Question

Position vectors of the four angular points of a tetrahedron $ABCD$ are $A(3,-2,1);$ $B(3,1,5);$ $C(4,0,3)$ and $D(1,0,0).$ Acute angle between the plane faces $ADC$ and $ABC$ is

• A. ${\tan }^{-1}\left(\frac{15}{2}\right)$
• B. ${\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)$
• C. ${\cos }^{-1}\left(\frac{4}{\sqrt{29}}\right)$
• D. ${\cot }^{-1}\left(\frac{3}{2}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)$

${\overrightarrow{v}}_1=\overrightarrow{b}-\overrightarrow{a}=(3-3)\hat{i}+(1+2)\hat{j}+(5-1)\hat{k}=0\hat{i}+3\hat{j}+4\hat{k}$ ${\overrightarrow{v}}_2=\overrightarrow{c}-\overrightarrow{a}=(4-3)\hat{i}+(0+2)\hat{j}+(3-1)\hat{k}=\hat{i}+2\hat{j}+2\hat{k}$ ${\overrightarrow{v}}_3=\overrightarrow{d}-\overrightarrow{a}=(1-3)\hat{i}+(0+2)\hat{j}+(0-1)\hat{k}=-2\hat{i}+2\hat{j}-\hat{k}$


${\overrightarrow{n}}_1={\overrightarrow{v}}_1\times {\overrightarrow{v}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 3& 4\\ 1& 2& 2\end{array}\right|$
$\Rightarrow {\overrightarrow{n}}_1=(-2)\hat{i}-(0-4)\hat{j}+(0-3)\hat{k}$
$\Rightarrow {\overrightarrow{n}}_1=-2\hat{i}+4\hat{j}-3\hat{k}$

${\overrightarrow{n}}_2={\overrightarrow{v}}_2\times {\overrightarrow{v}}_3=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ -2& 2& -1\end{array}\right|$
$\Rightarrow {\overrightarrow{n}}_2=(-2-4)\hat{i}-(-1+4)\hat{j}+(2+4)\hat{k}$
$\Rightarrow {\overrightarrow{n}}_2=-6\hat{i}-3\hat{j}+6\hat{k}=-3(2\hat{i}+\hat{j}-2\hat{k})$

$\cos \theta =\frac{{\overrightarrow{n}}_1\cdot {\overrightarrow{n}}_2}{\left|{\overrightarrow{n}}_1\right|\left|{\overrightarrow{n}}_2\right|}=\frac{-4+4+6}{\sqrt{29}\cdot 3}=\frac{2}{\sqrt{29}}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{2}{\sqrt{29}}\right)$