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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2014 · Mathematics · Three Dimensional Geometry

Paragraph : The vertices of a triangle A B C are A(2,3,1),...

Paragraph : The vertices of a triangle ABCA B C are A(2,3,1),B(−2,2,0)A(2,3,1), B(-2,2,0) and C(0,1,−1)C(0,1,-1).
Question : What is the area of the triangle?

Options

  1. A626 \sqrt{2} square unit
  2. B323 \sqrt{2} square unit
  3. C10310 \sqrt{3} square unit
  4. DNone of the above

Correct answer

B. 323 \sqrt{2} square unit

Step-by-step solution

a=BC=(2)2+(−1)2+(−1)2=4+1+1=6b=CA=(2)2+(2)2+(2)2=4+4+4=12c=AB=(4)2+(1)2+(1)2=16+1+1=18.\begin{aligned} & a=\mathrm{BC}=\sqrt{(2)^2+(-1)^2+(-1)^2}=\sqrt{4+1+1}=\sqrt{6} \\ & b=\mathrm{CA}=\sqrt{(2)^2+(2)^2+(2)^2}=\sqrt{4+4+4}=\sqrt{12} \\ & c=\mathrm{AB}=\sqrt{(4)^2+(1)^2+(1)^2}=\sqrt{16+1+1}=\sqrt{18}.\end{aligned}$\begin{aligned} \cos \mathrm{C} & =\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{6+12-18}{2 \sqrt{6} \times \sqrt{12}}=0 \\ \Rightarrow \quad \cos \mathrm{C} & =\cos 90^{{}^{\circ}} \Rightarrow \angle \mathrm{C}=90^{{}^{\circ}} \end{aligned}\thereforeItisrighttriangle.It is right triangle.\text{ Areaofof} △ABC=12× base × height =12×6×12=12×72=12×62=32 square unit. \begin{aligned} \triangle \mathrm{ABC} & =\frac{1}{2} \times \text{ base } \times \text{ height } \\ & =\frac{1}{2} \times \sqrt{6} \times \sqrt{12}=\frac{1}{2} \times \sqrt{72} \\ & =\frac{1}{2} \times 6 \sqrt{2}=3 \sqrt{2} \text{ square unit. } \end{aligned}$
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