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2300hw1SP11_Solutions - MATH 2300 Calculus III Spring 2011...

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——————————————————————————————————– 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 , 0 , 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 = ( 0 , 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 = ( 0 , 2 , 4 ) , we have: θ = 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:
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# » 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 proj # » BC # » BD = # » BC · # » BD | # » BC | 2 ( # » BC ) = ( 0 , 2 , 4 ) · ( 1 , 1 , 9 ) 20 ( 0 , 2 , 4 ) = 19 10 ( 0 , 2 , 4 ) = (bigg 0 , 19 5 , 38 5 )bigg
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