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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:


A
cos1(1935)
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B
cos1(7131)
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C
300
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D
900
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Solution

The correct option is A $$\displaystyle \cos ^{ -1 }{ \left( \dfrac { 19 }{ 35 }  \right)  } $$
Vector perpendicular to face $$OAB$$
$$=OA\times OB=\begin{vmatrix} i & j & k \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix}=5i-j-3k$$   ...$$(1)$$

Vector perpendicular to face $$ABC$$
$$=AB\times AC=\begin{vmatrix} i & j & k \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix}=i-5j-3k$$    ...$$(2)$$

Since the angle between the faces $$=$$ angle between their normals thus
$$\cos { \theta  } =\displaystyle\dfrac { 5+5+9 }{ \sqrt { 35 } \sqrt { 35 }  } =\dfrac { 19 }{ 35 } \Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 35 }  \right)  } $$

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