$\mathrm{A}(1,2,1), \mathrm{B}(2,3,2), \mathrm{C}(3,1,3)$ and $\mathrm{D}(2,1.3)$ are the vertices of a…

$\mathrm{A}(1,2,1), \mathrm{B}(2,3,2), \mathrm{C}(3,1,3)$ and $\mathrm{D}(2,1.3)$ are the vertices of a tetrahedron. If $\theta$ is the angle between the faces $A B C$ and $A B D$ then $\cos \theta=$
  1. $\frac{5}{\sqrt{14}}$
  2. $\frac{15}{8 \sqrt{7}}$
  3. $\frac{3}{\sqrt{14}}$
  4. $\frac{5}{2 \sqrt{7}}$

Solution

(d) $\mathrm{A}(1,2,1), \mathrm{B}(2,3,2), \mathrm{C}(3,1,3)$ and $\mathrm{D}(2,1,3)$
Vector perpendicular to face ABC $\begin{aligned} & =\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} \\ & =\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 2 \end{array}\right|=3 \hat{i}-3 \hat{k} \end{aligned}$
Vector perpendicular to face ABD $=\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 2 \end{array}\right|=3 \hat{i}-\hat{j}-2 \hat{k}$
Angle between $A B C$ and $A B D$ $\cos \theta=\frac{9+6}{\sqrt{18} \sqrt{14}}=\frac{15}{3 \sqrt{28}}=\frac{5}{2 \sqrt{7}}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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