32 +( 2) ( 4) ( 2)

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Bryan McDowell
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1 Math 241 Exam 1 Sample 2 Solutions 1. (a) If ā = 3î 2ĵ+1ˆk and b = 4î+0ĵ 2ˆk, find the sine and cosine of the angle θ between [10 pts] ā and b. We know that ā b = ā b cosθ and so cosθ = ā b ā b = (3)( 4)+( 2)(0)+(1)( 2) 32 +( 2) ( 4) ( 2) 2 We also know that ā b = ā b sinθ and so sinθ = ā b ā b = 4î+2ĵ 8ˆk 32 +( 2) ( 4) ( 2) = 2 +( 8) ( 2) ( 4) ( 2) 2 (b) Do the points (0,0,0), (0,3,4) and (5,2,1) form a right triangle? Justify. Solution 1: If we measure the three distances between the points we see: Distance from (0,0,0) to (0,3,4) is = 25. Distance from (0,0,0) to (5,2,1) is = 30. Distance from (0,3,4) to (5,2,1) is = 35. We then observe that there is no way to assign these as a,b,c so that the Pythagorean Theorem holds, so they do not form a right triangle. Solution 2: If we find the three vectors connecting the points: Vector from (0,0,0) to (0,3,4) is 0î+3ĵ+4ˆk. Vector from (0,0,0) to (5,2,1) is 5î+2ĵ+1ˆk. Vector from (0,3,4) to (5,2,1) is 5î 1ĵ 3ˆk. If we take the dot product of any two of these: [10 pts] (0î+3ĵ+4ˆk) (5î+2ĵ+1ˆk) 0 (0î+3ĵ+4ˆk) (5î 1ĵ 3ˆk) 0 (5î+2ĵ+1ˆk) (5î 1ĵ 3ˆk) 0 We never get zero, meaning no two of these are perpendicular, meaning no two of these form a right angle.

2 2. (a) Find the equation of the plane containing the line x 2 = 2 y 3 = z and the point (2, 1,3) and [10 pts] check whether this plane contains the origin. We need a normal vector for the plane. To get this we will take the cross product of two vectors which are parallel to the plane. If we first rewrite the line: x 0 2 = y 2 3 = z 0 1 we see that the direction vector for the line is 2î 3ĵ+1ˆk so this vector is parallel to the plane (since the line is in the plane). A point on the line is (0,2,0) and so another vector parallel to the plane is the vector from (0,2,0) to (2, 1,3) since both points are in the plane. This vector is 2î 3ĵ+3ˆk. The cross product of these is then (2î 3ĵ+1ˆk) (2î 3ĵ+3ˆk) = 6î 4ĵ+0ˆk and so the plane has equation 6(x 2) 4(y ( 1))+0(z 3) = 0. The point (0,0,0) does not satisfy this equation so the plane does not contain the origin. (b) Provide a piecewise smooth parametrization of the triangle with vertices (0, 0), (5, 2) and [10 pts] (4,7). We need to do this piece by piece. I don t mind if the values of t overlap between pieces, just do them independently! Each of these is basically the vector equation of a line: First part, (0,0) to (5,2): L = 5î + 2ĵ with (0,0) gives r(t) = (0 + 5t)î + (0 + 2t)ĵ for Second part, (5,2) to (4,7): L = 1î+5ĵ with (5,2) gives r(t) = (5 1t)î+(2+5t)ĵ for Third part, (4,7) to (0,0): L = 4î 7ĵ with (4,7) gives r(t) = (4 4t)î+(7 7t)ĵ for Note that since I didn t say anything about the direction, each of these would be fine if we had changed the order of the points.

3 3. (a) Find the tangent and normal vectors for the curve r(t) = t 2 î+tĵ at t = 2. [10 pts] We have r (t) = 2tî+1ĵ so r ( 2) = 2 2î+1ĵ and so T( 2) = 2 2î+1ĵ 2 2î+1ĵ = 2 2î+1ĵ Next note that and so T(t) = 2tî+1ĵ 4t2 +1 = 2t (4t 2 +1) 1/2 î+(4t2 +1) 1/2 ĵ T (t) = 2(4t2 +1) 1/2 2t 1/2(4t 2 +1) 1/2 (8t) 4t 2 +1 î 1 2 (4t2 +1) 3/2 (8t)ĵ so that T ( 2) = 2()1/2 16() 1/2 î 4 2() 3/2 ĵ = 6 16/3 î 4 2/27ĵ = 2/27î 4 2/27ĵ and so N( 2) = 2/27î 4 2/27ĵ 2/27î 4 2/27ĵ = 2/27î 4 2/27ĵ (2/27) 2 +( 4 2/27) 2 (b) Find the value of a so that the curve r(t) = tî+(2+3t)ĵ for 0 t a has length 7. Observe that r (t) = 1î+3ĵ and so [10 pts] r (t) = 1+ = 10 For the length to be 7 we need: a 0 10 dt = 7 a 10 t = a = 7 a = 7/ 10

4 4. (a) Sketch the sphere x 2 +y 2 6y = z 2 +2z. Mark at least two points with their coordinates. [5 pts] We need to get everything on one side and complete the square first: x 2 +y 2 6y +z 2 2z = 0 x 2 +y 2 6y ++z 2 2z +1 = +1 x 2 +(y 3) 2 +(z 1) 2 = 10 So the center is (0,3,1) and the radius is 10: (b) Find the equation of the sphere with center (1, 2,1) and which also contains the origin. Since the sphere contains the origin, the distance from the origin to the center must be the radius. This radius is then 1 2 +( 2) = 6 and so the equation of the sphere is (x 1) 2 +(y ( 2)) 2 +(z 1) 2 = ( 6) 2 [5 pts] (c) Find the distance between the plane containing (0,0,1), (1,2,3) and (0, 2,4) and the origin. [10 pts] The formula is dist = N PQ N for which we have P, we can use (0,0,1) and Q = (0,0,0). We just need N. To do this we form two vectors parallel to the plane: Vector from (0,0,1) to (1,2,3) is 1î+2ĵ+2ˆk. Vector from (0,0,1) to (0, 2,4) is 0î 2ĵ+3ˆk. We cross these to get N = 10î 3ĵ 2ˆk. Then PQ = 0î+0ĵ 1ˆk and so the distance is (10î 3ĵ 2ˆk) (0î+0ĵ 1ˆk) 2 = 102 +( 3) 2 +( 2)

5 5. (a) Sketch r(t) = 2sin(t)î+5cos(t)ĵ 1ˆk for 0 t π 2. Label the start and end points with [10 pts] their coordinates and indicate direction. (b) Let L be the line r(t) = (2 t)î+tĵ+(3t+1)ˆk and let P be the plane x+y +2z = 10. i. Find the point where L and P meet. [5 pts] In order to meet, the x,y,z from the line must satisfy the plane equation: (2 t)+t+2(3t+1) = 10 6t = 6 t = 1 The point is then r(1) = 1î+1ĵ+4ˆk or (1,1,4). ii. Show that the line is not perpendicular to the plane. The line being perpendicular to the plane would mean that the direction vector for the line and the normal vector for the plane would be parallel. The direction vector is 1î+1ĵ+3ˆk and the normal vector is 1î+1ĵ+2ˆk. Since these are not multiples of one another they are not parallel and so the line is not perpendicular to the plane. [5 pts]

Math 210, Exam 1, Practice Fall 2009 Problem 1 Solution

Math 210, Exam 1, Practice Fall 2009 Problem 1 Solution Math 20, Exam, Practice Fall 2009 Problem Solution. Let A = (,,2), B = (0,,), C = (2,,). (a) Find the vector equation of the plane through A, B, C. (b) Find the area of the triangle with these three vertices.