Question

A tetrahedron has vertices \(\mathrm{O}(0,0,0), \mathrm{A}(1,2,1)\), \(B(2,1,3), C(-1,1,2)\). Then the angle between the faces OAB and ABC will be

• A. \(\mathrm{cos}^{-1}(\frac{19}{35})\)
• B. \(\mathrm{cos}^{-1}(\frac{1}{35})\)
• C. \(\mathrm{cos}^{-1}(\frac{9}{35})\)
• D. \(\mathrm{cos}^{-1}(\frac{4}{35})\)

Hint:

Find the normals of the two faces using cross products and use the cosine of the angle between the normals.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The angle between two faces equals the angle between their normal vectors, found from \(\cos\theta=\frac{\vec n_1\cdot\vec n_2}{|\vec n_1||\vec n_2|}\).

Step 2: Normal of OAB
\(\vec{OA}=(1,2,1),\ \vec{OB}=(2,1,3)\). \(\vec n_1=\vec{OA}\times\vec{OB}=(5,-1,-3)\).

Step 3: Normal of ABC
\(\vec{AB}=(1,-1,2),\ \vec{AC}=(-2,-1,1)\). \(\vec n_2=\vec{AB}\times\vec{AC}=(1,-5,-3)\).

Step 4: Calculation
\(\vec n_1\cdot\vec n_2=5+5+9=19\) and \(|\vec n_1|=|\vec n_2|=\sqrt{35}\). So \(\cos\theta=\frac{19}{35}\), option (A).

Final Answer:
The angle between the faces is \(\cos^{-1}\left(\frac{19}{35}\right)\). \[ \boxed{\text{(A) }\cos^{-1}\left(\dfrac{19}{35}\right)} \]