A) 150 sq. unit
B) 145 sq. unit
C) \[\frac{\sqrt{155}}{2}\] sq. unit
D) \[\frac{155}{2}\] sq. unit
Correct Answer: C
Solution :
Area of triangle = \[\frac{1}{2}|\overrightarrow{AB}\,\times \,\overrightarrow{AC}|\] \[=\frac{1}{2}\left| \,\left| \,\begin{matrix} \mathbf{i} & j & \mathbf{k} \\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\ \end{matrix}\, \right|\, \right|\] Here, \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\equiv (1,\,2,\,3)\], \[({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}})\equiv (2,\,5,\,-1)\], \[({{x}_{3}},\,{{y}_{3}},\,{{z}_{3}})\equiv (-1,\,1,\,2)\] \[=\frac{1}{2}\left| \,\left| \,\begin{matrix} i & j & k \\ 1 & 3 & -4 \\ -2 & -1 & -1 \\ \end{matrix}\, \right|\, \right|\]\[=\frac{1}{2}|(-7\mathbf{i}+9\mathbf{j}+5\mathbf{k})|\] \[=\frac{1}{2}\sqrt{49+81+25}\] \[=\frac{\sqrt{155}}{2}\] sq. unit.
You need to login to perform this action.
You will be redirected in
3 sec
