math primer for cisc/cmpe 330 part 1: vector geometry · 2018-09-12 · defined by orthonormal...

Copyright © Gabor Fichtinger, 2018Laboratory for Percutaneous Surgery – The Perk Lab

Math primer for CISC/CMPE 330

part 1: Vector geometry


Examples


Location/position vector in 2D

x

yp=P(x,y) =

x = | p | *cos(a)

y= | p | *sin(a)

x

y

length (p) = | p | = sqrt(x2 + y2)

a

P

p

| p |


Location/position vector in 3D

x

y

zp =P(x,y,z) =

x

yx

y

length (p) = | p | = sqrt(x2 + y2 + z2)

_

P

z

zp


Unit vector

P(x,y,z)

x

y

length(p) = 1

sqrt(x2 + y2+ z2) = 1

p


Scaling up/down a vector

P(x,y,z)

x

y Q = s * P = (s*x,s*y,s*z)

s> 1 stretch

s< 1 shrink

s= 1 leave it alone

s= 0 probably wrong…


Unit vector

a

|v| =1

Normalize a by scaling down the vector by its own length

v = a / |a| or v = a /length(a)


Create unit vector from a vector (Example)

Q(3,4,0)

x

y

length(Q) =sqrt(3*3 + 4*4+0*0) = sqrt(25) = 5

v = (3/5, 4/5, 0)

v so that |v| =1


Sum of vectors

p1 (x1,y1,z1)

x

y

Concatenate the vectors

p2 (x2,y2,z2)

p1 + p2 = p3 (x1+x2, y1+y2, z1+z2)

p2p3


Subtraction of vectors

p1 (x1,y1,z1)

x

y

Reverse p2 and concatenate to p1

p3= p1 - p2 = p1 + (- p2)

p3 (x1-x2, y1-y2, z1-z2)

p2(x2,y2,z2)

-p2


Vector from A to B

x

y

AB = B – A = (x1-x2, y1-y2, z1-z2)

Subtract A from B

A (x2,y2,z2)

B (x1,y1,z1)


Distance between A and B

x

y

Distance = length( AB) = length(B – A) =

sqrt( (x1-x2)2 + (y1-y2)

2 + (z1-z2) 2 )

A (x2,y2,z2)

B (x1,y1,z1)


Vector equation of a line

v

P

L = P + t*vP is called holding point or fixed point

L is called the running point along the line

Equation broken to x,y,z components

Lx = Px + t*vx

Ly = Py + t*vy

Lz = Pz + t*vz

Infinite line:

t=(-inf,inf)

Ray:

t=(0,inf)

Line segment:

t=(tmin, tmax)

L

x

x

y


Unit direction vector of a line

v

P

L

x

Obtain the unit direction vector:

AB = B – A Subtract

v= AB / |AB| Normalize

x

x

A

B

Line defined by two points

v


Vector equation of a line

v

P

L = P + t*v

then

L - P = t*v

then

|L - P | = t|v |

Now if |v |=1 (v was normalized)

then

|L - P | = t

‘t’ measures the distance between the

holding point (P) and running point (L)

L

x

x

y


Cross product (or vector product) of u and v is

denoted as q = u x v, where u,v,q are 3D vectors

denoted as u(x1,y1,z1), v(x2,y2,z2), q(x3,y3,z3),

|q | = |u| * |v |* sin(a)

q is perpendicular to both u and v, and

x3=y1*z2-y2*z1

y3=x2*z1-x1*z2

z3=x1*y2-x2*y1

Cross product

v

u

a

Cross product = 0 if and only if

sin(a)=0

i.e. u and v are parallel

u

v

q

NOT COMMUTATIVE! ORDER MATTERS !


Area of a triangle

A= ½ |b|* h --- area of triangle

h=|c|*sin(a)

A= ½ |b|* |c| *sin(a)

A= ½ | b x c |

c

b

a

For non-zero b and c, the area is 0 if and only if sin(a) =0

i.e. c and b are parallel, so a=0

a

h


Is the point P on the line?

x

yP

S

v

P-S=c

P is on the line if and

only if

a=0

|v x c| =0

(if v and c are parallel

L

a


Ortho-Normal vector base from 3 points

A

C

B

x = AB

z

z = x x AC

y = z x x

x

y

Step 1 Ortho: Use cross products

to create three pairwise orthogonal

vectors, while keeping the right hand rule

Step 2 Normal: Normalize x, y, z

to make them unit length

The simplest test case is

A(0,0,0), B(1,0,0), C(0,1,0).

Yielding x(1,0,0), y(0,1,0) z(0,0,1)


Cartesian coordinate system

defined by orthonormal vector base and center

O

C

B

z

x

y

A

To fix a coordinate system, to the

ABC triangle, we create an

orthonormal vector base and

select a center.

The center can be anywhere, but

we must be consistent with the

choice. Commonly, we place it in

the Center of Gravity (COG),

which is the same as the

unweighted average of the points:

O= (A+B+C)/ 3


Dot product (or scalar product) = dot(u,v) = u*v

u(x1,y1,z1) v(x2,y2,z2)

dot product = u * v = |u| * |v| * cos(a) =(x1x2 + y1y2 + z1z2)

Dot product

u

v

a

Dot product = 0 if and only if cos(a)=0,

i.e. u and v are perpendicular u

v90

It is commutative, so order does not matter

Result is a scalar number, not a vector


Dot product and the length of a vector

Length = square root of the dot product with itself

v=(x,y,z)

length(v) = sqrt(x2 + y2+ z2) = sqrt( dot(v,v) )


Some More Dot Product Facts

ab = ba commutative

(ab)c != a (bc) not associative

(a + b )c = ac + bc distributive with addition


Angle between vectors

u

v

a

dot(u,v)=length(u) * length(v) * cos(a)

If u and v are unit vectors:

dot(u,v) =cos(a)

a = ...by inverse…

(length(cross(u,v)) =length(u) * length(v) * sin (a)

If u and v are unit vectors:

(length(cross(u,v)) = sin (a)

a = ...by inverse…

OR


Distance of a point from a line

v is a known unit vector, so length(v) =1

c=P-A -- this is a known vector

dot(v,c) = vc = |v| * |c |* cos(a) = |c|*cos(a) = |a|

a= v * |a| = v(vc)

d = c – a = c- v(vc)

dist = |d| = |c- v(vc)| or d2 = c2- (vc)2

P

A

_

v

a

d

c

a

Distance=|d|

90

P

A

_

v

d= ?

90


Intersection of two lines in 3D ?

v1_

P1

L1 = P1 + t1*v1

L2 = P2 + t2 *v2

----------------------

L1x = P1x + t1 *v1x

L1y = P1y + t1 *v1y

L1z = P1z + t1 *v1z

L2x = P2x + t2 *v2x

L2y = P2y + t2 *v2y

L2z = P2z + t2 *v2z

where

t=(-inf,inf)

u=(-inf,inf)

_

L1

x

v2

_

P2

_

L2

x

x

x

Write up the equations of the 2 lines


Intersection of two lines in 3D ?

In the intersection point :

L1=L2

or by components

L1x = L2x

L1y = L2y

L1z = L2z

v1

_

P1

_

L1

xv2

_

P2

_

L2

x

x

x


Intersection of two lines in 3D – IMPOSSIBLE

Generally, two lines avoid each other in 3D space.

In the intersection point L1 equals L2:

L1x –L2x= 0 = P1x - P2x + t1*v1x - t2 *v2x

L1y –L2y= 0 = P1y - P2y + t1*v1y - t2 *v2y

L1z –L2z= 0 = P1z - P2z + t1*v1z - t2 *v2z

where

t=(-inf,inf)

u=(-inf,inf)

Trouble: 3 eqs, 2 unknowns there is no guaranteed solution

The lines might intersect, but they do not have to. When they intersect, one of the

three eqs cancels out. In other words: a linear combination of any two yields the

third one. (We have to be extremely lucky for this to happen.)

We can approximate the solution: find the shortest distance between the

two lines and find the point in midway.


Intuition: Find the line (L3) that is perpendicular to both lines

L1L2

9090d/2

d/2

P2

ML3

x

P1

x

v1

v2

xv3

v1 x v2 = v3

REMINDER: If v1(x1,y1,z1), v2(x2,y2,z2),

and v3(x3,y3,z3), then the cross product is

x3=y1*z2-y2*z1

y3=x2*z1-x1*z2

z3=x1*y2-x2*y1

Approximate intersection of two lines in 3D


Write up the equation of each line

L1L2

9090

P2

L3x

P1

x

v1

v2

x

v3

L1 = P1 + t1*v1

L2 = P2 + t2 *v2

L3 = P3 + t3 *v3

P3


Derive conditions, make an equation system

=L1 L29090

P2

L3

x

P1x

v1

v2

v3

P3 = L2

L3 = L1

v1 x v2 = v3

[1] L1 = P1 + t1*v1

[2] L2 = P2 + t2 *v2

[3] L3 = P3 + t3 *v3

=P3

Conditions of

intersection

Equations


Solve the vector equation system

P3 = L2 -- replace P3 in [3]

L3 = L1 -- replace L3 in [3]

[1] L1 = P1 + t1*v1

[2] L2 = P2 + t2 *v2

[3] L3 = P3 + t3 *v3

where

[1] L1 = P1 + t1*v1 -- plug [1] in [3]

[2] L2 = P2 + t2 *v2 -- plug [1] in [3]

[3] L1 = L2 + t3 *v3

P1 + t1*v1 = P2 + t2 *v2 + t3 *v3 – arrange

P1 - P2 = -t1*v1 + t2 *v2 + t3 *v3 – break it up

P1x - P2x = -t1*v1x + t2 *v2x + t3 *v3x

P1y - P2y = -t1*v1y + t2 *v2y + t3 *v3y

P1z - P2z = -t1*v1z + t2 *v2z + t3 *v3z

• 3 linear equations

• 3 unknowns

• SOLVABLE !!!


Solve the linear equation system

P1x - P2x -v1x v2x v3x t1

P1y - P2y = -v1y v2y v3y t2

P1z - P2z -v1z v2z v3z t3

P1x - P2x t1

P1y - P2y = t2

P1z - P2z t3

-v1x v2x v3x

-v1y v2y v3y

-v1z v2z v3z

-1

M =( L1 + L2 ) /2

Error metric: distance of M from

either of the lines

L1 = P1 + t1*v1

L2 = P2 + t2 *v2

Plug t1 and t2 back

into L1 and L2 line

equations

Use any of the three methods:

1. Gaussian elimination

2. Substitution

3. Matrix inversion (shown below)


Intersection of N lines in 3D – proper way

v1_

P1

L1 = P1 + t1*v1

L2 = P2 + t2 *v2

…

Ln = Pn + tn *vn

_

L1

x

v2

_

P2

_

L2

x

x

Write up the equations of all lines

Pose the problem as a least square

optimization:

S = d12 + d2

2 + … dn2

Compute M where S is at the minimum

Mx

d2

d1

_

Ln

_

L2

dn

vn

_

Pnx

Always compute error metric:

RMS(d1 d2… dn)

Avg (d1 d2… dn)

STD(d1 d2… dn)

Look for outliers that “stand out”,

such as di >2*STD

Remove these “bad” lines and

recompute M and the error

metrics

Keep repeating until no outlier is

left (or you run out of lines :-)


Intersection of N lines in 3D – cheap but not too dirty

v1_

P1

L1 = P1 + t1*v1

L2 = P2 + t2 *v2

…

Ln = Pn + tn *vn

_

L1

x

v2

_

P2

_

L2

x

x

Compute pairwise approx. intersections for all

pairs (M12, M13… M1n, M2n...) then compute

M as the average of those. (i.e. Center Of

Gravity.)

If the input is “decent” (no outliers and all lines

come very close) simple average is almost as

good as the proper least square minimum.

M x

M2n

_

Ln

_

L2

vn

_

Pnx

M12

M1n

xx

x

d2

d1

dn


RMS(d1 d2… dn)

Avg (d1 d2… dn)

STD(d1 d2… dn)

Look for outliers, such as

|Avgd-di| > 2*STD

Remove these “bad apples” and

recompute M and the error

metrics


left (or you run out of input :-)


Plane representation

Normal vector and one point in plane (n,P)

x P

n


Plane normal from 3 points in the plane

x P

x

R

x

Q

___ __

n = PQ x PRn


Is point in the plane?

n

x

P

Q

Q is in the plane if and only if PQ and n are orthogonal:

dot(PQ, n) =0


Distance of point from plane

n

x

P

x Q

d (distance)

d = PQ * n

If d>0 Q is above

If d<0 Q is below

If d=0 Q is in plane


Intersection of 3 planes

n1

x

P1

Q

Q lays in each of the 3 planes

dot(P1Q, n1) = 0

dot(P2Q, n2) = 0

dot(P3Q, n3) = 0

These yield a linear equation system for Q, which is always solvable if the 3 planes are non-parallel

n2

x

P2


Intersection of line and plane

n

x

P

Q

Q is in the plane if and only if PQ and n are orthogonal:

dot(PQ, n) = 0 i.e. (A+vt-P)n= 0

Yields a linear equation for t, which is always solvable unless v and n

are orthogonal (i.e. line and plane is parallel)

vAx Q=A+vt


Equations of sphere & ellipsoid

(x-xo)2 + (y-yo)2 + (z-zo) 2 = R2 Sphere

x [xo,yo,zo]

R

(x-xo)2 (y-yo)2 (z-zo) 2

-------- + -------- + --------- = 1 Ellipsoid

a2 b2 c2

where a,b,c are principal axes.

Use case: represent anatomical organs like the orbit, cranium, etc.


Intersection of line and sphere

(x-xo)2 + (y-yo)2 + (z-zo) 2 = R2

Leads to a quadratic equation in “t”

There are 0,1, or 2 intersections

Check the determinant to find out how many exist

Follow the same method for ellipsoid

v

Px

LL = P+ vt

or

Lx = Px + t*vx

Ly = Py + t*vy

Lz = Pz + t*vzx=Lx

y=Ly

z=Lz

Plug into the sphere’s equation

Use cases: compute where a linear surgical tool penetrates the eye, cranium, etc.

x [xo,yo,zo]

R


Fitting points to line in 3D – by least square

Concept: Pose the problem as a

least square optimization. Write up

the cost function as sum of square

distances of all points from the object

S = d12 + d2

2 + … dn2

Minimize the cost function to find the

optimal (P,v)

v_

P

P1, P2…Pi… points observed

_ _

L = P + t*v eq. of the line

di2 = ci

2- (vci)2 sq. distance of Pi from line

x Pi

P1

xdix

P4x

P2

x

P3

x


RMS(d1 d2… dn)

Avg (d1 d2… dn)

STD(d1 d2… dn)


|Avg-di| > 2*STD or

|ARMS-di| > 2*STD or


recompute the optimization and

error metrics.



Use case: reconstruct the path of a

straight surgical tool from points observed

ci


Fitting points to line in 3D – by average

Concept: if the input is “decent” (lots

of points and no outliers, average is

as good as least square minimum.

Compute both average P and

average v.

P=Avg (P1 P2… Pn)

v=Avg (c1 c2… cn)

Note: normalize all c1 c2.. and flip

them in one direction!

v_

P

P1, P2…Pi… points observed

_ _

L = P + t*v eq. of the line

di2 = ci

2- (vci)2 sq. distance of Pi from line

x Pi

P1

xdix

P4x

P2

x

P3

x

ci

Use case: reconstruct the path of a

straight surgical tool from points observed


RMS(d1 d2… dn)

Avg (d1 d2… dn)

STD(d1 d2… dn)


|Avg-di| > 2*STD or




error metrics.




Fitting points to sphere in 3D – by least square

Concept: Pose the problem as a

least square optimization. Write up

the cost function as sum of square

distances of all points from the object

S = d12 + d2

2 + … dn2

Minimize the cost function to find the

optimal (R,xo,yo,zo)

Pi

P1

xdix

P4x

P2

x

P3

x


di = |R - length(Pi,O)|

R

O[xo,yo,zo]x

Use case: reconstruct the path of a pivoting

surgical tool from points observed.


RMS(d1 d2… dn)

Avg (d1 d2… dn)

STD(d1 d2… dn)


|Avg-di| > 2*STD or




error metrics.




Fitting points to sphere in 3D – by average

Concept: Compute an “average O”.

Ideally, any two (Pi,Pj) pair are on

the surface of the sphere. Place a

normal plane in the midpoint

between (Pi,Pj) then compute O as

the average intersection of all such

planes. Then compute average R as

R =avg ( length(PiO))

Pi

P1

xdix

P4x

P2

x

P3

x


di = |R - length(Pi,O)|

R

O[xo,yo,zo]x

Use case: reconstruct the path of a pivoting

surgical tool from points observed.


RMS(d1 d2… dn)

Avg (d1 d2… dn)

STD(d1 d2… dn)


|avg-di| > 2*STD or

|RMS-di| > 2*STD or



error metrics.



math primer for cisc/cmpe 330 part 1: vector geometry · 2018-09-12 · defined by orthonormal...

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