# A tetrahedron has vertices at $\dpi{100} O$ (0, 0, 0) , $\dpi{100} A$ (1,2,1), $\dpi{100} B$ (2,1,3) and $\dpi{100} C$ (-1,1,2). Then the angle between the faces $\dpi{100} OAB$ and $\dpi{100} ABC$ will be  Option 1) $\cos ^{-1}(17/31)\;$ Option 2) $\; 30^{\circ}\;$ Option 3) $\; 90^{\circ}\;$ Option 4) $\; \cos ^{-1}(19/35)\;$

P Plabita

As we learnt in

Ange between two lines in terms of direction cosines and direction ratios -

If two lines having direction ratios a1,b1,c1 and a2,b2,c2 then the angle between them is given by

$\cos \Theta = \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a{_{1}}^{2}+b{_{1}}^{2}+c{_{1}}^{2}}\sqrt{a{_{2}}^{2}+b{_{2}}^{2}+c{_{2}}^{2}}}$

If two lines have direction ratios as l1,m1,n1 and l2,m2,n2 then the angle betwee them is given by

$\cos \Theta = l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}$

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Vector perpendicular to face OAB= $\vec{OA}\times \vec{OB}$

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

Similarily, vector perpendicular to face ABC= $= \hat{i}- 5\hat{j}-3\hat{k}$

Angle between faces

$\cos \theta= \left | \frac{5+5+9}{\sqrt{35}\sqrt{35}} \right |= \frac{19}{35}$

(Using dot product)

Option 1)

$\cos ^{-1}(17/31)\;$

Incorrect option

Option 2)

$\; 30^{\circ}\;$

Incorrect option

Option 3)

$\; 90^{\circ}\;$

Incorrect option

Option 4)

$\; \cos ^{-1}(19/35)\;$

Correct option

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