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  • Help me understand! - A tetrahedron has verticesThe angle between the faces OPQ and PQR is : - Vector Algebra - JEE Main

A tetrahedron has vertices P(1,2,1),\: Q(2,1,3),\: R(-1,1,2)\: \: and \: \: O\left ( 0,0,0 \right ). The angle between the faces OPQ and PQR is : 

  • Option 1)

    \cos ^{-1}\left ( \frac{7}{31} \right )

     

     

     

  • Option 2)

    \cos ^{-1}\left ( \frac{9}{35} \right )

  • Option 3)

    \cos ^{-1}\left ( \frac{19}{35} \right )

  • Option 4)

    \cos ^{-1}\left ( \frac{17}{31} \right )

Answers (1)

best_answer

 

Vector Product of two vectors -

Vector product

- wherein

\vec{a}= (a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})

\vec{b}= (b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k})

 

Vector perpendicular to face OAB

= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1& 2 &1 \\ 2&1 &3 \end{vmatrix} = 5\hat i - \hat j -3 \hat k

Vector perpendicular to face ABC

= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2& 1 & -1 \\ 1&-1 &2 \end{vmatrix} = \hat i - 5\hat j -2 \hat k

ANgle between two faces 

\cos \theta = \left | \frac{5 + 5 + 4}{\sqrt {35}\cdot \sqrt{35}} \right | = \frac{19}{35} \\ \\ \\ \Rightarrow \theta = \cos^{-1}\left(\frac{19}{35} \right )

 


Option 1)

\cos ^{-1}\left ( \frac{7}{31} \right )

 

 

 

Option 2)

\cos ^{-1}\left ( \frac{9}{35} \right )

Option 3)

\cos ^{-1}\left ( \frac{19}{35} \right )

Option 4)

\cos ^{-1}\left ( \frac{17}{31} \right )

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