HW 08-solutions

# HW 08-solutions - suleimenov(bs26835 – HW 08 – rusin...

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Unformatted text preview: suleimenov (bs26835) – 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 don’t know why they use that language.) 001 10.0 points The distance d from P to the line ℓ in Q ℓ R D d a b P is given by d = | a × b | | a | where Q, R are points on ℓ and a = −−→ QR , b = −−→ QP . Use this formula to find the distance from P (1 , 4 , 7) to the line ℓ specified in parametric form by x = 3 + 2 t , y = 2 − t , z = 5 + 2 t . 1. distance = √ 103 2 2. distance = √ 104 3 correct 3. distance = √ 103 3 4. distance = √ 105 3 5. distance = √ 104 2 6. distance = √ 105 2 Explanation: It is convenient to take Q = ( x (0) , y (0) , z (0)) = (3 , 2 , 5) and R = ( x (1) , y (1) , z (1)) = (5 , 1 , 7) . 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 ellipsoid? 1. z 2 − x 2 − y 2 = 1 2. 2 x 2 + y 2 + 3 z 2 = 1 correct 3. z = x 2 + y 2 4. z = y 2 − x 2 suleimenov (bs26835) – HW 08 – rusin – (55565) 2 5. z 2 = x 2 + y 2 6. x 2 + y 2 − z 2 = 1 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 we’ve decided what the graphs of those plane slices should be. Now as slicing of shows, a slice of this ellipsoid by each of x = a , y = a and z = a will be an ellipse (or circle) so long as the slice intersects the ellipsoid. Only the graph of 2 x 2 + y 2 + 3 z 2 = 1 has this property....
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## This note was uploaded on 09/15/2011 for the course M 55565 taught by Professor Rusin during the Spring '11 term at University of Texas.

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HW 08-solutions - suleimenov(bs26835 – HW 08 – rusin...

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