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(1 point) let a=(2,4,−5)a=(2,4,−5), b=(−3,6,−5)b=(−3,6,−5), c=(−8,7,0)c=(−8,7,0), and d=(−3,5,0)d=(−3,5,0). find the area of the parallelogram determined by these four points, the area of the triangle abcabc, and the area of the triangle abdabd.

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Areas and volumes of parallelograms and parallelepipeds in 3 dimensions are often easily found by making use of the cross product of the direction vectors of their edges. For edge vectors v1 and v2 of a triangle, the area is ...

... A = (1/2)║v1 × v2║

that is, half the norm of the cross-product vector. The area of a parallelogram with those edge vectors is simply ...

... A = ║v1 × v2║

Here, direction vectors are ...

  • ab = (-5, 2, 0)
  • bc = (-5, 1, 5)
  • cd = (5, -2, 0)
  • da = (5, -1, -5)

We can see that ab = -cd and bc = -da, as required for a parallelogram.

The cross product ab × bc is (10, 25, 5), so the area of the parallelogram is

... ║(10, 25, 5)║ = √(10² +25² +5²) = √750

... Area = 5√30 ≈ 27.3861 . . . . square units (parallelogram area)

The areas of each of the mentioned triangles is half the area of the parallelogram, so is

... Area Δabc = Area ∆abd = (5/2)√30 ≈ 13.6931 . . . . square units (triangles)

User AbhishekDwivedi
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