2.14. Given the vertices of a tetrahedron (see Fig. 2.11) \( A(2,-1,2) \), \( B(5,5,5) \), \( C(3,2,0) \), \( D(4,1,4) \). Calculate: 1) the volume of the tetrahedron; 2) its height dropped from vertex \( D \); 3) the angle between edge \( AD \) and its face \( ABC \).

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-103.jpg?height=337&width=457&top_left_y=130&top_left_x=147)

Fig. 2.11
Solution. 1. The volume of the tetrahedron $V$ is one-sixth of the volume of the parallelepiped $A B E C D B_{1} E_{1} C_{1}$. The volume of the parallelepiped, by the geometric property of the mixed product, is equal to the modulus of the mixed product, for example, of the vectors $\overline{A B}, \overrightarrow{A C}, \overrightarrow{A D}$.

Thus, the desired volume is determined by the expression

$$
V=\frac{1}{6}|\overrightarrow{A B} \times \overrightarrow{A C} \cdot \overrightarrow{A D}|
$$

To compute the mixed product, we will find the coordinates of the vectors $A B, A C, A D$. We have

$$
\overrightarrow{A B}=\{3,6,3\}, \overrightarrow{A C}=\{1,3,-2\}, \overrightarrow{A D}=\{2,2,2\}
$$

Now, computing the mixed product using formula (2.40), we get:

$$
\left.V=\frac{1}{6}\left|\begin{array}{ccc}
3 & 6 & 3 \\
1 & 3 & -2 \\
2 & 2 & 2
\end{array} \|=\frac{1}{6} \cdot\right| 18-24+6-18+12-12 \right\rvert\,=3
$$

2. The volume of the tetrahedron is also determined by the formula

$$
V=\frac{1}{3} S h
$$

where $S$ is the area of the base (in this case, it is obviously the area of the triangle $A B C$); $h$ is the height of the pyramid, which needs to be found. The area $S$ of the triangle $A B C$ is found, guided by the geometric meaning of the modulus of the vector product, according to which

$$
\left.S=\frac{1}{2} \right\rvert\, \overrightarrow{A B} \times \overrightarrow{A C} \vec{A}
$$

We find $\overrightarrow{A B} \times \overrightarrow{A C}$ using formula (2.37):

$$
\begin{aligned}
& \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & 6 & 3 \\
1 & 3 & -2
\end{array}\right|=\vec{i}\left|\begin{array}{cc}
6 & 3 \\
3 & -2
\end{array}\right|-\vec{j}\left|\begin{array}{cc}
3 & 3 \\
1 & -2
\end{array}\right|+\vec{k}\left|\begin{array}{cc}
3 & 6 \\
1 & 3
\end{array}\right|= \\
& =\vec{i}(-12-9)-\vec{j}(-6-3)+\vec{k}(9-6)=-21 \vec{i}+9 \vec{j}+3 \vec{k}
\end{aligned}
$$

Then, determining the modulus of the vector product $|\overrightarrow{A B} \times \overrightarrow{A C}|$, we find

$$
|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{441+81+9}=\sqrt{531}
$$

Since $h=3 V / S$, and $S=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$, then

$$
h=\frac{6 V}{|\overline{A B} \times \bar{A} \bar{C}|}
$$

Substituting the values of $V$ and $|\overrightarrow{A B} \times \overrightarrow{A C}|$ obtained above, we find that

$$
h=\frac{18}{\sqrt{531}}=\frac{6}{\sqrt{59}}
$$

3. Let's find the angle between the edge $A B$ and the face $A B C$. This is the angle $\varphi$ between the line $A B$ and its orthogonal projection onto the plane to which the face $A B C$ belongs (Fig. 2.11). To find the angle $\varphi$, it is sufficient to find the cosine of the angle $\psi$ between the vectors $\overrightarrow{A D}$ and $\overrightarrow{A B} \times \overrightarrow{A C}$ (Fig. 2.11). Then, since $\psi=\varphi \pm \pi / 2$ (depending on whether the angle $\psi$ is acute or obtuse), $\cos \psi=\cos (\varphi \pm \pi / 2)=\mp \sin \varphi$. From this, it follows that $\sin \varphi=\mp \cos \psi$. The sign "−" corresponds to the case when $\cos \psi > 0$. In both cases, $\sin \varphi > 0$. Both cases are easily accounted for by the relation $\sin \varphi=|\cos \psi|$. Considering that

$$
\cos \psi=\frac{|\overrightarrow{A D} \times \overrightarrow{A B} \cdot \overrightarrow{A C}|}{|\overrightarrow{A D}| \cdot|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}, \text { we get } \sin \varphi=\left|\frac{|\overrightarrow{A D} \times \overrightarrow{A B} \cdot \overrightarrow{A C}|}{|| \overrightarrow{A D}|\cdot| \overrightarrow{A B}|\cdot| \overrightarrow{A C} \mid}\right|
$$

In the numerator of the obtained expression, the modulus of the mixed product of the vectors $\overrightarrow{A D}, \overrightarrow{A B}$, and $\overrightarrow{A C}$ is recorded, which can be computed using the method described above, knowing the coordinates of these vectors. However, there is no need for this, as we have already found the vector $\overrightarrow{A B} \times \overrightarrow{A C}$ when answering the second question of the problem. Therefore, the scalar product $\overrightarrow{A D} \cdot \vec{A} \vec{B} \times \overrightarrow{A C}$ is easier to find directly, through the coordinates of the vectors being multiplied. Determining the modulus of the vector $\overrightarrow{A D}$, we have ( $|\overrightarrow{A B} \times \overline{A C}|$ was found in the second part of the problem):

$$
\sin \varphi=\left|\frac{2 \cdot(-21)+2 \cdot 9+2 \cdot 3}{2 \sqrt{3} \cdot \sqrt{531}}\right|=\frac{3}{\sqrt{177}}
$$

from which $\varphi=\arcsin \frac{3}{\sqrt{177}} \approx 13.03^{\circ}$.