MHT CET · Maths · Vector Algebra
A tetrahedron has vertices \(O(0,0,0) A(1,2,1) B(2,1,3) C(-1,1,2)\).
Then the angle between the faces \(O A B\) and \(A B C\) will be
- A \(\cos ^{-1}\left(\frac{19}{35}\right)\)
- B \(\cos ^{-1}\left(\frac{1}{35}\right)\)
- C \(\cos ^{-1}\left(\frac{9}{35}\right)\)
- D \(\cos ^{-1}\left(\frac{4}{35}\right)\)
Answer & Solution
Correct Answer
(A) \(\cos ^{-1}\left(\frac{19}{35}\right)\)
Step-by-step Solution
Detailed explanation
Normal vector for OAB: \(\vec{n_1} = \vec{OA} \times \vec{OB} = (1,2,1) \times (2,1,3) = (5,-1,-3)\) Normal vector for ABC: \(\vec{n_2} = \vec{AB} \times \vec{AC} = (1,-1,2) \times (-2,-1,1) = (1,-5,-3)\)
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