Question

Срочно! В правильном тетраэдре ABCD с ребром 2 найдите косинус угла между ребром AD и плоскости грани ABC.Дам 10 баллов!

Answer 1

Ответ:

$\dfrac{\sqrt{3} }{3} .$

Объяснение:

Пусть дан  тетраэдр ABCD .  DО - высота тетраэдра.

Тогда углом между ребром АD и плоскостью грани АВС будет  

∠DАО= α.

$AO= R= \dfrac{a}{\sqrt{3} } ;\\AO= \dfrac{2}{\sqrt{3} }.$

 Рассмотрим Δ DOA - прямоугольный.

Косинусом острого угла прямоугольного треугольника называется отношение прилежащего катета к гипотенузе .

$\cos \alpha = \dfrac{AO}{AD } ;\\\\\cos \alpha = \dfrac{\dfrac{2}{\sqrt{3} } }{2 }=\dfrac{2}{\sqrt{3} }: \dfrac{2}{1} = \dfrac{2}{\sqrt{3} }\cdot \dfrac{1}{2}=\dfrac{1}{\sqrt{3} }=\dfrac{\sqrt{3} }{3} .$

Answer 2

Finding the Cosine of the Angle between Edge AD and the Plane of Face ABC in a Regular Tetrahedron

To find the cosine of the angle between edge AD and the plane of face ABC in a regular tetrahedron, we can use trigonometric principles.

A regular tetrahedron is a polyhedron with four equilateral triangular faces. Each face of the tetrahedron is an equilateral triangle, and all edges have the same length.

Given that the length of each edge of the tetrahedron is 2, we need to find the cosine of the angle between edge AD and the plane of face ABC.

To find the cosine of the angle, we can use the dot product of two vectors. In this case, we can consider vector AD and a normal vector to the plane of face ABC.

Let's denote the angle between edge AD and the plane of face ABC as θ.

To find the cosine of θ, we can use the formula:

cos(θ) = (AD · n) / (|AD| * |n|)

Where: - AD is the vector representing edge AD. - n is the normal vector to the plane of face ABC. - |AD| represents the magnitude (length) of vector AD. - |n| represents the magnitude (length) of vector n.

Since the tetrahedron is regular, the normal vector to the plane of face ABC can be found by taking the cross product of vectors AB and AC.

Let's calculate the cosine of the angle using the given information.

Calculation:

1. Calculate the normal vector to the plane of face ABC: - Vector AB: AB = B - A = (0, 0, 0) - (2, 0, 0) = (-2, 0, 0) - Vector AC: AC = C - A = (1, 1, 0) - (2, 0, 0) = (-1, 1, 0) - Normal vector n: n = AB x AC = (-2, 0, 0) x (-1, 1, 0) = (0, 0, 2)

2. Calculate the magnitude of vector AD: - |AD| = 2 (given)

3. Calculate the magnitude of vector n: - |n| = sqrt(n1^2 + n2^2 + n3^2) = sqrt(0^2 + 0^2 + 2^2) = sqrt(4) = 2

4. Calculate the dot product of vectors AD and n: - AD · n = AD1 * n1 + AD2 * n2 + AD3 * n3 = 2 * 0 + 0 * 0 + 0 * 2 = 0

5. Calculate the cosine of the angle: - cos(θ) = (AD · n) / (|AD| * |n|) = 0 / (2 * 2) = 0

Therefore, the cosine of the angle between edge AD and the plane of face ABC in the given regular tetrahedron is 0.

Please note that the cosine of an angle can range from -1 to 1, and in this case, the cosine is 0, indicating that the angle between edge AD and the plane of face ABC is a right angle (90 degrees).

I hope this helps! Let me know if you have any further questions.