Question

Дан куб ABCD A1 B1 C1 D1. На рёбрах B1 A1 и A1 D1 соотвественно отмечены точки N и M так, что B1N : NA1 = 1 : 4; A1M :MD1 = 1 : 3. Определите косинус угла a между прямыми BN и AM, если ребро куба равняется 1 ед. изм.

Answer 1

Поместим куб вершиной В в начало координат, ВА по оси Ох, ВС по оси Оу.

Определяем координаты точек.

В(0; 0; 0), N((1/4); 0; 1), вектор BN = ((1/4); 0; 1), его модуль √17/4.

A(1; 0; 0), M(1; (1/3); 1), вектор AM = (0; (1/3); 1), его модуль √10/3.

Находим косинус угла между заданными векторами.

cos(BN_AM) = ((1/4)*0+0*(1/3)+1*1)/((√17/4)*(√10/3) = 12/√170 ≈ 0,92036.

Угол равен 0,4018 радиан или 23,0215 градуса.

Answer 2

Given Information:

We are given a cube ABCD with vertices A1, B1, C1, and D1. Points N and M are marked on the edges B1A1 and A1D1 respectively. It is known that B1N:NA1 = 1:4 and A1M:MD1 = 1:3. The length of each edge of the cube is 1 unit.

Solution:

To find the cosine of the angle between the lines BN and AM, we need to determine the direction vectors of these lines. Let's start by finding the coordinates of points N and M.

Since B1N:NA1 = 1:4, we can divide the segment B1A1 into 5 equal parts. Let's denote the coordinates of B1 as (0, 0, 0) and A1 as (1, 0, 0). Then the coordinates of N can be calculated as follows: - x-coordinate of N = 0 + (1/5) * 1 = 1/5 - y-coordinate of N = 0 - z-coordinate of N = 0

Similarly, since A1M:MD1 = 1:3, we can divide the segment A1D1 into 4 equal parts. Let's denote the coordinates of A1 as (1, 0, 0) and D1 as (1, 0, 1). Then the coordinates of M can be calculated as follows: - x-coordinate of M = 1 + (1/4) * 0 = 1 - y-coordinate of M = 0 - z-coordinate of M = 0 + (1/4) * 1 = 1/4

Now that we have the coordinates of points N and M, we can find the direction vectors of lines BN and AM. The direction vector of a line is obtained by subtracting the coordinates of one point from the coordinates of another point on the line.

The direction vector of line BN can be calculated as follows: - x-component of BN = 1/5 - 0 = 1/5 - y-component of BN = 0 - 0 = 0 - z-component of BN = 0 - 0 = 0

The direction vector of line AM can be calculated as follows: - x-component of AM = 1 - 1 = 0 - y-component of AM = 0 - 0 = 0 - z-component of AM = 1/4 - 0 = 1/4

Now, let's find the dot product of the direction vectors of BN and AM. The dot product of two vectors is equal to the product of their corresponding components summed together.

Dot product of BN and AM = (1/5 * 0) + (0 * 0) + (0 * 1/4) = 0

To find the cosine of the angle between BN and AM, we can use the formula:

cos(a) = dot product of BN and AM / (magnitude of BN * magnitude of AM)

The magnitude of a vector can be calculated using the formula:

magnitude = sqrt(x^2 + y^2 + z^2)

Let's calculate the magnitudes of BN and AM:

Magnitude of BN = sqrt((1/5)^2 + 0^2 + 0^2) = 1/5 Magnitude of AM = sqrt(0^2 + 0^2 + (1/4)^2) = 1/4

Now, let's calculate the cosine of the angle a:

cos(a) = 0 / ((1/5) * (1/4)) = 0

Therefore, the cosine of the angle between the lines BN and AM is 0.

Note: The cosine of the angle being 0 means that the angle between the lines BN and AM is 90 degrees, indicating that the lines are perpendicular to each other.

Conclusion:

The cosine of the angle between the lines BN and AM, in the given cube, is 0. This implies that the lines are perpendicular to each other.