3c-spring2011-exam_1_solutions

# 3c-spring2011-exam_1_solutions - Math 3C — Exam#1...

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Unformatted text preview: Math 3C — Exam #1 Solutions Laney College, Spring 2011 Fred Bourgoin 1. Let vectorv = 2 vector i + 3 vector j- vector k and vectorw = vector i- vector j + 2 vector k . (14 points) (a) Compute vectorv × vectorv . Solution. Since vectorv is parallel to itself, vectorv × vectorv = vector 0, the zero vector. (b) Compute ( vectorv · vectorw ) vectorv . Solution. The dot product is vectorv · vectorw = 2- 3- 2 =- 3, so ( vectorv · vectorw ) vectorv =- 3(2 vector i + 3 vector j- vector k ) =- 6 vector i- 9 vector j + 3 vector k . (c) Compute ( vectorv × vectorw ) · vectorw . Solution. The cross product is vectorv × vectorw = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vector i vector j vector k 2 3- 1 1- 1 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 7 vector i- 5 vector j- 5 vector k , so ( vectorv × vectorw ) · vectorw = (7 vector i- 5 vector j- 5 vector k ) · ( vector i- vector j + 2 vector k ) = 7 + 5- 10 = 2. 2. Find an equation for the plane through the points (0 , , 2), (0 , 3 , 0), and (5 , , 0). (7 points) Solution. Let’s call the points A , B , and C , respectively. From A to C , y does not change, so the x-slope is m =- 2 5 =- 2 5 . From A to B , x does not change, so the y-slope is n =- 2 3 =- 2 3 . Using these two slopes and point A , we get z = 2- 2 5 x- 2 3 y , or 2 5 x + 2 3 y + z = 2 ....
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3c-spring2011-exam_1_solutions - Math 3C — Exam#1...

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