Example 1. Given the vertices of the tetrahedron \( A(2,3,1), B(4,1,-2) \), \( C(6,3,7), D(-5,-4,8) \). Find:

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-046.jpg?height=330&width=366&top_left_y=612&top_left_x=78)

Fig. 4.11

1) the length of the edge \( A B \)
2) the angle between the edges \( A B \) and \( A D \);
3) the angle between the edge \( A D \) and the plane \( A B C \)
4) the volume of the tetrahedron \( A B C D \)
5) the equation of the edge \( A B \)
6) the equation of the plane \( A B C \);
7) the equation of the height dropped from \( D \) to \( A B C \)
8) the projection \( O \) of point \( D \) onto the base \( A B C \)
9) the height \( D O \).
Solution. The condition of the problem is satisfied by the constructed drawing (Fig. 4.11).

1) $AB$ is calculated by the formula

$$
\begin{aligned}
d & =\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\
AB & =\sqrt{(4-2)^{2}+(1-3)^{2}+(-2-1)^{2}}=\sqrt{17}
\end{aligned}
$$

2) The angle $\varphi=(\widehat{\overrightarrow{AB}, \overrightarrow{AD}})$ is calculated by the formula (Ch. III, § 2)

$$
\cos \varphi=\frac{\overrightarrow{AB} \cdot \overrightarrow{AD}}{|\overrightarrow{AB}| \cdot|\overrightarrow{AD}|}
$$

a) $\overrightarrow{AB}=\{2,-2,-3\}, AB=\sqrt{17}, \overrightarrow{AD}=\{-7,-7,7\}, AD=7 \sqrt{3}$;

b) $\overrightarrow{AB} \cdot \overrightarrow{AD}=-14+14-21=-21$

c) $\cos \varphi=-\frac{21}{\sqrt{17} \cdot 7 \sqrt{3}} \approx-0.42, \quad \varphi \approx 180^{\circ}-65^{\circ}=115^{\circ}$.

3) The sine of the angle $\theta$ between the edge $AD$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-046.jpg?height=164&width=392&top_left_y=1740&top_left_x=78)

Fig. 4.12 and the plane $ABC$ is equal to the cosine of the angle between the edge $AD$ and the normal vector $N$ of the plane $ABC$ (Fig. 4.12). The vector $\vec{N}$ is collinear with the vector product $\overrightarrow{AB} \times \overrightarrow{AC}$ (Ch. III, § 3)
a) $\overrightarrow{AB}=\{2,-2,-3\}, \overrightarrow{AC}=2\{2,0,3\}$

$$
\overrightarrow{AB} \times \overrightarrow{AC}=2 \cdot\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & -2 & -3 \\
2 & 0 & 3
\end{array}\right|=4 \cdot\{-3,-6,2\}
$$

We take $\vec{N}=\{-3,-6,2\}$.

b) $\overrightarrow{AD}=7\{-1,-1,1\}, \quad \sin \theta=\frac{3+6+2}{\sqrt{3} \cdot 7} \approx 0.92 ; \quad \theta \approx 67^{\circ}$.

4) The volume of the tetrahedron $ABCD$ is equal to $1 / 6$ of the modulus of the mixed product of the vectors $(\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD}$ (Ch. III, § 4).

We have: $\overrightarrow{AB}=\{2,-2,-3\}, \quad \overrightarrow{AC}=2\{2,0,3\}, \quad \overrightarrow{AD}=-7\{1,1,-1\};$

$$
\left|\begin{array}{ccc}
2 & -2 & -3 \\
2 & 0 & 3 \\
1 & 1 & -1
\end{array}\right|=2 \cdot\left|\begin{array}{cc}
0 & 3 \\
1 & -1
\end{array}\right|+2 \cdot\left|\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right|-3\left|\begin{array}{cc}
2 & 0 \\
1 & 1
\end{array}\right|=-22
$$

The desired volume is: $V=\frac{2 \cdot 7 \cdot 22}{6}=\frac{154}{3}$.

5) The equations of the line passing through two points have the form (§ 2) $\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p} ;\{m, n, p\}$ - the direction vector $\vec{l}$ of the line. We take $\vec{l}=\overrightarrow{AB}=\{2,-2,-3\}$. Then

$$
(AB): \quad \frac{x-2}{2}=\frac{y-3}{-2}=\frac{z-1}{-3}
$$

6) The equation of the plane passing through three given points (§ 1):

$$
\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|=0
$$

We have

$$
(ABC):\left|\begin{array}{ccc}
x-2 & y-3 & z-1 \\
2 & -2 & -3 \\
2 & 0 & 3
\end{array}\right|=0
$$

or, after expanding the determinant: $3 x+6 y-2 z-22=0$.

7) As the direction vector $\vec{l}$ of the line $DO$, we can take the vector $\vec{l}=\vec{N}=\vec{N}(ABC)=\{3,6,-2\}(\S 2)$,

$$
(DO): \frac{x+5}{3}=\frac{y+4}{6}=\frac{z-8}{-2}
$$

or

$$
x=-5+3 t, y=-4+6 t, z=8-2 t
$$

8) The projection of $D$ on $ABC$ is the point $O$ (the intersection point of $DO$ with $ABC$). We substitute the values of $x, y$ and $z$, expressed in terms of the parameter $t$, into the equation of $ABC$. We find the value of $t$ and substitute it back into the expressions for $x, y$ and $z(\oint 3)$.

We get $3(3 t-5)+6(6 t-4)-2(8-2 t)-22=0,49 t=77, t=\frac{11}{7}$.

Then $x=-5+\frac{33}{7}=-\frac{2}{7}, y=\frac{38}{7}, z=\frac{34}{7}, O\left(-\frac{2}{7}, \frac{38}{7}, \frac{34}{7}\right)$

9) The height $DO$ can be calculated as the distance between $D$ and $O$, or as the distance from $D$ to the plane, or using the formula for the volume of the tetrahedron (§ 1).

In any case, we get $H=OD=\sqrt{\frac{33^{2}}{49}+\frac{66^{2}}{49}+\frac{22^{2}}{49}}=11$.