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100hw2soln

100hw2soln - t changes from 0 to 40 5 ±or instance the...

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Math 100 Homework 2 fall 2010 1. Let P be a point on the given line, then there is a number t such that P = (3 t, 7 t, t ). If P is on the given plane also, we have 2(3 t ) - 7 t + t +1 = 0 or 1 = 0!!! which is impossible. Therefore the given line and the given plane do not intersect. 2. P 1 - P 2 = ( - 3 , 1 , 0) is the direction of the line joining P 1 and P 2 . Hence ( - 3 , 1 , 0) × (4 , - 1 , 3) = (3 , 9 , - 1) is perpendicular to both the line joining P 1 , P 2 and (4 , - 1 , 3) (Note: this is a normal to the given plane). Thus, (3 , 9 , - 1) is a normal to the plane we want. This plane would have defining equation 3 x + 9 y - z + d = 0 for a certain number d . Since this plane passes through (1 , 0 , 4), we have 3 + 0 - 4 + d = 0 or d = 1. Consequently, the plane we want is defined by 3 x + 9 y - z + 1 = 0. 3. After rearrangement, the given surface is defined by - 4 9 ( x + 1) 2 - 1 9 ( y - 1) 2 + 1 9 ( z - 2) 2 = 1 . It is a hyperboloid with two sheets centered at ( - 1 , 1 , 2) and its axis of symmetry is defined by x = - 1, y = 1. 4. When t changes from 0 to 40, 0 . 25 t has changed by 10. The given helix has made [40
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Unformatted text preview: t changes from 0 to 40. 5. ±or instance, the curves r 1 ( t ) = (0 , t ) and r 2 ( t ) = (0 , 1 + t ) are indeed the same straight line. They intersect at inFnitely many points. But the equation r 1 ( t ) = r 2 ( t ) does not have any solution. 6. Since r (0) = (1 , 1 , 0) and r (0) = (-1 , , 3), the line tangent to r at (1 , 1 , 0) is the line through (1 , 1 , 0) and is along the direction (-1 , , 3). This line has vector representation (1 , 1 , 0) + t (-1 , , 3) . 7. Since r 1 (0) = r 2 (-1) = (2 , 1 , 3), the curves r 1 and r 2 intersect at (2 , 1 , 3). Now, r 1 (0) = (-2 , , 0) r 2 (-1) = (-1 ,-2 , 3) are the directions of the lines tangent to r 1 , r 2 respectively at (2 , 1 , 3). The angle between them is cos-1 r (0) · r 2 (-1) || r 1 (0) || || r 2 (-1) || = cos-1 1 √ 14 . 1...
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