Using Vectors in Geometry and Physics
Date: 07/10/98 at 04:33:31 From: Jenna Subject: vectors Dear Dr Math, I'm hoping you can help me to solve the following problems. Question 1 ---------- Triangle OAB has sides OA = a and OB = b. Show that the two medians, one from A and the other from B, divide one another in the ratio 2:1. Question 2 ---------- The length of the side of a cube is 5cm. A triangular pyramid is removed from one corner of the cube. The sides that are left after the pyramid is removed consist of edges smaller in length than 5cm. The removal of the pyramid only affects three edges which are primarily joined at the same vertex. The edges are now 2cm, 1cm, and 3cm long. Use vector methods to find the surface area of the truncated cube. I know it is difficult to explain as a picture can't be shown. Question 3 ---------- A rocket is lauched from a deep underground silo and its altitude in metres is recorded electronically each second for the first 7 seconds after the launch. The data are shown below. Using any acceptable method and showing all work, calculate the velocity and the acceleration of the rocket at the end of seven seconds. Time Altitude 1 -33 2 -6 3 153 4 588 5 1491 6 3102 7 5709 Your help would be excellent, as I don't even know where to begin. Jenna
Date: 07/10/98 at 08:01:55 From: Doctor Anthony Subject: Re: vectors For Question 1: Take the origin at O and let a = VECTOR OA, b = VECTOR OB. The median from A to OB will meet OB at position vector b/2. The median from B to OA will meet OA at position vector a/2. Now any point on the median from A has position vector: a + p(b/2 - a) = a(1-p) + b(p/2) where p is a scalar fraction. Any point on the median from B has position vector: b + q(a/2 - b) = a(q/2) + b(1-q) Where these two lines intersect we can equate the position vectors, so: a(1-p) + b(p/2) = a(q/2) + b(1-q) This gives two equations: 1-p = q/2 which implies 2-2p = q (Equation 1) p/2 = 1-q (Equation 2) Using (1) in (2) we get: p/2 = 1 - (2-2p) p/2 = 1 - 2 + 2p p/2 = -1 + 2p p = -2 + 4p 2 = 3p So p = 2/3 and similarly q = 2/3. It follows that the medians intersect at 2/3 the distance along the median to the opposite side. So this point divides the medians in the ratio 2:1. For Question 2: The area of the faces of original cube is easily calculated by subtracting areas of the triangular portions cut away. So we get: 25 - (1/2 x 2 x 3) = 22 25 - (1/2 x 2 x 4) = 21 25 - (1/2 x 4 x 3) = 19 ------- Total = 62 The areas of other 3 faces = 3 x 25 = 75 Thus, the area of cube (not including triangular face at the cut) is: 62 + 75 = 137 The area of the triangular face is found as follows using vector methods. Take origin at the corner cut away. The triangle's corners have position vectors: A = (2, 0, 0) B = (0, 4, 0) C = (0, 0, 3) Then vector AB = (-2, 4, 0) and vector AC = (-2, 0, 3). The area of the triangle is found by (1/2) vector product of AB and AC: (1/2) AB x AC = (1/2)|i j k| |-2 4 0| |-2 0 3| = (1/2)[12i + 6j + 8k] = [6i + 3j + 4k] The magnitude of this vector will be the area of triangle ABC: Area = sqrt[36 + 9 + 16] = sqrt(61) Then the total surface area of truncated cube is: 137 + sqrt(61) = 144.81 For Question 3: Find average velocity in 2nd second, 3rd second, ... , 7th second. For example: In the 2nd second, average velocity = (33-6) = 27 m/sec In the 3rd second, average velocity = (6+153) = 159 m/sec In the 4th second, average velocity = (588-153) = 435 m/sec Plot an average velocity - time graph. The ordinates should be plotted at 1.5 secs, 2.5 secs, 3.5 secs, ... , 6.5 secs with the values I have calculated, plus the other ones which you can easily work out. Read off the velocity at 7 secs, and also find the slope of the graph at 7 seconds. The slope gives the acceleration at that time. - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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