Engineering Calculus Notes 72

Engineering Calculus Notes 72 - An immediate corollary of...

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60 CHAPTER 1. COORDINATES AND VECTORS Conversely, given a nonzero vector −→ N and a basepoint P 0 ( x 0 ,y 0 ,z 0 ), we can write an equation for the plane through P 0 perpendicular to −→ N in vector form as −→ N · −→ p = −→ N · −→ p 0 or equivalently −→ N · ( −→ p −→ p 0 ) = 0 . For example an equation for the plane through the point P 0 (3 , 1 , 5) perpendicular to −→ N = 4 −→ ı 6 −→ + 2 −→ k is 4( x 3) 6( y + 1) + 2( z + 5) = 0 or 4 x 6 y + 2 z = 8 . Note that the point P 0 (2 , 1 , 3) also lies in this plane. If we used this as our basepoint (and kept −→ N = 4 −→ ı 6 −→ + 2 −→ k ) the equation −→ N · ( −→ p −→ p 0 ) = 0 would take the form 4( x 2) 6( y 1) + 2( z 3) = 0 which, you should check, is equivalent to the previous equation.
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Unformatted text preview: An immediate corollary of Remark 1.5.1 is Corollary 1.5.2. The planes given by two linear equations A 1 x + B 1 y + C 1 z = D 1 A 2 x + B 2 y + C 2 z = D 2 are parallel (or coincide) precisely if the two normal vectors −→ N 1 = A 1 −→ ı + B 1 −→ + C 1 −→ k −→ N 2 = A 2 −→ ı + B 2 −→ + C 2 −→ k are (nonzero) scalar multiples of each other; when the normal vectors are equal (i.e., the two left-hand sides of the two equations are the same) then the planes coincide if D 1 = D 2 , and otherwise they are parallel and non-intersecting....
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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