Nomenclature 244 Posted December 7, 2017 Report Share Posted December 7, 2017 I'm reviewing for my Calc III course and I came across a problem that's stumped me and I can't easily find any example of on the web. Given point A (0, 2, 2) and B (2, 2, 2), find the set of all points P such that vector AP is perpendicular to vector BP. My first reaction was thinking this was easy, and I just set up an equation with one unknown P, where I let the dot product = 0 (as the vectors should be perpendicular). This gives me a quadratic equation and thus two values for P. However, I'm afraid this is wrong as my options for the surface (I also have to write the surface's equation) are plane/line/sphere/cone/parabaloid/hyperboloid. Thanks. Reply Link to post Share on other sites

kw0573 1,253 Posted December 7, 2017 Report Share Posted December 7, 2017 For ease of computation I'm going to translate A--> A'(0,0,0), B --> B' (2,0,0) (a translation of (0, -2, -2)). Translation does not distort geometry. A'P = (x, y, z), B'P = (x-2, y, z). set dot product = 0 x^2 - 2x + y^2 + z^2 = 0. (x - 1)^2 + y^2 + z^2 = 1 This is a sphere, where AB forms the diameter of the sphere. 1 Reply Link to post Share on other sites

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