# Scalar Equation of Plane.pdf - Let P be a plane with vector...

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Let P be a plane with vector equation ~ q + t~u + s~v A normal vector for P is a non-zero vector ~n such that ~n · ~u = ~n · ~v = 0. Now let ~x, ~ y P . This means that ~x = ~ q + t 1 ~u + s 2 ~v, ~ y = ~ q + t 1 ~u + s 1 ~v for some scalars t 1 , s 1 , t 2 , s 2 R . Then we have that ~x - ~ y = ( t 1 - t 2 ) ~u + ( s 1 - s 2 ) ~v In particular we see that ~n · ( ~x - ~ y ) = ~n · (( t 1 - t 2 ) ~u + ( s 1 - s 2 ) ~v ) = ( t 1 - t 2 ) ~n · ~u + ( s 1 - s 2 ) ~n · ~v = 0 because ~n · ~u = ~n · ~v = 0. So obtain an equivalent definition of a normal vector. Let P be a plane. A non-zero vector ~n is a normal vector for P if and only if for all ~x, ~ y P ~n · ( ~x - ~ y ) = 0 Example 1. Let ~u = 1 0 5 and ~v = 1 1 1 and ~ q = 1 - 1 3 . Let P be the plane with vector equation ~ q + t~u + s~v . To find a normal vector ~n for P we may take the cross-product ~u × ~v or work by inspection to note that 5 - 4 - 1 is perpendicular to both ~u,~v . Now let ~x P . Then we have that there are t, s R with ~x = ~ q + t~u + s~v = t + s + 1 s - 1 5 t + s + 3 Now consider 5 - 4 - 1 · t + s + 1 s - 1 5 t + s + 3 = 5 t + 5 s + 5 - 4 s + 4 - 5 t - s - 3 = 5 + 4 - 3 = 6 On the other hand, 5 - 4 - 1 · 1 - 1 3 = 5 + 4 - 3 = 6 So what we have shown is that for any ~x P we have 1
~n · ~x = ~n · ~ q Now suppose that ~x = x 1 x 2 x 3 and ~n · ~x = 6 . Then we have an solution to the equation 5 x 1 - 4 x 2 - x 3 = 6 Thus we have that x 3 = 5 x 1 - 4 x 2 - 6 So we have ~x = x 1 x 2 5 x 1 - 4 x 2 - 6 = x 1 1 0 5 + x 2 0 1 - 4 + 0 0 - 6 Now notice that 0 0 - 6 = 1 - 1 3 + 1 1 1 - 2 1 0 5 = - 2 ~u + ~v + ~ q So 0 0 - 6 P . On the other hand, we have that 0 1 - 4 = - 1 0 5 + 1 1 1 = - ~u + ~v Putting this together shows that ~x = x 1 ~u + x 2 0 1 - 4 +
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