![](https://cdn.mathpix.com/cropped/2024_05_06_e53884ac984930b8f3cbg-42.jpg?height=123&width=2056&top_left_y=2580&top_left_x=1)

In space, the points \( A(-1 ; 2 ; 0), B(5 ; 2 ;-1), C(2 ;-1 ; 4) \) and \( D(-2 ; 2 ;-1) \) are given. Find:

a) the distance from the vertex \( D \) of the tetrahedron \( A B C D \) to the point of intersection of the medians of the base \( A B C \);

b) the equation of the plane \( A B C \);

c) the height of the tetrahedron drawn from the vertex \( D \);

d) the angle between the lines \( B D \) and \( A C \);

e) the angle between the faces \( A B C \) and \( A C D \);

f) the distance between the lines \( B D \) and \( A C \).
a) The coordinates of point $M$, the intersection of the medians of triangle $ABC$, are equal to the arithmetic mean of the coordinates of the vertices of the triangle, i.e., $x_{0}=\frac{-1+5+2}{3}=2, y_{0}=\frac{2+2-1}{3}=1, z_{0}=\frac{0-1+t}{3}=1$. Therefore,

$$
D M=\sqrt{(2-(-2))^{2}+(1-2)^{2}+(1-(-1))^{2}}=\sqrt{16+1+4}=\sqrt{21}.
$$

b) Let's find the coordinates of the vectors:

$$
\overrightarrow{A B}=(5-(-1); 2-2; -1-0)=(6; 0; -1), \overrightarrow{A C}=(2-(-1); -1-2; 4-0)=(3; -3; 4)
$$

Let $\vec{n}(a; b; c)$ be a non-zero vector perpendicular to the plane $ABC$ (the normal vector). Then $\overrightarrow{A B} \cdot \vec{n}=0$ and $\overrightarrow{A C} \cdot \vec{n}=0$, or

$$
\left\{\begin{array}{l}
6 a - c = 0 \\
3 a - 3 b + 4 c = 0
\end{array}\right.
$$

Let $a=1$. Then $c=6$ and $b=\frac{1}{3}(3 a + 4 c)=9$. The equation of the plane through point $A(-1; 2; 0)$ and the normal vector $\vec{n}(a; b; c)$ is

$$
a(x+1) + b(y-2) + c(z-0) = 0 \Leftrightarrow 1(x+1) + 9(y-2) + 6(z-0) = 0 \Leftrightarrow x + 9 y + 6 z - 17 = 0
$$

c) The height $D H$ of the pyramid $A B C D$ can be found using the formula for the distance from a point to a plane:

$$
D H=\frac{\left|a x_{0} + b y_{0} + c z_{0} + d\right|}{\sqrt{a^{2} + b^{2} + c^{2}}}=\frac{|1 \cdot(-2) + 9 \cdot 2 + 6 \cdot(-1) - 17|}{\sqrt{1^{2} + 9^{2} + 6^{2}}}=\frac{7}{\sqrt{118}}
$$

d) The angle between the lines $B D$ and $A C$ is equal to either the angle between the vectors $\overrightarrow{B D}=(-2-5; 2-2; -1-(-1))=(-7; 0; 0)$ and $\overrightarrow{A C}=(3; -3; 4)$, or it complements this angle to $180^{\circ}$, so

$$
\cos \alpha=\frac{|\overrightarrow{B D} \cdot \overrightarrow{A C}|}{|\overrightarrow{B D}| \cdot|\overrightarrow{A C}|}=\frac{|-7 \cdot 3 + 0 \cdot(-3) + 0 \cdot 4|}{\sqrt{(-7)^{2} + 0^{2} + 0^{2}} \cdot \sqrt{3^{2} + (-3)^{2} + 4^{2}}}=\frac{21}{7 \sqrt{34}}=\frac{3}{\sqrt{34}}
$$

e) Let $\overrightarrow{n_{1}}=(a 1; b 1; c 1)$ be the normal vector of the plane $A C D$. Then $\overrightarrow{A D} \cdot \overrightarrow{n_{1}}=0$ and $\overrightarrow{A C} \cdot \overrightarrow{n_{1}}=0$, where $\overrightarrow{A D}=(-2-(-1); 2-2; -1-0)=(-1; 0; -1)$ and $\overrightarrow{A C}=(3; -3; 4)$. From the system

$$
\left\{\begin{array}{l}
-a_{1} - c_{1} = 0 \\
3 a_{1} - 3 b_{1} + 4 c_{1} = 0
\end{array}\right.
$$

assuming $a 1=3$ and $c 1=-3$, we find that $b 1=-1$. Therefore, the vector $\overrightarrow{n_{1}}=(3; -1; -3)$ can be taken as the normal vector of the plane $A C D$.

The angle between the planes $A B C$ and $A C D$ is equal to either the angle between their normal vectors, or it complements this angle to $180^{\circ}$. Therefore,

$$
\begin{aligned}
& \cos \beta=\frac{\left|\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}\right|}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{a_{1} a + b_{1} b + c_{1} c}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \cdot \sqrt{a^{2} + b^{2} + c^{2}}}= \\
& =\frac{|3 \cdot 1 + (-1) \cdot 9 + (-3) \cdot 6|}{\sqrt{9 + 1 + 9} \cdot \sqrt{1 + 81 + 36}}=\frac{24}{\sqrt{19} \cdot \sqrt{118}}=\frac{24}{\sqrt{2242}}
\end{aligned}
$$

f) Through the line $A C$, we draw a plane parallel to the line $B D$. The equation of this plane can be found using point $A(-1; 2; 0)$ and the vector $\overrightarrow{n_{2}}=(a 2; b 2; c 2)$, which is perpendicular to the vectors $\overrightarrow{A C}=(3; -3; 4)$ and $\overrightarrow{B D}=(-7; 0; 0)$. The coordinates of this vector can be found from the system

$$
\begin{cases}3 a_{2} - 3 b_{2} + 4 c_{2} & =0 \\ -7 a_{2} & =0\end{cases}
$$

As such a vector, we can take the vector $\overrightarrow{n_{2}}=(0; 4; 3)$. The equation of the desired plane is $0(x+1) + 4(y-2) + 3(z-0) = 0$, or $4 y + 3 z - 8 = 0$.

The distance between the lines $B D$ and $A C$ is equal to the distance from any point on the line $B D$ (for example, from point $D(-2; 2; -1)$), parallel to $A C$, to the drawn plane, i.e.,

$$
d=\frac{\left|a_{2} \cdot(-2) + b_{2} \cdot 2 + c_{2} \cdot(-1) + d_{2}\right|}{\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}=\frac{|0 \cdot(-2) + 4 \cdot 2 + 3 \cdot(-1) - 8|}{\sqrt{0^{2} + 4^{2} + 3^{2}}}=\frac{3}{5}.
$$

## Answer

m) $\sqrt{21}$; $\left.{ }^{\perp}\right) x + 9 y + 6 z - 17 = 0$; y) $\frac{7}{\sqrt{118}}$; d) $\arccos \frac{3}{\sqrt{34}}$; e) $\arccos \frac{24}{\sqrt{2242}}$; f) $\frac{3}{5}$.