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MHT CET · Maths · Three Dimensional Geometry

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

  1. A \(\cos ^{-1}\left(\frac{17}{35}\right)\)
  2. B \(\cos ^{-1}\left(\frac{19}{31}\right)\)
  3. C \(\cos ^{-1}\left(\frac{19}{35}\right)\)
  4. D \(\cos ^{-1}\left(\frac{17}{31}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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
\(\theta=\cos ^{-1}\) \(\left(\frac{5 \times 1+(-1) \times(-5)+(-3) \times(-3)}{\sqrt{5^2+(-1)^2+(-3)^2} \cdot \sqrt{1^2+(-5)^2+(-3)^2}}\right) \)
\( \Rightarrow \theta=\cos ^{-1} \frac{19}{\sqrt{35} \cdot \sqrt{35}}=\cos ^{-1}\left(\frac{19}{35}\right)\)