Point Euclidean spaces

$\rho (A, B) = \rho (B, A)$ for any points $A$ and $B$

$\rho (A, B) \leqslant \rho (A, C) + \rho (B, C)$ for any points $A, B, C$

for any point $C$ such that $\overrightarrow {AC} = \lambda \overrightarrow {AB}$, the equality $\rho (A, C) = \left|\lambda \right| \rho (A, B)$ holds.

$A(- 1, 0, - 1, 2), \quad B(0, 2, 0, 3), \quad C(2, 1, 1, 2)$

$A(1, 2, 2, - 1), B(3, 0, 3, - 1), \quad C(2, 1, 1, 0)$

$A(0, 1, - 1, 2, - 1), \quad B(4, 1, 1, 2, 3), C(3, 4, 2, 5, - 1)$.

$A_{0}(4, - 2, - 1, - 1), A_{1}(1, 1, 2, 2), A_{2}(3, 1, 0, 0)$, $A_{3}(0, 2, 3, - 1), \quad A_{4}(1, - 5, - 4, 2)$

$A_{0}(3, 3, 1, - 1), A_{1}(1, 3, 3, 1), A_{2}(0, 3, 4, - 1)$, $A_{3}(2, 1, 2, 3), \quad A_{4}(2, 3, 0, 1)$.

the distance between the point $A$ and its orthogonal projection onto $\mathrm{m}$

the length of the orthogonal component of the vector $\overrightarrow {AB}(B$ is an arbitrary point of $\mathrm{m}$) with respect to the directing subspace of the plane $\mathrm{m}$.

the vector with coordinate column $\left(a_{1}, \ldots , a_{n}\right)^{T}$ is orthogonal to m

the distance from the point $A\left(y_{1}, \ldots , y_{n}\right)$ to m equals

$A(9, 2, - 3, 1), \quad \mathrm{m}: 3x_{1} + x_{2} - x_{3} - 5x_{4} + 3 = 0$

$A(1, - 3, 0, - 2, 4), \mathrm{m}: 2x_{1} - 5x_{2} + x_{3} + 3x_{4} + 5x_{5} - 7 = 0$.

$\mathrm{m}: 5x_{1} + 2x_{2} - 4x_{3} + 2x_{4} = 3, \quad d = 2$

$\mathrm{m}: x_{1} - 4x_{2} + 2x_{3} + 2x_{4} = 4, \quad d = 5$

$\mathrm{m}: 2x_{1} - x_{2} - x_{3} + x_{4} + 3x_{5} = - 5, \quad d = 3$.

$A(7, - 1, 6, 1), \quad \mathrm{m}: 3x_{1} - x_{2} + 2x_{3} + x_{4} = 5$

$A(1, 2, 8, - 2)$, m: $2x_{1} - 2x_{3} + x_{4} = 11$

$A(3, 0, - 1, 2, 6)$, m: $5x_{1} + 3x_{2} - 2x_{3} - x_{4} + 4x_{5} = - 16$.

$A(- 3, 0, 1, 3), \quad B(5, 2, 2, 3), \quad \mathrm{m}: 2x_{1} + x_{2} - x_{4} = 3$

$A(3, 3, - 8, - 3, 4), \quad B(3, 2, - 1, - 2, 2), \mathrm{m}: \quad x_{1} + x_{2} - - 2x_{3} + x_{4} + x_{5} = 5$.

$A(5, 5, 3, 3), \quad \mathrm{m}: 2x_{1} + 3x_{2} + x_{3} + 2x_{4} + 2 = 0$

$A(3, 5, - 3, 5), \mathrm{m}: x_{1} - 3x_{2} + 4x_{3} - 5x_{4} - 2 = 0$

$A(3, 6, 3, 8, 1), \mathrm{m}: x_{1} - x_{2} - 2x_{4} + 2x_{5} - 3 = 0$.

$A(1, - 5, 2, 0)$; l: $x_{1} = 4 + t, \quad x_{2} = 3 + 2t, \quad x_{3} = - 3 - t$, $x_{4} = 7 + 3t$

$A(- 2, 1, 4, 2)$; l: $x_{1} = - 3 + 2t, \quad x_{2} = 3 - t, \quad x_{3} = - 1 + t$, $x_{4} = - 3 + t$

$A(2, 4, 3, - 1, 1)$; l: $x_{1} = 2 - 2t, \quad x_{2} = - 1 + 3t, \quad x_{3} = - 1 + 2t, \quad x_{4} = 2 + t, \quad x_{5} = - t$.

$A(1, - 3, - 2, 4)$; l: $x_{1} = 4 + 3t, \quad x_{2} = 2 + t, \quad x_{3} = 3 + t$, $x_{4} = - 1 - t$

$A(1, - 3, - 1, 3)$; l: $x_{1} = 2 + t, \quad x_{2} = 1 - 2t, \quad x_{3} = - 1 + 2t, \quad x_{4} = t$

$A(4, 0, 1, 1, 1)$; l: $x_{1} = t, \quad x_{2} = 3 - 2t, \quad x_{3} = - 2 + t$, $x_{4} = - 3 + 2t, \quad x_{5} = t$.

$A(4, 1, - 1, - 1)$, $\mathrm{l}$ is the line of problem $34. 15, 1)$

$A(2, 5, - 3, - 2)$, $\mathrm{l}$ is the line of problem $34. 15, 2)$.

$\mathbf{a} = \left[\begin{smallmatrix} 0\\ 1\\ 0\\ 1\end{smallmatrix}\right], \quad \mathrm{~ m}: 3x_{1} - x_{2} + x_{3} - 5x_{4} = 2$

$\mathbf{a} = \left[\begin{smallmatrix} 1\\ - 1\\ 1\\ 1\end{smallmatrix}\right], \quad \mathrm{~ m}: 3x_{1} - x_{2} + 2x_{3} + 2x_{4} = 5$

$\mathbf{a} = \left[\begin{smallmatrix} 1\\ - 3\\ 2\\ - 1\\ - 1\end{smallmatrix}\right], \quad \mathrm{~ m}: x_{1} + x_{2} - 2x_{3} + 3x_{4} - x_{5} = 1$.

$l_{1}: x_{1} = 4 + t, x_{2} = - 2t, x_{3} = 1 - t, x_{4} = 2; l_{2}: x_{1} = 3$, $x_{2} = t, \quad x_{3} = 5 + t, \quad x_{4} = - 1$

$\mathrm{l}_{1}: x_{1} = 1 + t, \quad x_{2} = 2 + t, \quad x_{3} = 3 + t, \quad x_{4} = 2t, \quad x_{5} = 1 - t; \quad \mathrm{l}_{2}: x_{1} = t, \quad x_{2} = 5, \quad x_{3} = - 1 + t, \quad x_{4} = 3 - 2t, \quad x_{5} = 2t$.

$A(0, 3, 2, - 5)$; l: $x_{1} = 1 + t, \quad x_{2} = - t, x_{3} = 2 + 2t$, $x_{4} = - 2 + 2t$

$A(2, - 2, 1, 5); \quad \mathrm{l}: x_{1} = 3 + t, \quad x_{2} = - 1 + t, \quad x_{3} = 2 + t$, $x_{4} = - t$

$A(3, 3, 1, 0, 0); \quad \mathrm{l}: \quad x_{1} = 2 + 3t, \quad x_{2} = 1 + 2t, \quad x_{3} = - t$, $x_{4} = 1 + t, \quad x_{5} = - 1 - t$

$A(1, - 1, - 1, 1)$; l: $x_{1} + x_{2} + 2x_{3} + 1 = 0, \quad 3x_{2} + 2x_{3} - - x_{4} - 1 = 0, x_{1} - x_{2} + x_{3} + x_{4} + 2 = 0$.

if $a_{1}$ and $a_{2}$ are not collinear, then the square of the distance between $l_{1}$ and $l_{2}$ equals $\operatorname {det} \Gamma \left(\overrightarrow {A_{1} A_{2}}, a_{1}, a_{2}\right) / \operatorname {det} \Gamma \left(a_{1}, a_{2}\right)$

if $a_{1}$ and $a_{2}$ are collinear, then the square of the distance between $l_{1}$ and $l_{2}$ equals $\operatorname {det} \Gamma \left(\overrightarrow {A_{1} A_{2}}, a_{1}\right) / \left|a_{1}\right|^{2}$.

$\mathrm{l}_{1}: \quad x_{1} = 1 + t, \quad x_{2} = - 1, \quad x_{3} = - t, \quad x_{4} = - 2 + t$; $\mathrm{l}_{2}: x_{1} = 4 + t, \quad x_{2} = 2t, \quad x_{3} = 1 + t, \quad x_{4} = t$

$\mathrm{l}_{1}: \quad x_{1} = 2 + t, \quad x_{2} = - 1 - 2t, \quad x_{3} = 2 + 2t, \quad x_{4} = 1 - t$; $l_{2}: x_{1} = 3 - t, \quad x_{2} = 1 + 2t, \quad x_{3} = - 1 - 2t, \quad x_{4} = 2 + t$

$\mathrm{l}_{1}: \quad x_{1} = 3 + t, \quad x_{2} = 2, \quad x_{3} = t, \quad x_{4} = 3 + t, \quad x_{5} = - t$; $l_{2}: x_{1} = 1 + 2t, \quad x_{2} = 2t, \quad x_{3} = 1 - t, \quad x_{4} = t, \quad x_{5} = 2$

$l_{1}: x_{1} = 1 + t, \quad x_{2} = 2t, \quad x_{3} = 1 - t, \quad x_{4} = - 1 + t, \quad x_{5} = t$; $l_{2}: x_{1} = 3 + t, \quad x_{2} = - 2t, \quad x_{3} = - 1 - t, \quad x_{4} = 1 + t, \quad x_{5} = 2 + t$

$\mathrm{l}_{1}: \quad x_{1} = 1 - 2t, \quad x_{2} = 0, \quad x_{3} = t, \quad x_{4} = 1 + t, \quad x_{5} = 2$; $l_{2}: x_{1} = - 1 + t, \quad x_{2} = - 1 + t, \quad x_{3} = 0, \quad x_{4} = 1, \quad x_{5} = - 2 - t$.

$A(3, 7, - 2, 1); \mathrm{m}: x_{1} = 2 + t_{1}, \quad x_{2} = 2 + t_{2}, \quad x_{3} = t_{1} + t_{2}, \quad x_{4} = - t_{2}$

$A(- 3, - 1, 4, 7, - 3); \mathrm{m}: x_{1} = t_{1} + t_{3}, \quad x_{2} = 2 + t_{2} + t_{3}$, $x_{3} = 2 + t_{1}, \quad x_{4} = 1 + t_{2} + t_{3}, \quad x_{5} = 1 + t_{2}$.

$A(- 3, 2, 2, - 2); \mathrm{m}: x_{1} = 2 + t_{1} + t_{2}, \quad x_{2} = 4 + 2t_{1}$, $x_{3} = t_{1}, \quad x_{4} = - t_{2}$

$A(3, 2, 1, 4, - 1)$; $\mathrm{m}: x_{1} = 1 + t_{1}, \quad x_{2} = - 1 + t_{2}, \quad x_{3} = 2 + t_{1} + t_{2}, \quad x_{4} = - 2 - t_{1}, \quad x_{5} = t_{2}$

$A(0, - 1, 5, 1, - 2)$; m: $x_{1} = 1 + t_{1}, \quad x_{2} = t_{3}, \quad x_{3} = 1 + t_{1} + t_{2}, \quad x_{4} = - 2 + t_{3}, \quad x_{5} = - 1 + t_{2}$.

$A(5, 3, - 1, - 1); \mathrm{m}: \quad x_{1} = 1 + t_{1}, \quad x_{2} = t_{2}, \quad x_{3} = - 2 + t_{2}, \quad x_{4} = - 1 + t_{1}$

$A(3, 5, 0, 2, 2); \mathrm{m}: x_{1} = t_{1}, \quad x_{2} = 2 + t_{2}, \quad x_{3} = - 3 + t_{1}$, $x_{4} = 3 - t_{1} - t_{2}, \quad x_{5} = 1$.

in problem $34. 24, 1)$

in problem $34. 24, 2)$.

$A(1, 2, 1, 1); \mathrm{m}: \quad x_{1} = - 2t_{1} + 4t_{2}, \quad x_{2} = - 1 + t_{1} - t_{2}$, $x_{3} = - t_{3}, \quad x_{4} = t_{1} - t_{2}$

$A(3, 1, 1, 0)$; $\mathrm{m}: x_{1} = - 2 + t_{1}, \quad x_{2} = - t_{1} + 2t_{2}, \quad x_{3} = t_{1} - t_{2}, \quad x_{4} = 1 - t_{1} - t_{2}$

$A(1, 2, 1, 3, 0)$; $\mathrm{m}: \quad x_{1} = 1 + t_{1}, \quad x_{2} = - t_{1} + t_{2}, \quad x_{3} = 1 + t_{2}, \quad x_{4} = - 1 - t_{2}, \quad x_{5} = t_{1}$.

$A(1, 0, 0, 1)$; m: $x_{1} + 2x_{2} + 2x_{3} - 3x_{4} = 7, \quad x_{1} - 2x_{3} + 2x_{4} = - 6$

$A(1, 2, 0, 0)$; m: $x_{1} + x_{2} - x_{3} - x_{4} = 1, \quad 2x_{2} - 3x_{3} + x_{4} = 2, \quad 2x_{1} + x_{3} - 3x_{4} = 0$.

$A(1, 2, 2, 3), \quad B(4, 0, 0, 2); \mathrm{m}: x_{1} = 1 + t_{1}, \quad x_{2} = 2 + t_{2}, \quad x_{3} = - t_{1} + t_{2}, \quad x_{4} = 3$

$A(0, 1, - 1, 0, 1), B(3, 1, 0, 1, 2); \mathrm{m}: x_{1} = t_{1} + t_{2}, x_{2} = 5$, $x_{3} = - t_{2}, \quad x_{4} = - t_{1} + t_{2}, \quad x_{5} = 2 + t_{1}$

$A(- 1, - 1, 1, 0, 1), \quad B(2, 1, 1, 1, 0)$; m: $x_{1} = t_{1} + t_{3}$, $x_{2} = 2 + t_{2}, \quad x_{3} = 1 - t_{2}, \quad x_{4} = - t_{1} + t_{3}, \quad x_{5} = - 2t_{3}$.

l: $x_{1} = 2 - 2t, \quad x_{2} = 4 + t, \quad x_{3} = 1 + t, \quad x_{4} = 0$; $\mathrm{m}: x_{1} = 1 - 2t_{1}, \quad x_{2} = 1 + 2t_{1} + 3t_{2}, \quad x_{3} = 1 + t_{1}, \quad x_{4} = 1 + 2t_{1} + 2t_{2}$

l: $x_{1} = 3 + t_{1} + 2t_{2}, \quad x_{2} = - t_{1}, \quad x_{3} = 1 + t_{1} - t_{2}$, $x_{4} = - t_{1} - t_{2}; \quad \mathrm{m}: \quad x_{1} = 2t_{1} + t_{2}, \quad x_{2} = 1 - 3t_{1} + t_{2}$, $x_{3} = - 8 - t_{2}, x_{4} = 1 + t_{1} - t_{2}$

l: $x_{1} = 2 + t, \quad x_{2} = 2t, \quad x_{3} = 1, \quad x_{4} = t, \quad x_{5} = 0$; $\mathrm{m}: \quad x_{1} = 0, \quad x_{2} = 1 + t_{1} + t_{2}, \quad x_{3} = 3 + 2t_{2}, \quad x_{4} = 2t_{1}, \quad x_{5} = 1 + t_{1} - t_{2}$

l: $x_{1} = 1 + t_{2}, \quad x_{2} = t_{2}, \quad x_{3} = t_{1}, \quad x_{4} = t_{1}, \quad x_{5} = 2t_{1}$; $\mathrm{m}: x_{1} = t_{2} + 2t_{3}, \quad x_{2} = 2 + t_{1} + 2t_{3}, \quad x_{3} = x_{4} = 1 + t_{1} - t_{2} + t_{3}$, $x_{5} = 2 + t_{1} - t_{2} + 2t_{3}$

l: $2x_{1} - x_{3} + 3x_{4} = 0, \quad 2x_{1} - 2x_{2} + 3x_{3} - 3x_{4} = 8$; $\mathrm{m}: x_{2} = - 3x_{3} - 2x_{4} = 2, \quad x_{1} - x_{3} - x_{4} = 0$

l: $x_{1} + x_{2} - x_{3} = 1, \quad 2x_{1} + x_{2} - x_{4} = 4$; m: $x_{1} + x_{2} + x_{3} = - 1, \quad x_{1} + x_{3} + x_{4} = 1, \quad 2x_{1} - x_{2} - x_{4} = 0$.

there exists a unique plane of dimension $n - - k_{1} - k_{2}$, orthogonal to $\mathrm{m}_{1}$ and to $\mathrm{m}_{2}$ and intersecting each of these planes

there exists a unique line, orthogonal to $\mathrm{m}_{1}$ and to $\mathrm{m}_{2}$ and intersecting each of these planes.

$\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ are the lines of problem $34. 23, 1$)

$\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ are the lines of problem $34. 23, 3$)

$\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ are the planes of problem $34. 33, 3$).

between the faces $A_{0} A_{1} A_{2}$ and $A_{0} A_{3} A_{4}$

between the faces $A_{0} A_{1} A_{2}$ and $A_{0} A_{3} A_{4} A_{5}$

between the faces $A_{0} A_{1} A_{2}$ and $A_{1} A_{2} A_{3} A_{4} A_{5}$.