1 k0 1

2D Shear (yaxis): 1 0

k 1

3D Scaling: xk 0 00 yk 0

0 0 zk

3D Rotation aboutzaxis:

cos sin 0 sin cos 00 0 1

2D Translation(Homogeneous Coords):

1 0 dx0 1 dy

0 0 1

Rotation and Translation


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Rotation and Translation

Applying Geometric Transformations


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pp y g

When applying geometric transformations to objects

Rotate the point coordinate and vectors (normals etc.)Simply do matrix-vector multiplication

For polygons and Boundary representations only need to applytransformation to the point (geometric) data the topology isunchanged

Compound Geometric Transformations


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p

It is a simple task to create compound transformations (e.g.A rotationfollowed by a translation:

Perform Matrix-Matrix multiplications,e.g. T.RNote in general T.R= R.T (Basic non-commutative law of

multiplication)

One problem exists in that a 2D Translation cannotbe representedas a 2D matrix it must be represented a 3D matrix:

T=

1 0 dx0 1 dy

0 0 1

So how do we combine a 2D rotation R. = cos sin sin cos

with a translation?

We need to defineHomogeneous Coordinatesto deal with thisissue.

Homogeneous Coordinates


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g

To work in homogeneous coordinates do the following:

Increase the dimension of the space by 1, i.e. 2D3D, 3D4D.We can now accommodate the translation matrix.

Convert all standard other 2D transformations via the following:

X X 0X X 0

0 0 1

where

X XX X

is the regular 2D transformation

Perform compound transformation by matrix-matrixmultiplication in homogeneous matrices

Homogeneous Coordinates Example: 2D Rotation


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g p

The transformation for a 2D rotation (angle of rotationis:

cos sin sin cos

The correspondinghomogenous formis:

cos sin 0 sin cos 00 0 1

Compound Geometric Transformations MATLAB Examples


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Simply create the matrices for individual transformations inMATALB (2D, 3D or homogeneous)

Assemble compound via matrix multiplication.

Apply compound transform to coordinates/vectors etc. viamatrix-vector multiplication.

2D Geometric Transformation MATLAB Demo Code

3D Transformations
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3D Rotation aboutxaxis:

1 0 00 cos sin 0 sin cos

3D Rotation abouty axis:

cos 0 sin 0 1 0

sin 0 cos

3D Rotation aboutzaxis:

cos sin 0

sin cos 0

0 0 1

3D Scaling:

xk 0 00 yk 00 0 zk

3D Translation(Homogeneous Coords (4D):

1 0 0 dx0 1 0 dy0 0 1 dz0 0 0 1

Need to put all 3D transformation into homogenous form forcompound transformations.


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Least Squares FittingWe have already seen that we need 2 points to define a line and 3

points to define circle and a plane the minimum number of freeparameters.However, in many data processing applications we will have many

more samples than the minimum number required.

Errors in the positions of these points mean that in practice they

do not all lie exactly on the expected line/curve/surface.

Least squares approximation methods find the surface of the giventype which best represents the data points.

Minimise some error function


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Least Squares Fit of a Line

We use the general explicit equation of a line here:

y=mx+c

We can define an error,ias the distance from the line:

i=yimxic

Least Squares Fit of a Line (Cont.)


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The best fitting plane to a set ofnpoints{Pi}occurs when

2 =

2i =

(yimxic)2is a minimum, where this and all subsequent sums are to be taken

over all points,Pi, i= 1, . . . , n.This occurs when

2

m = 0

2

c = 0

Least Squares Fit of a Line (Cont.)


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Now

2

m

=

2(yi

mxi

c)xi= 0

2

c = 2

(yimxic) = 0

which leads to

yimxi

nc = 0xiyim

x2ic

xi = 0

which can be solved formandcto give:

m = nxiyi

xi yin

x2i(

xi)2

c =

yi

x2i

xi

xiyi

n

x2i(

xi)2

Least Squares Fit of a Line (MATLAB Code)


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The MATLAB code to perform least squares line fitting is,line fit.m

function [m c] = line_fit(x,y)

[n1 n] = size(x);

sumx = sum(x);

sumy = sum(y);

sumx2 = sum(x.*x);

sumxy = sum(x.*y);

det = n* sumx2 - sumx*sumx;

m = ( n*sumxy - sumx*sumy)/det;

c = (sumy*sumx2 - sumx*sumxy)/det;

Seeline fit demo.mfor example call.
http://lastpage/http://close/http://www.cs.cf.ac.uk/Dave/CM0268/Lecture_Examples/Computational_Geometry/line_fit.mhttp://www.cs.cf.ac.uk/Dave/CM0268/Lecture_Examples/Computational_Geometry/line_fit_demo.mhttp://www.cs.cf.ac.uk/Dave/CM0268/Lecture_Examples/Computational_Geometry/line_fit_demo.mhttp://www.cs.cf.ac.uk/Dave/CM0268/Lecture_Examples/Computational_Geometry/line_fit.mhttp://close/http://lastpage/


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Least Squares Fit of a Plane (Cont.)


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The best fitting plane to a set ofn points {Pi} occurs when 2 = 2iis a minimum, where this and all subsequent sums are to be takenover all points,Pi, i= 1, . . . , n.

This occurs when2

a = 0 (1)

and2

b = 0 (2)

and and when2

c = 0 (3)



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Now 2

a

= 0gives:

xizia

x2ib

xiyic

xi= 0,

and 2

b = 0gives:yizia

xiyib

y2ic

yi= 0,

and 2

c = 0gives

zia

xib

yinc= 0.

Simplifying a Least Squares Fit of a Plane


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Now, some of these sums can become quite large and any attempt tosolve the equations directly can lead to poor results due to rounding

errors.

One common method (for any least squares approximation) ofreducing the rounding errors is to re-express the coordinates ofeach point relative to the centre of gravity of the set of points,(xg, yg, zg).

All subsequent calculations are performed under this translationwith the final results requiring the inverse translation to return tothe original coordinate system.

The translation also simplifies our minimisation problem since nowi

xi=i

yi=i

zi= 0.



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Therefore, using the above fact in previous equations and solving fora, bandcgives:

a = xizi

y2i

yizi

xiyi

x2

i

y2

i(xiyi)2 , (4)

b =

yizi

x2i

xizi

xiyi

x2i

y2i(

xiyi)2 , (5)

c = 0. (6)

All we now need to do is to translate back.


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12 cm0268 geometric computing2

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