homogeneous coordinates equationmisc point(w ; x, y, z)r w = S 0 where S 0 = xi + yj + zk  3 points in 3D space

How to define a plane? suppose we are given S and r 1 where: r 1 is a vector to some point on the plane (units of L) S is the direction vector  to the plane (dimensionless) let r be a vector to any arbitrary point on the plane Can we come up with some general expression for r in terms of S and r 1 ?

(r – r 1 )  S = 0 will write as r  S + D 0 = 0 where D 0 = -r 1  S equation of the plane coordinates of the plane: [D 0 ; A, B, C] where S = Ai + Bj + Ck

Determine the coordinates of a plane, i.e. [D 0 ; A, B, C] when given 3 points on the plane. sample problem (i) P(1 ; 3, 4, 1) m Q(1 ; -1, 2, 4) R(1 ; 3, 2, 2)

Given the coordinates of a plane, i.e. [D 0 ; A, B, C]: 1.can I calculate a point on the plane? 2.can I find the point on the plane that is closest to the origin?

homogeneous coordinates equationmisc point(w ; x, y, z)r w = S 0 where S 0 = xi + yj + zk  3 points in 3D space plane[D 0 ; A, B, C] r  S + D 0 = 0  3 planes in 3D space

a line can be defined by two points, r 1 and r 2 S = r 2 – r 1 (dimensionless direction vector) (r –r 1 )  S = 0 r  S = r 1  S let S 0L = r 1  S then r  S = S 0L is the equation of the line where {S ; S 0L } are the Plücker coordinates of the line Note that S × S 0 L = 0

{S ; S 0L } {L, M, N ; P, Q, R}

Given the coordinates of a line, i.e. {S ; S 0L }: 1.can I calculate a point on the line? 2.can I find the point on the line that is closest to the origin?

Figure 1.7: Conceptualization of the Line at Infinity, {0 ; S}

homogeneous coordinates equationmisc point(w ; x, y, z)r w = S 0 where S 0 = xi + yj + zk  3 points in 3D space plane[D 0 ; A, B, C] r  S + D 0 = 0  3 planes in 3D space line{S ; S 0L } where S × S 0L = 0 r  S = S 0L  4 lines in 3D space

Fig The Plane Through a Pair of Parallel Lines

Figure 1.13: A Line and a Plane Determine a Point

Figure 1.14: Closest Point on a Line to a Given Point

1.15

Numerical example: given:{S 1 ; S 0L1 } = {1, 2, 1 ; -2, 1, 0} dimensionless ; meters {S 2 ; S 0L2 } = {-3, 1, 0 ; 1, 3, 5} find:for your choice of a 12 find  12 and a 12

1.16

1.17 given: {S 1 ; S 0L1 }, {S 2 ; S 0L2 } where |S 1 | = |S 2 | = 1, S 1  S 0L2 + S 2  S 0L1 = 0, and S 1 is not parallel to S 2 find: point of intersection, r 0 will show two solution approaches

Approach 1 The point r 0 must lie on both lines. Thus, r 0  S 1 = S 0L1 (1.155) r 0  S 2 = S 0L2 (1.156) performing a cross product of (1.155) with S 2 gives S 2  (r 0  S 1 ) = S 2  S 0L1 (S 2  S 1 ) r 0 – (r 0  S 2 ) S 1 = S 2  S 0L1 (1.158) performing a scalar product with S 1 gives (r 0  S 1 )(S 2  S 1 ) – (S 1  S 1 )(r 0  S 2 ) = S 1  (S 2  S 0L1 )(1.159) substituting (S 1  S 1 ) = 1 and multiplying by -1 gives r 0  [S 2 – (S 2  S 1 ) S 1 ] = (S 1  S 0L1 )  S 2 (1.160) Now we can solve (1.155) and (1.160) for r 0. - form cross product of (1.155) with [S 2 – (S 2  S 1 ) S 1 ]. expand, and solve for r 0

given: Plucker coordinates of 2 non-parallel intersecting lines {3, 2, 4 ; 8, 0, -6} {dimensionless ; meters} {-1, 3, 2 ; 4, 0, 2} note that the mutual moment of the 2 lines is zero find: point of intersection

Approach 2 The point r 0 must lie on both lines. Thus, r 0  S 1 = S 0L1 (1.155) r 0  S 2 = S 0L2 (1.156) performing a cross product gives (r 0  S 1 )  (r 0  S 2 ) = S 0L1  S 0L2 (1.165) expanding this expression gives [(r 0  S 1 )  S 2 ] r 0 – [(r 0  S 1 )  r 0 ] S 2 = S 0L1  S 0L2 (1.166) substitute [(r 0  S 1 )  r 0 ] = 0 and (r 0  S 1 ) = S 0L1 to get

Comparison of Approaches Equation (1.167) will yield 0/0 if the first line passes through the origin the two lines lie in a plane that passes through the origin (because S 0L1 will be perpendicular to the plane and S 2 is in the plane ; also S 0L1 and S 0L2 will be parallel)