2019-03-15T00:45:19-04:00
Calculate the area of a triangle whose vertices are given by (3,-1,2),(1,-1,2),(4,-2,1)
1
2019-03-15T12:45:45-0400
Let
A = ( 3 , − 1 , 2 ) , B = ( 1 , − 1 , 2 ) , C = ( 4 , − 2 , 1 ) A=(3,-1,2),\quad B=(1,-1,2), \quad C=(4,-2,1) A = ( 3 , − 1 , 2 ) , B = ( 1 , − 1 , 2 ) , C = ( 4 , − 2 , 1 ) So
A B → = ( 1 − 3 , − 1 − ( − 1 ) , 2 − 2 ) = ( − 2 , 0 , 0 ) A C → = ( 4 − 3 , − 2 − ( − 1 ) , 1 − 2 ) = ( 1 , − 1 , − 1 ) \overrightarrow{AB}=(1-3,-1-(-1),2-2)=(-2,0,0)\\
\overrightarrow{AC}=(4-3,-2-(-1),1-2)=(1,-1,-1) A B = ( 1 − 3 , − 1 − ( − 1 ) , 2 − 2 ) = ( − 2 , 0 , 0 ) A C = ( 4 − 3 , − 2 − ( − 1 ) , 1 − 2 ) = ( 1 , − 1 , − 1 )
The vector product
[ A B → × A C → ] = ∣ i ^ j ^ k ^ − 2 0 0 1 − 1 − 1 ∣ = − 2 j ^ + 2 k ^ [\overrightarrow{AB}\times \overrightarrow{AC}]=\begin{vmatrix}
\hat i & \hat j & \hat k\\
-2 & 0 & 0\\
1 & -1 & -1
\end{vmatrix}=-2\hat j+2\hat k [ A B × A C ] = ∣ ∣ i ^ − 2 1 j ^ 0 − 1 k ^ 0 − 1 ∣ ∣ = − 2 j ^ + 2 k ^ The area of the triangle
A = 1 2 ∣ [ A B → × A C → ] ∣ = 1 2 ( − 2 ) 2 + 2 2 = 2 A=\frac{1}{2}|[\overrightarrow{AB}\times \overrightarrow{AC}]|=\frac{1}{2}\sqrt{(-2)^2+2^2}=\sqrt{2} A = 2 1 ∣ [ A B × A C ] ∣ = 2 1 ( − 2 ) 2 + 2 2 = 2
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