Unformatted text preview: S E C T I O N 13.6 A Survey of Quadric Surfaces (ET Section 12.6) 397 or Â¿ 2 3 , âˆ’ 1 3 , 2 3 Ã€ Â· h x , y , z i = 4 , in normal form. 71. Let n = âˆ’â†’ O P , where P = ( x , y , z ) is a point on the sphere x 2 + y 2 + z 2 = r 2 , and let P be the plane with equation n Â· h x , y , z i = r 2 . Show that the point on P nearest the origin is P itself and conclude that P is tangent to the sphere at P (Figure 13). FIGURE 13 The terminal point of n lies on the sphere of radius r . SOLUTION First notice that the terminal point P of n lies on the plane P , since substituting h x , y , z i = âˆ’â†’ O P = n in the equation of the plane gives n Â· h x , y , z i = n Â· âˆ’â†’ O P = n Â· n = k n k 2 = r 2 Since the vector âˆ’â†’ O P = n is orthogonal to the plane, that is, the radius O P is perpendicular to the plane, we conclude that P is the point on P closest to the origin and that P is tangent to the sphere of radius r centered at the origin. Of course, P is the tangency point....
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 Winter '08
 GANGliu
 Hyperboloid

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