Question

В правильной четырёхугольной пирамиде SABCD, все ребра которой равны 1, найдите косинус угла между плоскостями SAD и SBC

Answer 1

В правильной четырехугольной пирамиде в основании лежит квадрат. В нашем случае это квадрат со стороной, равной 1. Тогда апофема (высота грани пирамиды) равна по Пифагору √(AS²-AH²), где АН - половина АD. Нам дано, что ВСЕ ребра пирамиды =1. Значит апофема равна √(1-1/4)=√3/2.
Угол между плоскостями SAD и SBC - это угол между двумя противоположными гранями пирамиды, а значит это угол между их апофемами. Найдем синус угла между высотой пирамиды SO и апофемой грани SH. Он равен отношению половины стороны основания HO (противолежащий катет) к апофеме SH (гипотенузе), то есть 1/2:√3/2=1/√3=√3/3. Но это синус половины искомого угла. Косинус искомого угла находится по формуле:
Cos2α=1-2*Sin²α. В нашем случае Coc2α=1-2*(√3/3)²=1-3/9=1-2/3=1/3.
Ответ: косинус между плоскостями SAD и SBC равен 1/3.

Answer 2

Finding the Cosine of the Angle between the Planes SAD and SBC in the Pyramid SABCD

To find the cosine of the angle between the planes SAD and SBC in the pyramid SABCD, we need to determine the normal vectors of these planes and then calculate the cosine of the angle between the two normal vectors.

Let's start by finding the normal vectors of the planes SAD and SBC.

The plane SAD is defined by the points S, A, and D. Since all the edges of the pyramid SABCD are equal to 1, we can determine the coordinates of these points as follows:

- Point S: (0, 0, 0) - Point A: (1, 0, 0) - Point D: (0, 1, 0)

To find the normal vector of the plane SAD, we can calculate the cross product of the vectors formed by the points A and S, and A and D. The cross product of two vectors gives us a vector that is perpendicular to both of them, which is the normal vector of the plane they lie on.

The vector AS can be calculated as follows: AS = A - S = (1, 0, 0) - (0, 0, 0) = (1, 0, 0)

The vector AD can be calculated as follows: AD = D - A = (0, 1, 0) - (1, 0, 0) = (-1, 1, 0)

Now, let's calculate the cross product of AS and AD to find the normal vector of the plane SAD.

AS x AD = (1, 0, 0) x (-1, 1, 0) = (0, 0, 1)

Therefore, the normal vector of the plane SAD is (0, 0, 1).

Similarly, we can find the normal vector of the plane SBC.

The plane SBC is defined by the points S, B, and C. Using the same approach, we can determine the coordinates of these points as follows:

- Point S: (0, 0, 0) - Point B: (0, 1, 0) - Point C: (1, 1, 0)

The vector BS can be calculated as follows: BS = B - S = (0, 1, 0) - (0, 0, 0) = (0, 1, 0)

The vector BC can be calculated as follows: BC = C - B = (1, 1, 0) - (0, 1, 0) = (1, 0, 0)

Now, let's calculate the cross product of BS and BC to find the normal vector of the plane SBC.

BS x BC = (0, 1, 0) x (1, 0, 0) = (0, 0, -1)

Therefore, the normal vector of the plane SBC is (0, 0, -1).

Now that we have the normal vectors of the planes SAD and SBC, we can calculate the cosine of the angle between them using the dot product formula:

cos(theta) = (N1 · N2) / (||N1|| ||N2||)

where N1 and N2 are the normal vectors of the planes SAD and SBC, respectively, and ||N1|| and ||N2|| are their magnitudes.

The magnitude of a vector (a, b, c) can be calculated as follows: ||N|| = sqrt(a^2 + b^2 + c^2)

Let's calculate the cosine of the angle between the planes SAD and SBC:

N1 = (0, 0, 1) N2 = (0, 0, -1)

||N1|| = sqrt(0^2 + 0^2 + 1^2) = 1 ||N2|| = sqrt(0^2 + 0^2 + (-1)^2) = 1

(N1 · N2) = 0 * 0 + 0 * 0 + 1 * (-1) = -1

cos(theta) = (-1) / (1 * 1) = -1

Therefore, the cosine of the angle between the planes SAD and SBC in the pyramid SABCD is -1.

Please note that the search results did not provide direct information on the cosine of the angle between the planes SAD and SBC in the given pyramid. The calculation was done based on the provided information about the pyramid's structure and the mathematical principles involved in finding the cosine of the angle between two planes.