Question

The sine of angle formed by the lateral face ADC and plane of the base ABC of the tetrahedron ABCD where $a\equiv (3,-2,1);B\equiv (3,1,5);C\equiv (4,0,3)andD\equiv (1,0,0)is$

• A. $\frac{2}{\sqrt{29}}$
• B. $\frac{5}{\sqrt{29}}$
• C. $\frac{3\sqrt{3}}{\sqrt{29}}$
• D. $\frac{-2}{\sqrt{29}}$

Answer 1

Answer: (B)

$\frac{5}{\sqrt{29}}$

$\overrightarrow{AD}=-2\hat{i}+2\hat{j}-\hat{k},\overrightarrow{Ac}=\hat{i}+2\hat{j}+2\hat{k},\overrightarrow{AB}=3\hat{j}+4\hat{k}\overrightarrow{n_1}=\overrightarrow{AD}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 2& -1\\ 1& 2& 2\end{array}\right|=6\hat{i}+3\hat{j}-6\hat{k}=3(2\hat{i}+\hat{j}-2\hat{k})\overrightarrow{n_2}=\overrightarrow{AC}\times \overrightarrow{AB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 0& 3& 4\end{array}\right|=2\hat{i}-4\hat{j}+3\hat{k}|\overrightarrow{n_1}\times \overrightarrow{n_2}|=3\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -2\\ 2& -4& 3\end{array}\right|=3(5\hat{i}-10\hat{j}-10\hat{k})\sin \theta =\frac{5}{\sqrt{29}}(\because \sin \theta =\frac{|\overrightarrow{n_1}\times \overrightarrow{n_2}|}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|})$