Question 974425: Find the area of the triangle having verties at A(1,3,2) ,B(2,-1,1) ,C(-1,2,3).
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Find the area of the triangle having vertices at A(1,3,2) ,B(2,-1,1) ,C(-1,2,3).
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Vector AB = i - 4j - k
Vector AC = -2i - j + k
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The area is 1/2 the magnitude of the cross product.
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+i +j +k
+1 -4 -1
-2 -1 +1
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i*(-4-1) - j*(1-2) + k*(-1-8)
= -5i + j - 9k
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Mag = sqrt(5^2 + 1^1 + 9^2) = sqrt(107)
Area = sqrt(107)/2
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