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A tetrahedron has vertices $\mathrm{P}(1,2,1), \mathrm{Q}(2,1,3), \mathrm{R}(-1,1,2)$ and $\mathrm{O}(0,0,0)$. Then the angle between the faces OPQ and PQR is
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Verified Answer
The correct answer is:
$\cos ^{-1}\left(\frac{19}{35}\right)$
Equation of OPQ is $\left|\begin{array}{lll}x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right|=0$
$\Rightarrow 5 x-y-3 z=0$
equation of PQR is $\left|\begin{array}{ccc}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{array}\right|=0$
$\Rightarrow x-5 y-3 z+12=0$
Angle between the planes
$\begin{aligned} & \theta=\cos ^{-1}\left(\frac{5 \times 1+(-1) \times(-5)+(-3) \times(-3)}{\sqrt{5^2+(-1)^2+(-3)^2} \cdot \sqrt{1^2+(-5)^2+(-3)^2}}\right) \\ & \Rightarrow \theta=\cos ^{-1} \frac{19}{\sqrt{35} \cdot \sqrt{35}}=\cos ^{-1}\left(\frac{19}{35}\right)\end{aligned}$
$\Rightarrow 5 x-y-3 z=0$
equation of PQR is $\left|\begin{array}{ccc}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{array}\right|=0$
$\Rightarrow x-5 y-3 z+12=0$
Angle between the planes
$\begin{aligned} & \theta=\cos ^{-1}\left(\frac{5 \times 1+(-1) \times(-5)+(-3) \times(-3)}{\sqrt{5^2+(-1)^2+(-3)^2} \cdot \sqrt{1^2+(-5)^2+(-3)^2}}\right) \\ & \Rightarrow \theta=\cos ^{-1} \frac{19}{\sqrt{35} \cdot \sqrt{35}}=\cos ^{-1}\left(\frac{19}{35}\right)\end{aligned}$
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