[ Angles between lines and planes ]  $\left[\begin{array}{l}\text { Equation of a plane } \\ {[\text { d }}\end{array}\right]$

Given points $A(1 ; 0 ; 1), B(-2 ; 2 ; 1), C(2 ; 0 ; 3)$ and $D(0 ; 4 ;-2)$. Find the angle between the line $A B$ and the plane $B C D$.
Let $\vec{n}=(a ; b ; c)-$ be a non-zero vector perpendicular to the plane $B C D$. We will find the coordinates of the vectors $\overrightarrow{B C}$, $\overrightarrow{B D}_{\text {and }} \overrightarrow{A B}$:

$$
\begin{aligned}
& \overrightarrow{B C}=(2-(-2) ; 0-2 ; 3-1)=(4 ;-2 ; 2) \\
& \overrightarrow{B D}=(0-(-2) ; 4-2 ;-2-1)=(2 ; 2 ;-3)
\end{aligned}
$$

$$
\overrightarrow{A B}=(-2-1 ; 2-0 ; 1-1)=(-3 ; 2 ; 0)
$$

Since $\vec{n} \cdot \overrightarrow{B C}=0$ and $\vec{n} \cdot \overrightarrow{B D}=0$, we have the system

$$
\left\{\begin{array}{l}
4 a-2 b+2 c=0 \\
2 a+2 b-3 c=0
\end{array}\right.
$$

Adding these equations term by term, we get $6 a-c=0$. Let $c=6$. Then $a=1, b=\frac{1}{2}(4 a+2 c)=2 a+c=8$. Thus, the vector $\vec{n}=(1 ; 8 ; 6)$ is perpendicular to the plane $B C D$. Let $\alpha$ be the angle between the line $A B$ and the plane $B C D$, and $\phi$ be the angle between the vectors $\vec{n}=(1 ; 8 ; 6)$ and $\overrightarrow{A B}=(-3 ; 2 ; 0)$. Then

$$
\sin \alpha=|\cos \phi|=|| \frac{\overline{\bar{n}} \cdot \overrightarrow{A B}}{\overrightarrow{\vec{q}}|\cdot| \overrightarrow{A B \mid}||} \left\lvert\,=\frac{\mid 1 \cdot(-3)+8 \cdot 2+6 \cdot 0)}{\sqrt{\left(1^{2}+8^{2}+6^{2}\right)\left(3^{2}+2^{2}+0^{2}\right)}}=\frac{13}{\sqrt{101 \cdot 13}}=\frac{\sqrt{13}}{\sqrt{101}} .\right.
$$

## Answer

$\arcsin \frac{\sqrt{13}}{\sqrt{101}}$