Problem 11. Find the distance $A_{1} H$ from the point $A_{1}(1 ; 6 ;-7)$ to the plane $A B_{1} K$, defined by the points $A(-1 ; 2 ;-3), B_{1}(5 ; 8 ;-3)$ and $K(-1 ; 8 ;-6)$ (Fig. 2).
The first solution. Let $P(0 ; y ; z)$ be such a point that $\overline{A P} \cdot \overline{A B_{1}}=0$ and $\overline{A P} \cdot \overline{A K}=0$. Since $\overline{A P}=(1 ; y-2$; $z-3), \overline{A B_{1}}=(6 ; 6 ; 0), \overline{A K}=(0 ; 6 ;-3)$, the last two equations take the form:

$$
6+6(y-2)=0 ; 6(y-2)-3(z+3)=0
$$

Solving this system of equations, we get $y=1, z=-5$. Thus, $\overline{A P}=(1 ;-1 ;-2)$.

Let $\overline{A M}=\overline{H A}_{1}$. Then $\overline{M A}_{1} \cdot \overline{A M}=0$.

If $\overline{A M}=n \overline{A P}=(n,-n,-2 n)$, then $\overline{M A}_{1}=\overline{A A}_{1}-$ $-\overline{A M}=(2-n, \quad 4+n, \quad-4+2 n) \quad$ and $\quad(2-n) n+$ $+(4+n) n-(-4+2 n) 2 n=0$. From this, $n=1, \overline{H A_{1}}=$ $=(1,-1,-2)$ and $H A_{1}=\sqrt{6}$.

The second solution. Let $\overline{A_{1} B^{\prime}}=\overline{A B}_{1}$ and segment $M T$ be the common perpendicular to lines $A_{1} B^{\prime}$ and $A K$ (Fig. 3). Then $A_{1} H=M T$.

Let $\overline{A_{1} M}=m \overline{A_{1} B_{1}^{\prime}}=m \overline{A B_{1}}=(6 m ; 6 m ; 0)$ and $\overline{A T}=$ $=k \overline{A K}=(0 ; 6 k ;-3 k)$. Obviously,

$$
\begin{aligned}
& \overline{A_{1} B_{1}^{\prime}} \cdot \overline{T M}=0, \\
& \overline{A K} \cdot \overline{T M}=0 .
\end{aligned}
$$

We find $\overline{T M}=\overline{A A_{1}}+\overline{A_{1} M}-\overline{A T}=(2+6 m ; 4+6 m-$ $-6 k ;-4+3 k$ ). After this, equations (1) and (2) transform into:

$$
\begin{gathered}
6(2+6 m)+6(4+6 m-6 k)=0 \\
6(4+6 m-6 k)-3(-4+3 k)=0
\end{gathered}
$$

Solving this system of equations, we get:

$m=-1: 6, k=2: 3, \overline{T M}=(1 ;-1 ;-2)$ and $A_{1} H=\sqrt{6}$.

The third solution. Let $\overline{A H}=n \overline{A B_{1}}+k \overline{A K}$; $\overline{H A_{1}} \cdot \overline{A B_{1}}=0 ; \overline{H A_{1}} \cdot \overline{A K}=0$ (Fig. 4). Then

$$
\begin{aligned}
& \left(\overline{A A_{1}}-\overline{A H}\right) \overline{A B_{1}}=0, \\
& \left(\overline{A A_{1}}-\overline{\overline{A H})} \overline{A K}=0 .\right.
\end{aligned}
$$

Obviously, $\overline{H A_{1}}=\overline{A A_{1}}-\overline{A H}=\overline{A A_{1}}-n \overline{A B_{1}}-k \overline{A K}=$ $=(2-6 n, 4-6 n-6 k,-4+3 k) ; \quad \overline{A B_{1}}=(6,6,0)$; $\overline{A K}=(0,6,-3)$.

After this, equations (1) and (2) take the form:

$$
\begin{gathered}
6(2-6 n)+6(4-6 k)=0 \\
6(4-6 n-6 k)-3(-4+3 k)=0
\end{gathered}
$$

Fig. 4
![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-127.jpg?height=358&width=806&top_left_y=252&top_left_x=728)

Fig 5

Solving this system of equations, we get:

$$
k=2: 3, n=1: 6, \overline{H A_{1}}=(1,-1,-2) \text { and } H A_{1}=\sqrt{6}
$$

The fourth solution. Let the line $A_{1} H$ intersect the plane $X O Y$ at point $P(x, y, 0)$. Then

$$
\begin{aligned}
& \overline{A_{1} P} \cdot \overline{A K}=0 \\
& \overline{A_{1} P} \cdot \overline{A B}=0
\end{aligned}
$$

Since $\overline{A_{1} P}=(x-1, y-6,7) ; \overline{A K}=(0,6,-3)$; $\overline{A B}_{1}=(6,6,0)$, equations (1) and (2) transform into:

$$
6(y-6)-21=0,6(x-1)+6(y-6)=0
$$

Solving this system of equations, we find $P(-2,5 ; 9,5$; $0)$. From the right triangle $A H A_{1}$ we get (Fig. 5 ) :

$$
\begin{aligned}
& A_{1} H=\left|A_{1} A \cos <A A_{1} H\right|=\left|A_{1} A \frac{\overline{A_{1} A} \cdot \overline{A_{1} P}}{\left|\overline{A_{1} A \mid} \cdot\right| \overline{A_{1} H \mid}}\right|= \\
& =\frac{\left|\overline{A_{1} A} \cdot \overline{A_{1} P \mid}\right|}{\mid \overline{A_{1} P \mid}}=\left|\frac{(-2 ;-4 ; 4) \cdot(-3,5 ; 3,5 ; 7)}{\sqrt{3,5^{2}+3,5^{2}+7^{2}}}\right|=\sqrt{6}
\end{aligned}
$$

The fifth solution. Let $\overline{H M}=\overline{A_{1} H}$ and $M(x, y, z)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-127.jpg?height=69&width=1353&top