Mathematics › Three Dimensional Geometry › Basics of point in 3 dimension
A tetrahedron has vertices at $\mathrm{P}(2,1,3)$, $\mathrm{Q}(-1,1,2), \mathrm{R}(1,2,1)$ and $\mathrm{O}(0…
A tetrahedron has vertices at $\mathrm{P}(2,1,3)$, $\mathrm{Q}(-1,1,2), \mathrm{R}(1,2,1)$ and $\mathrm{O}(0,0,0)$, then angle between the faces OPQ and PQR is
$\cos ^{-1}\left(\frac{5}{7 \sqrt{59}}\right)$ $\cos ^{-1}\left(\frac{\sqrt{25}}{\sqrt{59} \cdot \sqrt{35}}\right)$ $\cos ^{-1}\left(\frac{5}{413}\right)$ $\quad \cos ^{-1}\left(\frac{25}{\sqrt{59} \sqrt{35}}\right)$
Solution
$\begin{aligned} O P \times O Q & =\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & 3 \\ -1 & 1 & 2\end{array}\right| \\ & =-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\end{aligned}$
$\begin{aligned} P Q \times P R & =\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -3 & 0 & -1 \\ -1 & 1 & -2\end{array}\right| \\ & =\hat{\mathrm{i}}-5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\end{aligned}$
Angle between the faces OPQ and PQR is
$\cos \theta=\frac{(-\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})}{\sqrt{(-1)^2+(-7)^2+3^2} \sqrt{1^2+(-5)^2+(-3)^2}}$
$=\frac{-1+35-9}{\sqrt{59} \sqrt{35}}$
$\begin{aligned} & \Rightarrow \cos \theta=\frac{25}{\sqrt{59} \sqrt{35}} \\ & \Rightarrow \theta=\cos ^{-1}\left(\frac{25}{\sqrt{59} \sqrt{35}}\right)\end{aligned}$
Asked in: MHT CET 2023 (13 May Shift 2)
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