Get Answers to all your Questions

header-bg qa

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)


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}




 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

Posted by


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE