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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

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

Option 2)
$\; 30^{\circ}\;$

Option 3)
$\; 90^{\circ}\;$

Option 4)
$\; \cos ^{-1}(19/35)\;$

Answers (1)

best_answer

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,n₁and l₂,m₂,n₂ then the angle betwee them is given by

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

-

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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