Vector Algebra Question 68
Question: A tetrahedron has vertices at $ O(0,,0,,0) $ , $ A(1,,2,,1),B(2,,1,,3) $ and $ C(-1,,1,,2) $ . Then the angle between the faces $ OAB $ and $ ABC $ will be
[MNR 1994; UPSEAT 2000; AIEEE 2003]
Options:
A) $ {{\cos }^{-1}}( \frac{19}{35} ) $
B) $ {{\cos }^{-1}}( \frac{17}{31} ) $
C) $ 30{}^\circ $
D) $ 90{}^\circ $
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Answer:
Correct Answer: A
Solution:
- Angle between two plane faces is equal to the angle between the normals $ {{\mathbf{n}}_1} $ and $ {{\mathbf{n}}_2} $ to the planes. $ {{\mathbf{n}}_1} $ the normal of face OAB is given by $ \overrightarrow{OA}\times \overrightarrow{OB}=| ,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \\ \end{matrix}, |=5\mathbf{i}-\mathbf{j}-3\mathbf{k} $ …..(i) $ {{\mathbf{n}}_2} $ the normal of face ABC is given by $ \overrightarrow{AB}\times \overrightarrow{AC} $ $ 2-1,1-2,3-1 $ and $ -1-1,1-2,2-1 $ i.e., $ 1,,-1,2 $ and $ -2,-1,1. $
$ \therefore ,{{\mathbf{n}}_2}=| ,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \\ \end{matrix}, |=\mathbf{i}-5\mathbf{j}-3\mathbf{k} $ ?..(ii) If q be the angle between $ {{\mathbf{n}}_1} $ and $ {{\mathbf{n}}_2} $ , then $ \cos \theta =\frac{{{\mathbf{n}}_1},.,{{\mathbf{n}}_2}}{|{{\mathbf{n}}_1}|.|{{\mathbf{n}}_2}|}=\frac{5,+5+9}{\sqrt{35},.,\sqrt{35}}=\frac{19}{35} $
$ \Rightarrow \theta ={{\cos }^{-1}}( \frac{19}{35} ) $ .
BETA
