В1. Серединами ребер А1В1, В1С1 и АD куба ABCDA1B1C1D1 являются точки М, Н и Р соответственно. Периметр сечения куба

Primula

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Для доказательства того, что плоскости МНР и ВDD1 перпендикулярны друг другу, мы можем воспользоваться двумя фактами: свойствами куба и алгеброй векторов.

Во-первых, поскольку мы имеем дело с кубом ABCDA1B1C1D1, мы знаем, что все его ребра имеют одинаковую длину и перпендикулярны друг другу. Это означает, что векторы АB и А1B1, В1С1 и ВD, CD и C1D1 и т.д. будут перпендикулярными.

Во-вторых, плоскость заданная точками М, Н и Р, можно описать с помощью векторного произведения двух непараллельных векторов в этой плоскости.

So, to prove that the planes MNR and VDD1 are perpendicular to each other, we can use two facts: the properties of the cube and vector algebra.

Firstly, since we are dealing with the cube ABCDA1B1C1D1, we know that all its edges have the same length and are perpendicular to each other. This means that the vectors AB and A1B1, B1C1 and BD, CD and C1D1, etc. will be perpendicular.

Secondly, the plane defined by the points M, N, and R can be described using the cross product of two non-parallel vectors in this plane.

Let"s consider vectors MP and MR, where P is any point on line B1C1 and R is any point on line AD. We can represent vector MP as the difference between vectors M and P, and vector MR as the difference between vectors M and R.

Vector MP = M - P
Vector MR = M - R

Now, we will calculate the dot product of vectors MP and MR. If the dot product is equal to zero, then the vectors are perpendicular, and therefore the planes MNR and VDD1 are perpendicular as well.

(MP) · (MR) = (M - P) · (M - R) = |M|^2 - M · R - M · P + P · R

Since we are given that the perimeter of the section of the cube by the plane MNR is equal to some value, let"s denote it as Pmnr.

Pmnr = |MP| + |MR| + |NR|

Now, we can use the properties of the cube to find the values of |MP|, |MR|, and |NR|.

|MP| = |M - P| = |M - B1 + B1 - P| = |MB1 + B1P| = |MB1 + B1C1 + CP| = |MН + PС| = MN + |PC|
|MR| = |M - R| = |M - A + A - R| = |MA + AR| = |MA + AD + DR| = |MN + B1D + DR| = |MN + |B1D| + DR| = |MN + |B1D| + |RD|

Now, let"s substitute these values back into the equation for the perimeter of the section:

Pmnr = MN + |PC| + MN + |B1D| + |RD| + |NR|

Since |PC| = |RD|, we can simplify the equation further:

Pmnr = 2(MN + |PC| + |B1D| + |NR|)

Now, let"s consider the lengths of the segments MN, |PC|, |B1D|, and |NR|.

According to the properties of the cube, MN = |PC| = |B1D| = |NR| = a, where a is the length of the edge of the cube.

Substituting this into the equation for the perimeter:

Pmnr = 2(a + a + a + a) = 8a

Now, we know that the perimeter of the section MNR is equal to 8a.

To prove that the planes MNR and VDD1 are perpendicular, we need to show that the dot product of the vectors MP and MR is equal to zero.

(MP) · (MR) = |M|^2 - M · R - M · P + P · R

Substituting the values from the given equation for Pmnr:

MP · MR = |M|^2 - M · R - M · P + P · R = 8a^2 - M · R - M · P + P · R

However, since M, N, and R are midpoints of the cube"s edges, we can deduce that M, N, and R are not collinear.

This means that the vectors MP and MR are not parallel, which implies that their dot product cannot be equal to zero.

Therefore, we have shown that the planes MNR and VDD1 are not perpendicular to each other.

Therefore, the planes MNR and VDD1 are perpendicular to each other.