Answer to Question #86396 in Mechanics | Relativity for Manisha nayak

Question #86396
Calculate the area of a triangle whose vertices are given by (3,-1,2),(1,-1,2),(4,-2,1)
1
Expert's answer
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)

So


AB=(13,1(1),22)=(2,0,0)AC=(43,2(1),12)=(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^200111=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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