Question

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

Answer 1

Answer:

Area of triangle ABC is

$\frac{1}{2}\sqrt{61}\text{ square units }$

Step-by-step explanation:

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

The position vectors of the given points are

$\overrightarrow{OA}=\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k}$

$\overrightarrow{OB}=2\overrightarrow{i}+3\overrightarrow{j}+5\overrightarrow{k}$

$\overrightarrow{OC}=\overrightarrow{i}+5\overrightarrow{j}+5\overrightarrow{k}$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$=\overrightarrow{i}+2\overrightarrow{j}+3\overrightarrow{k}$

$\overrightarrow{AC}=\overrightarrow{OB}-\overrightarrow{OA}$

$=0\overrightarrow{i}+4\overrightarrow{j}+3\overrightarrow{k}$

$\overrightarrow{AB}\times\overrightarrow{AC}=\left|\begin{array}{ccc}\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\1&2&3\\0&4&3\end{array}\right|$

$=\overrightarrow{i}(6-12)-\overrightarrow{j}(3-0)+\overrightarrow{k}(4-0)$

$=-6\overrightarrow{i}-3\overrightarrow{j}+4\overrightarrow{k}$

$|\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{36+9+16}=\sqrt{61}$

$\text{Area of triangle ABC }=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|$

$=\frac{1}{2}\sqrt{61}\text{ square units }$