University of Texas M 408D Rusin HW 08

# University of Texas M 408D Rusin HW 08 - byrne (cmb3744) HW...

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Unformatted text preview: byrne (cmb3744) HW 08 rusin (55565) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Note: here, the trace of a graph in a plane just means the intersection of the graph with the plane. (I dont know why they use that language.) 001 10.0 points The distance d from P to the line in Q R D d a b P is given by d = | a b | | a | where Q, R are points on and a = QR , b = QP . Use this formula to find the distance from P (0 , 1 , 7) to the line specified in parametric form by x = 2 2 t , y = 3 t , z = 5 2 t . 1. distance = 106 2 2. distance = 105 2 3. distance = 106 3 4. distance = 104 3 correct 5. distance = 105 3 6. distance = 104 2 Explanation: It is convenient to take Q = ( x (0) , y (0) , z (0)) = (2 , 3 , 5) and R = ( x (1) , y (1) , z (1)) = (0 , 2 , 3) . With this choice of Q and R , a = ( 2 , 1 , 2 ) , b = ( 2 , 2 , 2 ) , while a b = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 2 1 2 2 2 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 1 2 2 2 vextendsingle vextendsingle vextendsingle vextendsingle i vextendsingle vextendsingle vextendsingle vextendsingle 2 2 2 2 vextendsingle vextendsingle vextendsingle vextendsingle j + vextendsingle vextendsingle vextendsingle vextendsingle 2 1 2 2 vextendsingle vextendsingle vextendsingle vextendsingle k = 6 i + 8 j + 2 k . Consequently, d = | 6 i + 8 j + 2 k | |( 2 , 1 , 2 )| = 104 3 . keywords: distance, distance from line, cross product, vector product, vector, length, 002 10.0 points For which of the following quadratic rela- tions is its graph a one-sheeted hyperboloid? 1. 2 x 2 + y 2 + 3 z 2 = 1 2. z 2 x 2 y 2 = 1 3. z 2 = x 2 + y 2 4. x 2 + y 2 z 2 = 1 correct byrne (cmb3744) HW 08 rusin (55565) 2 5. z = x 2 + y 2 6. z = y 2 x 2 Explanation: The graphs of each of the given quadratic relations is a quadric surface in standard posi- tion. We have to decide which quadric surface goes with which equation - a good way of do- ing that is by taking plane slices parallel to the coordinate planes, i.e. , by setting respec- tively x = a , y = a and z = a in the equations once weve decided what the graphs of those plane slices should be. Now as slicing of shows, slices of this one-sheeted hyperboloid by z = a are circles, while the slices by x = 0 and y = 0 are hyperbolas opening left and right. Only the graph of x 2 + y 2 z 2 = 1 has these properties....
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## This note was uploaded on 05/08/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas at Austin.

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University of Texas M 408D Rusin HW 08 - byrne (cmb3744) HW...

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