This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: byrne (cmb3744) HW 08 rusin (55565) 1 This printout should have 27 questions. Multiplechoice 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 onesheeted 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 onesheeted 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....
View
Full
Document
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.
 Spring '09
 Chu

Click to edit the document details