2300hw1SP11_Solutions

2300hw1SP11_Solutions - MATH 2300 - Calculus III Spring...

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Unformatted text preview: MATH 2300 - Calculus III Spring 2011 Homework 1 - SOLUTIONS 1. Find the equation of the sphere centered at (3 , 1 , 2), which is tangent to: (a) the yz-plane SOLUTION: In order to be tangent to the yz-plane, the radius must be exactly the distance from the center to the yz-plane, or 3 units. Therefore the equation is ( x 3) 2 + ( y + 1) 2 + ( z + 2) 2 = 9 . (b) the plane y = 7 SOLUTION: The radius will be the distance from the center to the plane y = 7 , or r = 7 ( 1) = 8 . Thus, the equation is ( x 3) 2 + ( y + 1) 2 + ( z + 2) 2 = 64 . 2. Consider the parallelogram shown below, with vertices at A (2 , 1 , 4), B (1 , , 1), C (1 , 2 , 3), and D . Find the following: (a) the coordinates of D SOLUTION: The point D ( x,y,z ) can be used to form the vector # AD = ( x 2 ,y + 1 ,z 4 ) , which is equal to the vector # BC = ( , 2 , 4 ) on the opposite side of the parallelogram. Thus, we find that D has coordinates (2 , 1 , 8) . (b) the interior angle at B (in degrees) SOLUTION: Using the vectors # BA = ( 1 , 1 , 5 ) and # BC = ( , 2 , 4 ) , we have: = cos 1 parenleftBigg # BA # BC | # BA || # BC | parenrightBigg = cos 1 parenleftbigg 18 27 20 parenrightbigg Therefore, the angle at B is approximately 39.2 . (c) a vector of length 8 in the direction of # AC SOLUTION: To find a vector of length 8 in the direction of # AC , we start by finding a unit vector parallel to the vector: # AC | # AC | = ( 1 , 3 , 1 ) 11 = (bigg 1 11 , 3 11 , 1 11 )bigg Now, we can multiply the unit vector by the desired length: 8 (bigg 1 11 , 3 11 , 1 11 )bigg = (bigg 8 11 , 24 11 , 8 11 )bigg (d) proj # BC # BD SOLUTION: The vector projection, is given by...
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2300hw1SP11_Solutions - MATH 2300 - Calculus III Spring...

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