Engineering Calculus Notes 81

# Engineering Calculus Notes 81 - x y z = 1 However in...

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1.5. PLANES 69 so R (0 , 4 , 1) works, with −→ w = −→ PR = 4 −→ ı 4 −→ + −→ k . This leads to the parametrization −→ p ( s,t ) = 4 −→ ı + s (2 −→ + −→ k ) + t ( 4 −→ ı 4 −→ + −→ k ) or x = 4 4 t y = 2 s 4 t z = s + t. A diferent parametrization results From setting coordinates equal to zero in pairs: this yields the same basepoint, P (4 , 0 , 0), but two new points, Q (0 , 6 , 0), R (0 , 0 , 3). Then −→ p ( s,t ) = 4 −→ ı + s ( 4 −→ ı 6 −→ ) + t ( 4 −→ ı + 3 −→ k ) or x = 4 4 s 4 t y = 6 s z = 3 t. The converse problem—given a parametrization oF a plane, to ±nd an equation describing it—can sometimes be solved easily: For example, the plane through −→ ı , −→ and −→ k easily leads to the relation
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Unformatted text preview: x + y + z = 1. However, in general, it will be easier to handle this problem using cross products ( Â§ 1.6 ). Exercises for Â§ 1.5 Practice problems: 1. Write an equation For the plane through P perpendicular to âˆ’â†’ N : (a) P (2 , âˆ’ 1 , 3), âˆ’â†’ N = âˆ’â†’ Ä± + âˆ’â†’ + âˆ’â†’ k (b) P (1 , 1 , 1), âˆ’â†’ N = 2 âˆ’â†’ Ä± âˆ’ âˆ’â†’ + âˆ’â†’ k (c) P (3 , 2 , 1), âˆ’â†’ N = âˆ’â†’ 2. Â²ind a point P on the given plane, and a vector normal to the plane: (a) 3 x + y âˆ’ 2 z = 1 (b) x âˆ’ 2 y + 3 z = 5...
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