Question #222313

Find the area of the triangle determined by the points (2,-1,0),(2,0,4) and (0,2,3)


1
Expert's answer
2021-08-12T15:26:03-0400

Let u=(22,0(1),40)=(0,1,4)\overline {u}=(2-2,0-(-1),4-0)=(0,1,4) be the vector joining (2,1,0)(2,-1,0) and (2,0,4).(2,0,4). Let v=(02,2(1),30)=(2,3,3)\overline {v}=(0-2,2-(-1),3-0)=(-2,3,3) be the vector joining (2,1,0)(2,-1,0) and (0,2,3).(0,2,3).

Let us find the vector product


u×v=ijk014233=9i8j+2k.\overline{u}\times\overline{v}= \begin{vmatrix}i & j & k\\ 0 & 1 & 4\\ -2 & 3 & 3 \end{vmatrix} =-9i-8j+2k.

It follows that the area of the triangle determined by the points(2,1,0),(2,0,4)(2,-1,0),(2,0,4) and (0,2,3)(0,2,3) is equal to


A=12u×v=12(9)2+(8)2+22=121496.1A=\frac{1}{2}|\overline{u}\times\overline{v}|=\frac{1}{2}\sqrt{(-9)^2+(-8)^2+2^2}=\frac{1}{2}\sqrt{149}\approx 6.1 (square units)



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