Engineering Calculus Notes 70

# Engineering Calculus Notes 70 - p = D − D = 0 Thus...

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58 CHAPTER 1. COORDINATES AND VECTORS is a plane in space. Using the dot product, we can extract a good deal of geometric information about this plane from Equation ( 1.21 ). Let us form a vector from the coeFcients on the left of ( 1.21 ): −→ N = A −→ ı + B −→ + C −→ k . Using −→ p = x −→ ı + y −→ + z −→ k as the position vector of P ( x,y,z ), we see that ( 1.21 ) can be expressed as the vector equation −→ N · −→ p = D. (1.22) In the special case that D = 0 this is the condition that −→ N is perpendicular to −→ p . In general, for any two points P 0 and P 1 satisfying ( 1.21 ), the vector −−−→ P 0 P 1 from P 0 to P 1 , which is the di±erence of their position vectors −−−→ P 0 P 1 = −→ p = −→ p 1 −→ p 0 lies in the plane, and hence satis²es −→ N · −→ p = −→ N · ( −→ p 1 −→ p 0 ) = −→ N · −→ p 1 −→ N · −→
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Unformatted text preview: p = D − D = 0 . Thus, letting the second point P 1 be an arbitrary point P ( x,y,z ) in the plane, we have Remark 1.5.1. If P ( x ,y ,z ) is any point whose coordinates satisfy ( 1.21 ) Ax + By + Cz = D then the locus of Equation ( 1.21 ) is the plane through P perpendicular to the normal vector −→ N := A −→ ı + B −→ + C −→ k . This geometric characterization of a plane from an equation is similar to the geometric characterization of a line from its parametrization: the normal vector −→ N formed from the left side of Equation ( 1.21 ) (by analogy with the direction vector −→ v of a parametrized line) determines the “tilt” of...
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