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© 2005 Baylor University Slide 1 Fundamentals of Engineering Analysis EGR 1302 Unit 2, Lecture G Approximate Running Time - 17 minutes Distance Learning / Online Instructional Presentation Presented by Department of Mechanical Engineering Baylor University Procedures: 1.Select “Slide Show” with the menu: Slide Show|View Show (F5 key), and hit “Enter” 2.You will hear “CHIMES” at the completion of the audio portion of each slide; hit the “Enter” key, or the “Page Down” key, or “Left Click” 3.You may exit the slide show at any time with the “Esc” key; and you may select and replay any slide, by navigating with the “Page Up/Down” keys, and then hitting “Shift+F5”.

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© 2005 Baylor University Slide 2 Plane: 1 Line: L1 3-D Equations for the Plane and the Line Substitute to get

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© 2005 Baylor University Slide 3 Find the Equation for the Plane Given A:(1,-2,1) B:(2,6,2) C:(-1,-2,3) Use the Three Point Form: Simplify:

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© 2005 Baylor University Slide 4 Now Find the Equation for the Line Given D:(-3,6,10) E:(0,2,8) Vector Parametric Form Cartesian Form

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© 2005 Baylor University Slide 5 Now Find the That Satisfies Both Equations The Plane: The Line: Substitute into the Plane equation And solve for Substituting back into the Line equation Are the of the point of intersection

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© 2005 Baylor University Slide 6 The Unit Vector

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© 2005 Baylor University Slide 7 The Basis Vectors

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© 2005 Baylor University Slide 8 Basis Vectors Simplify Vector Math Add components independently

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© 2005 Baylor University Slide 9 Position Vectors in Basis Vector Notation

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© 2005 Baylor University Slide 10 The Unit Vector of

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© 2005 Baylor University Slide 11 This concludes Unit 2, Lecture G You are now Ready to Take the Unit 2 Exam

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X = 2 + t y = -3 + 2t t = x – 2 t = (y + 3)/2 x – 2 = y + 3 2 2x – 4 = y + 3 y – 2x + 7 = 0 Finding the Cartesian Equation from a vector equation x = 2.

X = 2 + t y = -3 + 2t t = x – 2 t = (y + 3)/2 x – 2 = y + 3 2 2x – 4 = y + 3 y – 2x + 7 = 0 Finding the Cartesian Equation from a vector equation x = 2.

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