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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 cosine of angle ABC ?

Options

  1. A13\frac{1}{\sqrt{3}}
  2. B12\frac{1}{\sqrt{2}}
  3. C26\frac{2}{\sqrt{6}}
  4. DNone of the above

Correct answer

A. 13\frac{1}{\sqrt{3}}

Step-by-step solution

<br/><br/>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{B} & =\frac{a^2+c^2-b^2}{2 a c}=\frac{6+18-12}{2 \sqrt{6} \times \sqrt{18}} \\ & =\frac{12}{2 \sqrt{108}}=\frac{12}{2 \times 6 \sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}
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