Question

# The tetrahedron has vertices $$0\left ( 0,0,0 \right ),A\left ( 1,2,1 \right ),B\left ( 2,1,3 \right )$$ and $$C\left ( -1,1,2 \right )$$, then  the angle between the faces $$OAB$$ and $$ABC$$ will be

A
cos11731
B
300
C
900
D
cos11935

Solution

## The correct option is D $$\displaystyle \cos ^{-1}\frac{19}{35}$$Concept using the angle between the  phases is equal to their normals.$$\therefore$$ vector $$\perp$$ to the face $$OAB$$ is $$\overline{OA}\times \overline{OB}=5\hat{i}-\hat{j}-3\hat{k}$$and vector $$\perp$$ to the face $$ABC$$ is $$\overline{AB}\times \overline{AC}=\hat{i}-5\hat{j}-3\hat{k}$$$$\therefore$$ Let $$\theta$$ be the angle between the faces $$OAB$$ and $$ABC$$ $$\displaystyle \therefore \cos \theta =\frac{\left ( 5\hat{i}-\hat{j}-3\hat{k} \right )\left ( \hat{i}-5\hat{j}-3\hat{k} \right )}{\left | 5\hat{i}-\hat{j}-3\hat{k} \right |\left | \hat{i}-5\hat{j}-3\hat{k} \right |}$$$$\displaystyle \therefore \cos \theta =\frac{19}{35}$$Mathematics

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