Vector AnalysisDate: 7/6/96 at 23:22:39 From: Anonymous Subject: Vector Analysis Hi, Please help on my vector analysis problems: 1) Given the three points A(2,-1,2), B(-1,1,4), and C(4,3,-1), find (a) the angle between R(AB) and R(AC); (b) the (scalar) area of triangle ABC; (c) a unit vector perpendicular to ABC. 2) Given two points, M(2,5,-3) and N(-3,1,4); (a) find their separation; (b) find the distance from the origin to the midpoint of the line MN; (c)find the unit vector in the direction of R(MN); (d) find the point of intersection of the line MN and the x=0 plane. Date: 7/23/96 at 15:33:29 From: Doctor Jim Subject: Re: Vector Analysis Hi Ivan, Question 1: For part (a) remember that the cosine of the angle between the two vectors is given by dividing the dot product of the vectors by the product of their magnitudes and that R(AB)=B-A and R(AC)=C-A. I think the easiest way to do part (b) is Hero's Formula: given a triangle with sides of lengths a, b, and c, the area is given by the square root of s(s-a)(s-b)(s-c) where s is half of the perimeter. Remember that the distance between two points is the square root of (x1-x2)^2+(y1-y2)^2+z1-z2)^2. In part (c), to find a unit vector perpendicular to ABC, you just need to find a vector perpendicular to all three of the vectors in the triangle, at the same time. That is, find a vector so that its dot product with R(AB) and R(AC) and R(BC) is 0. Once you know that vector, divide by its magnitude to make it a unit vector. Question 2: Part (a): the distance between two points is the square root of (x1-x2)^2+(y1-y2)^2+z1-z2)^2. Part (b): the midpoint of a segment with endpoints (x1,y1.z1) and (x2,y2,z2) is ((x1-x2)/2, (y1-y2)/2,(z1-z2)/2), then use the distance formula again. Part (c): divide the vector R(MN) by its magnitude. -Doctor Jim, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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