Let u=(2−2,0−(−1),4−0)=(0,1,4) be the vector joining (2,−1,0) and (2,0,4). Let v=(0−2,2−(−1),3−0)=(−2,3,3) be the vector joining (2,−1,0) and (0,2,3).
Let us find the vector product
u×v=∣∣i0−2j13k43∣∣=−9i−8j+2k.
It follows that the area of the triangle determined by the points(2,−1,0),(2,0,4) and (0,2,3) is equal to
A=21∣u×v∣=21(−9)2+(−8)2+22=21149≈6.1 (square units)
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