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Question

Mathematics Question on introduction to three dimensional geometry

A tetrahedron has vertices at O(0,0,0),A(1,2,1)B(2,1,3) O (0, 0, 0), A(1, 2, 1) B(2, 1, 3 ) and C(1,1,2)C(-1, 1, 2). Then the angle between the faces OAB and ABC will be

A

cos1(1935)\cos^{-1}\left(\frac{19}{35}\right)

B

cos1(1731)\cos^{-1}\left(\frac{17}{31}\right)

C

30^\circ

D

90^\circ

Answer

cos1(1935)\cos^{-1}\left(\frac{19}{35}\right)

Explanation

Solution

Perpendicular to face OABOAB OA×OB\overrightarrow{OA} \times\overrightarrow{OB} =(i^+2j^+k^)×(2i^+j^+3k^)=\left(\hat{i} +2\hat{j}+\hat{k}\right)\times \left(2 \hat{i}+\hat{j}+3\hat{k}\right) =(5i^j^3k^)=\left(5 \hat{i}-\hat{j}-3\hat{k}\right) Vector perpendicular to face ABCABC AB×AC\overrightarrow{AB} \times\overrightarrow{AC} =(i^j^+2k^)×(2i^j^+k^)=\left(\hat{i}-\hat{j}+2 \hat{k}\right)\times\left(-2 \hat{i}-\hat{j}+\hat{k}\right) =i^5j^3k^=\hat{i}-5 \hat{j}-3 \hat{k} Since angle between face equals angle between their normals. Therefore cosθ=5×1+(1)×(5)+(3)×(3)52+(1)2+(3)212+(5)2+(3)cos\, \theta=\frac{5\times1+\left(-1\right)\times\left(-5\right)+\left(-3\right)\times\left(-3\right)}{\sqrt{5^{2}+\left(-1\right)^{2}+\left(-3\right)^{2}}\sqrt{1^{2}+\left(-5\right)^{2}+\left(-3\right)}} =5+5+93535=1935=\frac{5+5+9}{\sqrt{35}\sqrt{35}}=\frac{19}{35} θ=cos1(1935)\theta=cos^{-1}\left(\frac{19}{35}\right)