Engineering Calculus Notes 71

Engineering Calculus Notes 71 - the third one must be. If...

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1.5. PLANES 59 the plane, and then the right-hand side D picks out from among the planes perpendicular to −→ N (which are, of course, all parallel to one another) a particular one by, in eFect, determining a point that must lie in this plane. ±or example, the plane P determined by the equation 2 x 3 y + z = 5 is perpendicular to the normal vector −→ N = 2 −→ ı 3 −→ + −→ k . To ²nd an explicit point P 0 in P , we can use one of many tricks. One such trick is to ²x two of the values x y and z and then substitute to see what
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Unformatted text preview: the third one must be. If we set x = 0 = y, then substitution into the equation yields z = 5 , so we can use as our basepoint P (0 , , 5) (which is the intersection of P with the z-axis). We could nd the intersections of P with the other two axes in a similar way. Alternatively, we could notice that x = 1 y = 1 means that 2 x 3 y = 5 , so z = 0 and we could equally well use as our basepoint P (1 , 1 , 0) ....
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