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 planes faces $ADC$ and $ABC$ is

• A. ${\tan }^{-1}(\frac{5}{2})$
• B. ${\cos }^{-1}(\frac{1}{5})$
• C. ${\text{cosec}}^{-1}(\frac{5}{2})$
• D. ${\cot }^{-1}(\frac{3}{2})$

Answer 1

The correct option is A ${\tan }^{-1}(\frac{5}{2})$

We have $A(3,-2,1);B(3,1,5);C(4,0,3)$ and $D(1,0,0)$
So,

$\overrightarrow{AB}=3\hat{j}+4\hat{k}$
$\overrightarrow{AC}=\hat{i}+2\hat{j}+2\hat{k}$
$\overrightarrow{AD}=-2\hat{i}+2\hat{j}-\hat{k}$
The normal to plane $ABC$ and $ADC$ are $\overrightarrow{AC}\times \overrightarrow{AB}$ and $\overrightarrow{AC}\times \overrightarrow{AD}$ respectively.

Thus, $(\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k})\times (3\overrightarrow{j}+4\overrightarrow{k})=2\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k}$
$(\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k})\times (-2\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k})=-6\overrightarrow{i}-3\overrightarrow{j}+6\overrightarrow{k}$

Angle between normals will be the same as angle between the planes.
So, $\cos \theta =\frac{18}{9\times \sqrt{29}}=\frac{2}{\sqrt{29}}$

So, $\tan \theta =\frac{5}{2}$