question_answer

Using vectors, find the area of triangle with vertices A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).

Answer:

Let \[\overrightarrow{OA},\] \[\overrightarrow{OB}\] and \[\overrightarrow{OC}\] be the position vectors of points A, B and C respectively. Then, \[\overrightarrow{OA}=2\hat{i}+3\hat{j}+5\hat{k},\] \[\overrightarrow{OB}=3\hat{i}+5\hat{j}+8\hat{k}\] and \[\overrightarrow{OC}=2\hat{i}+7\hat{j}+8\hat{k}\] Then, \[\overrightarrow{AB}=\] Position vector of \[B-\] Position vector of A \[=\overrightarrow{OB}-\overrightarrow{OA}=3\hat{i}+5\hat{j}+8\hat{k}-(2\hat{i}+3\hat{j}+5\hat{k})\]            \[=\hat{i}+2\hat{j}+3\hat{k}\] and \[\overrightarrow{AC}=\] Position vector of \[C-\] Position vector           of A                  \[=\overrightarrow{OC}-\overrightarrow{OA}=2\hat{i}+7\hat{j}+8\hat{k}-(2\hat{i}+3\hat{j}+5\hat{k})\]                        \[=4\hat{j}+3\hat{k}\] Now, area of \[\Delta ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\,\,=\frac{1}{2}\left| \left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & 2 & 3  \\    0 & 4 & 3  \\ \end{matrix} \right| \right|\]             \[=\frac{1}{2}|i(6-12)-\hat{j}(3-0)+\hat{k}(4-0)|\]             \[=\frac{1}{2}|-\,6\hat{i}-3\hat{j}+4\hat{k}|\]             \[=\frac{1}{2}\sqrt{{{(-\,6)}^{2}}+{{(-\,3)}^{2}}+{{(4)}^{2}}}\]             \[=\frac{1}{2}\sqrt{36+9+16}=\frac{1}{2}\sqrt{61}\] Hence, area of \[\Delta ABC\] is  \[\frac{1}{2}\sqrt{61}\] sq. units