question_answer

A tetrahedron has vertices at O(0, 0, 0), A(1, -2, 1), B (-2, 1, 1) and C (1, -1, 2). Then the angle between the faces OAB and ABC will be

A) \[{{\cos }^{-1}}\left( \frac{1}{2} \right)\]                               

B) \[{{\cos }^{-1}}\left( \frac{-1}{6} \right)\]

C) \[{{\cos }^{-1}}\left( \frac{-1}{3} \right)\]             

D) \[{{\cos }^{-1}}\left( \frac{1}{4} \right)\]

Correct Answer: C

Solution :

Vector perpendicular to face \[\frac{N}{{{N}_{0}}}=\frac{60}{100}={{\left( \frac{1}{2} \right)}^{n}}\]                   \[\Rightarrow \]                \[{{2}^{n}}=\frac{10}{6}\]                \[\Rightarrow \]               \[n\times 0.3=1-0.778=0.22\]                \[n=\frac{0.222}{0.3}=0.74\] Vector perpendicular to face \[ABC={{n}_{2}}\] \[=AB\times AC\] \[=(-3\hat{i}+3\hat{j})\times (\hat{j}+\hat{k})\] \[=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    -3 & 3 & 0  \\    0 & 1 & 1  \\ \end{matrix} \right|\]\[=3\hat{i}+3\hat{j}-3\hat{k}\] Since, angle between faces equals to angle between their normals. \[t=n{{T}_{1/2}}=0.74\times 4.47\times {{10}^{8}}\]          \[t=3.3\times {{10}^{8}}yr\]                \[Y\left( n,\alpha  \right)\]                 \[\alpha -\] \[_{Z}{{Y}^{A}}{{+}_{0}}{{n}^{1}}{{\to }_{3}}L{{i}^{7}}{{+}_{2}}H{{e}^{4}}\]          \[\Rightarrow \]