Question

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

• A. ${\cos }^{-1}\left(\frac{19}{35}\right)$
• B. ${\cos }^{-1}\left(\frac{17}{31}\right)$
• C. ${30}^o$
• D. ${90}^o$

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{19}{35}\right)$
The normal of plane OAB is $OA\times OB=(i+2j+k)\times (2i+j+3k)=5i-j-3k$
The normal of plane ABC is $CA\times CB=(OA-OC)\times (OB-OC)=(2i+j-k)\times (3i+k)=i-5j-3k$
Angle between planes $\theta ={\cos }^{-1}\left[\frac{(5i-j-3k).(i-5j-3k)}{\sqrt{(5^2+1^2+3^2)(1^2+5^2+3^2)}}\right]={\cos }^{-1}\left(\frac{19}{35}\right)$
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