Question #86396

Calculate the area of a triangle whose vertices are given by (3,-1,2),(1,-1,2),(4,-2,1)

Expert's answer

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)

So


AB→=(1−3,−1−(−1),2−2)=(−2,0,0)AC→=(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)


The vector product


[AB→×AC→]=∣i^j^k^−2001−1−1∣=−2j^+2k^[\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

The area of the triangle

A=12∣[AB→×AC→]∣=12(−2)2+22=2A=\frac{1}{2}|[\overrightarrow{AB}\times \overrightarrow{AC}]|=\frac{1}{2}\sqrt{(-2)^2+2^2}=\sqrt{2}


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