(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.

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 ...

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)

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step-by-step explanation: