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# 5 - homework 05 – KERR KELSEY – Due Feb 1 2008 11:00 pm...

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Unformatted text preview: homework 05 – KERR, KELSEY – Due: Feb 1 2008, 11:00 pm 1 Question 1, chap 3, sect 4. part 1 of 3 10 points Consider two vectors vector A and vector B shown on the following isometric diagram: A B x y z On this diagram, the coordinate axes x , y , and z are shown as blue lines with ticks, each tick denoting one unit of distance. The vectors vector A and vector B are shown as black lines with arrow; both vectors begin at the origin of the coor- dinate system. The green dotted lines help locate the end points of vectors in 3D. For each vector, one dotted line is horizontal and projects onto the z components of the vector; the other dotted line is vertical and projects onto the green dot in the ( x, y ) plane, from which two more dotted line project onto the x and y components of the vector. Find the vertical component of vector A × vector B . Correct answer: 7 units 2 (tolerance ± 1 %). Explanation: See Eq. 2 in the next part. ( vector A × vector B ) z = bracketleftBig A x B y − A y B x bracketrightBig ˆ k = bracketleftBig (5 units) (3 units) − ( − 4 units) ( − 2 units) bracketrightBig ˆ k = 7 units 2 ˆ k . Question 2, chap 3, sect 4. part 2 of 3 10 points What is the magnitude of the vector prod- uct of these two vectors? Correct answer: 11 . 7473 units 2 (tolerance ± 1 %). Explanation: Basic Concept: vector A × vector B = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ˆ ı ˆ ˆ k A x A y A z B x B y B z vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle (1) = vextendsingle vextendsingle vextendsingle vextendsingle A y A z B y B z vextendsingle vextendsingle vextendsingle vextendsingle ˆ ı + vextendsingle vextendsingle vextendsingle vextendsingle A z A x B z B x vextendsingle vextendsingle vextendsingle vextendsingle ˆ + vextendsingle vextendsingle vextendsingle vextendsingle A x A y B x B y vextendsingle vextendsingle vextendsingle vextendsingle ˆ k = ( A y B z − A z B y )ˆ ı + ( A z B x − A x B z )ˆ + ( A x B y − A y B x ) ˆ k Let : A x = 5 units , A y = − 4 units , A z = − 1 units , B x = − 2 units , B y = 3 units , and B z = 2 units . Solution: Using Eq. 1, we have vector A × vector B = bracketleftBig A y B z − A z B y bracketrightBig ˆ ı + bracketleftBig A z B x − A x B z bracketrightBig ˆ + bracketleftBig A x B y − A y B x bracketrightBig ˆ k (2) = bracketleftBig ( − 4 units) (2 units) − ( − 1 units) (3 units) bracketrightBig ˆ ı + bracketleftBig ( − 1 units) ( − 2 units) − (5 units) (2 units) bracketrightBig ˆ + bracketleftBig (5 units) (3 units) − ( − 4 units) ( − 2 units) bracketrightBig ˆ k = ( − 5 units 2 )ˆ ı + ( − 8 units 2 )ˆ homework 05 – KERR, KELSEY – Due: Feb 1 2008, 11:00 pm 2 + (7 units 2 ) ˆ k bardbl vector A × vector B bardbl = braceleftBig ( − 5 units 2 ) 2 + ( − 8 units 2 ) 2 + (7 units 2 ) 2 bracerightBig 1 / 2 = 11 . 7473 units 2 ....
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