Question

Find the area of the triangle with vertices \( A(1,1,2),B(2,3,5),and \space C(1,5,5).\)

Answer 1

The vertices of triangle ABC are given as \(A(1,1,2),B(2,3,5),and \space C(1,5,5).\)
The adjacent sides \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) of \(△ABC\) are given as:

\(\overrightarrow{AB}\)\(=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=\hat{i}+2\hat{j}+3\hat{k}\)
\(\overrightarrow{BC}=(1-2)\hat{i}+(5-3)\hat{j}+(5-5)\hat{k}=-\hat{i}+2\hat{j}\)
Area of \(△ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{BC}|\)
\(\overrightarrow{AB}\times\overrightarrow{AC}\)=\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 2 & 3\\-1&2&0 \end{vmatrix}\)=\(\hat{i}(-6)-\hat{j}(3)+\hat{k}(2+2)=-6\hat{i}+3\hat{j}+4\hat{k}\)
|\(\overrightarrow{AB}\times\overrightarrow{AC}\)|\(=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}\)\(=\sqrt{36+9+16}=\sqrt{61}\)
Hence,the area of \(△ABC\) is \(\sqrt{\frac{61}{2}}\)square units.