brainyhouse5856 brainyhouse5856
  • 02-09-2019
  • Mathematics
contestada

Find an equation for the plane that passes through the following points.

(2, −1, 3), (0, 0, 4), and (4, 5, −1)

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sqdancefan
sqdancefan sqdancefan
  • 02-09-2019

Answer:

  5x +3y +7z = 28

Step-by-step explanation:

Call the given points A, B, and C. Then the cross product AB×AC will be normal to the plane, so will tell what the coefficients in the plane's equation might be.

  AB = (-2, 1, 1)

  AC = (2, 6, -4)

  AB×AC = (-10, -6, -14)

Dividing a factor of -2 from these coefficients, and using point B to determine the constant, we have the equation for the plane ...

  5x + 3y + 7z = 5·0 +3·0 +7·4

  5x + 3y + 7z = 28

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