# h2s - 18.02 HOMEWORK#2 DUE BJORN POONEN 1 Part A(After Sept...

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18.02 HOMEWORK #2, DUE SEPTEMBER 23, 2010 BJORN POONEN 1. Part A (After Sept. 16) 1E-1, 1E-2, 1E-3, 1E-6, 1E-7* (15 pts.) (assume that the lines are not parallel) Solution. Suppose that the ﬁrst line consists of the points r 1 ( t ) := a 1 + t v 1 for t R , where a 1 and v 1 are ﬁxed vectors with v 1 6 = 0, and that the second line is similarly parameterized as r 2 ( u ) := a 2 + u v 2 for u R . There are now several ways of ﬁnishing the problem. METHOD 1: The vector a 2 - a 1 goes from a point on the ﬁrst line to a point on the second line. The distance between the lines is the absolute value of the component of this vector in a direction perpendicular to both lines, i.e., perpendicular to v 1 and v 2 . This direction can be obtained by taking v 1 × v 2 (which is nonzero since v 1 and v 2 are not parallel) and dividing by its length to get a unit vector u := v 1 × v 2 | v 1 × v 2 | . Then the answer is | ( a 2 - a 1 ) · u | . METHOD 2: Geometry shows that the distance | r 1 ( t ) - r 2 ( u ) | is minimized when t and u are such that r 1 ( t ) - r 2 ( u ) is perpendicular to both lines, i.e., perpendicular to both v 1 and v 2 . (This can also be seen analytically by setting the partial derivatives of the function ( r 1 ( t ) - r 2 ( u )) · ( r 1 ( t ) - r 2 ( u )) equal to 0.) This leads to the system (( a 1 + t v 1 ) - ( a 2 + u v 2 )) · v 1 = 0 (( a 1 + t v 1 ) - ( a 2 + u v 2 )) · v 2 = 0 , which when expanded, is a system of two equations in two unknowns t and u . It can also be written in matrix form as ± v 1 · v 1 - v 1 · v 2 v 1 · v 2 - v 2 · v 2 ²± t u ² = ± ( a 1 - a 2 ) · v 1 ( a 1 - a 2 ) · v 2 . ² If θ is the angle between v 1 and v 2 , then the determinant of the ﬁrst matrix is -| v 1 | 2 | v 2 | 2 + ( | v 1 || v 2 | cos θ ) 2 = -| v 1 | 2 | v 2 | 2 sin 2 θ, which is nonzero since the lines are not parallel. Thus the system has a unique solution. Once t and u are obtained, the minimum distance is the value | r 1 ( t ) - r 2 ( u ) | . ± (After Sept. 16) p. 650: 34* (15 pts.) Solution. The point C is moving along a circle of radius a - b centered at O . After it has moved t radians counterclockwise, its position vector is C = (( a - b ) cos t, ( a - b ) sin t ) . 1

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Let B be the point of tangency of the two circles. Let θ be the measure of angle PCB . The arc length of the big circle from A to B is at (radius times arc measure), and since there is no slipping, this should equal the arc length of the smaller circle from B to P , which is , so θ = at/b . But P is clockwise from B , so the direction of
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h2s - 18.02 HOMEWORK#2 DUE BJORN POONEN 1 Part A(After Sept...

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