math20c Meesue Yoo fall 09 quiz2

math20c Meesue Yoo fall 09 quiz2 - the angle between two...

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Quiz 2 Solution for MATH 20C, 2009 1. Consider three points P = (1 , 1 , 5), Q = (3 , 4 , 3), R = (1 , 5 , 7) in three space. (a) Use the cross product to find the area of the triangle with vertices P , Q and R . Ans. -→ PQ = < 2 , 3 , - 2 > , -→ PR = < 0 , 4 , 2 > . -→ PQ × -→ PR = ± ± ± ± ± ± i j k 2 3 - 2 0 4 2 ± ± ± ± ± ± = 14 i - 4 j + 8 k. Area of 4 ABC = || -→ PQ × -→ PR || 2 = 196 + 16 + 64 2 = 276 2 = 69 . (b) Find the equation of the plane passing through the three points P , Q and R . Ans. We can take the normal vector of the plane as -→ n = -→ PQ × -→ PR , so the plane equation is 14( x - 1) - 4( y - 1) + 8( z - 5) = 0 or, 7 x - 2 y + 4 z = 25 . 2. We define the angle between two planes by the angle between their normal vectors. Compute the angle between two planes x + z = 3 and x - y - z = 5 . Ans. The normal vectors of the given planes are -→ n 1 = < 1 , 0 , 1 > and -→ n 2 = < 1 , - 1 , - 1 > . We take the dot product of the normal vectors to get the cosine of
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Unformatted text preview: the angle between two planes ;- n 1 - n 2 = 1-1 = 0 . Since the dot product is zero, they are orthogonal, i.e., the angle between two planes is / 2. 3. Find the parametrization of the intersection curve of the plane y = 1 2 and the sphere x 2 + y 2 + z 2 = 1. Ans. Since y is given by 1 / 2, the second surface becomes x 2 + z 2 = 3 / 4 which is an equation of a circle centered at the origin with radius 3 / 2. So we parametrize the circle by x = 3 2 cos t , z = 3 2 sin t and so, the parametrization of the intersection curve is r ( t ) = * 3 2 cos t, 1 2 , 3 2 sin t + ....
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This note was uploaded on 01/26/2010 for the course MATH Math 20C taught by Professor Lunasin during the Fall '08 term at UCSD.

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