1) a straight line is given by the path vector R(t) = (1+t)i hat + (3t) j hat +

Question

1) a straight line is given by the path vector R(t) = (1+t)i hat + (3t) j hat + (1-t) k hat

a. find a non-zero vector parallel to this line

b. for what value of c and d will this line be perpendicular to the plane x+cy+dz=0?

c. find ds/dt (speed) for a point moving on this line.

2) An isoscelese triangle in plane M has sides of length a, a and b. Projecting onto plane M' forms

an equilateral triangle with sides b. If the angle theta is 30 degrees (pi/6), find a.

3) a. simplify "A hat * derivative of A hat"

b. simplify "vector A * ((vector B x vector A) x vector A)

c. find a vector that bisects the angle between vector A= 4 i hat + 3 j hat, vector B = i hat + 2 j hat (hint: use unit vector)

Explanation / Answer

FOLOW THIS his note describes the technique and gives the solution to finding the shortest distance from a point to a line or line segment. The equation of a line defined through two points P1 (x1,y1) and P2 (x2,y2) is P = P1 + u (P2 - P1) The point P3 (x3,y3) is closest to the line at the tangent to the line which passes through P3, that is, the dot product of the tangent and line is 0, thus (P3 - P) dot (P2 - P1) = 0 Substituting the equation of the line gives [P3 - P1 - u(P2 - P1)] dot (P2 - P1) = 0 Solving this gives the value of u Substituting this into the equation of the line gives the point of intersection (x,y) of the tangent as x = x1 + u (x2 - x1) y = y1 + u (y2 - y1) The distance therefore between the point P3 and the line is the distance between (x,y) above and P3. Notes The only special testing for a software implementation is to ensure that P1 and P2 are not coincident (denominator in the equation for u is 0) If the distance of the point to a line segment is required then it is only necessary to test that u lies between 0 and 1. The solution is similar in higher dimensions. Contributed implementations C source from Damian Coventry: C source code VBA from Brandon Crosby: VBA source code Dephi from Graham O'Brien: Delphi version "R" version from Gregoire Thomas: pointline.r JAVA version from Pieter Iserbyt: DistancePoint.java LabView implementation from Chris Dancer: Pointlinesegment.vi.zip Right side distance by Orion Elenzil: rightside VBA VB6 by Thomas Ludewig: vbavb6.txt Minimum Distance between a Point and a Plane Written by Paul Bourke March 1996 Let Pa = (xa, ya, za) be the point in question. A plane can be defined by its normal n = (A, B, C) and any point on the plane Pb = (xb, yb, zb) Any point P = (x,y,z) lies on the plane if it satisfies the following A x + B y + C z + D = 0 The minimum distance between Pa and the plane is given by the absolute value of (A xa + B ya + C za + D) / sqrt(A2 + B2 + C2) . . . 1 To derive this result consider the projection of the line (Pa - Pb) onto the normal of the plane n, that is just ||Pa - Pb|| cos(theta), where theta is the angle between (Pa - Pb) and the normal n. This projection is the minimum distance of Pa to the plane. This can be written in terms of the dot product as minimum distance = (Pa - Pb) dot n / ||n|| That is minimum distance = (A (xa - xb) + B (ya - yb) + C (za - zb)) / sqrt(A2 + B2 + C2) . . . 2 Since point (xb, yb, zb) is a point on the plane A xb + B yb + C zb + D = 0 . . . 3 Substituting equation 3 into equation 2 gives the result shown in equation 1. Intersection point of two lines i

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