A tetrahedron has vertices

Question:

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

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

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

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

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


Correct Option: 1

Solution:

$\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OQ}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$5 \hat{\mathrm{i}}-\hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

$\overrightarrow{P Q} \times \overrightarrow{P R}=(\hat{i}-\hat{j}+2 \hat{k}) \times(-2 \hat{i}-\hat{j}+\hat{k})$

$\hat{\mathrm{i}}-5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

$\cos \theta=\frac{5+5+9}{(\sqrt{25+9+1})^{2}}=\frac{19}{35}$

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