Mathematics | VITEEE 2016 | Discussion Page

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

$120^{0}$

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

$30^{0}$

$90^{0}$

Option A

$\overrightarrow{AO}= \hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{AC}= -2\hat{i}-\hat{j}+\hat{k}$

Angle between faces OAB and ABC = Angle between $\overrightarrow{AO}$ and $\overrightarrow{AC}$ .

If Q be the angle between $\overrightarrow{AO}$ and $\overrightarrow{AC}$ , then

$\cos \theta= \frac{\overrightarrow{AO}.\overrightarrow{AC}}{|\overrightarrow{AO}||\overrightarrow{AC}|}$

$=\frac{1\times(-2)+2\times(-1)+1\times1}{\sqrt{1+4+1}\sqrt{4+1+1}}$
$\frac{-3}{6}=-\frac{1}{2}=\cos 120^{0}$

$\theta=120^{0}$

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