Question 764696
find the area of a triangle wit vertices at ( 3, -1, 2 ) ( 1,-1,-3) and (( 4,-3,-1)
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Find the length of each side.
{{{s = sqrt(diffx^2 + diffy^2 + diffz^2)}}}
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Then use Heron's Law to find the area.
I looked at that, it's possible but tedious.
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Another approach:
Find the lengths of 2 sides, a & b, and the angle between them.
Area = (a*b*sine(angle))/2
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Label the point.
A(3,-1,2) B(1,-1,-3) and C(4,-3,-1)
Find 2 vectors for 2 of the sides
U = AB = (2,0,5)
V = AC = (-1,2,3)
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Get UxV the cross product
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|+i +j +k|
|+2 +0 +5|
|-1 +2 +3|
--> i*(0-10) - j*(6+5) + k*(4 +0)
UxV = -10i - 11j + 4k
Find the magnitude of UxV
{{{M = sqrt(10^2 + 11^2 + 4^2) = sqrt(227)}}}
{{{Area = sqrt(227)/2}}}