Question

ДАЮ 50 БАЛЛОВ!! С объяснением, пожалуйста.. Точка М делит сторону AD параллелограмма ABCD в отношении 1:2, считая от точки А, отрезок ВМ пересекает диагональ АС в точке N. Разложите вектор MN по векторам АВ и AD.​

Answer 1

Точка М делит сторону AD параллелограмма ABCD в отношении 1:2, считая от точки А, отрезок ВМ пересекает диагональ АС в точке N. Разложите вектор MN по векторам АВ и AD.​

Объяснение:

1) Чтобы выразить вектор МN найдем в каком отношении точка N  делит отрезок ВМ.

ΔАМN~ΔCBN по 2 углам :∠ANM=∠CNM как вертикальные , ∠А=∠С как накрест лежащие при ВС║АD , АС-секущая. Тогда сходственные стороны пропорциональны АМ:ВС=МN:BN . Но по условию АМ=1/3*AD=1/3*BC, по свойству сторон параллелограмма. Тогда  (1/3ВС):ВС=МN:BN ⇒1/3=МN:BN . Т.е а отрезок ВМ приходится 4 части , при чем на МN приходится 1 часть , на BN приходится 3 части.

2)Вектор   $\displaystyle \vec{MN}$ = $\displaystyle \frac{1}{4} \vec{MB}$   .  По правилу разности векторов  $\displaystyle \vec{MN}=\displaystyle \frac{1}{4} \vec{MB}=\frac{1}{4} (\displaystyle \vec{AB}- \vec{AM})$   . Учтем , что длина АМ=1/3*AD , тогда

$\displaystyle \vec{MN}=\displaystyle \frac{1}{4} \vec{MB}=\frac{1}{4} (\displaystyle \vec{AB}- \vec{AM})=\frac{1}{4} ( \vec{AB}- \frac{1}{3} \vec{AD})=\frac{1}{4} \vec{AB}- \frac{1}{12} \vec{AD}$.

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Answer 2

Problem Analysis

We are given a parallelogram ABCD, where point M divides side AD in the ratio 1:2, counting from point A. The segment VM intersects diagonal AC at point N. We need to express vector MN in terms of vectors AB and AD.

Solution

To express vector MN in terms of vectors AB and AD, we can use the properties of parallelograms and the concept of vector addition.

Let's break down the solution step by step:

1. Since M divides AD in the ratio 1:2, we can say that AM is one-third of AD, and MD is two-thirds of AD. 2. Since ABCD is a parallelogram, we know that opposite sides are parallel and equal in length. Therefore, AB is parallel and equal in length to CD, and AD is parallel and equal in length to BC. 3. Since AB is parallel to CD, we can say that vector AB is equal to vector CD. 4. Similarly, since AD is parallel to BC, we can say that vector AD is equal to vector BC. 5. Now, let's express vector MN in terms of vectors AB and AD. We can write vector MN as the sum of vector MA and vector AN. 6. Vector MA can be expressed as vector AB minus vector BM. Similarly, vector AN can be expressed as vector AD minus vector DN. 7. Substituting the values of vector AB, vector AD, vector BM, and vector DN, we can express vector MN in terms of vectors AB and AD.

Detailed Solution

Let's go through the solution step by step:

1. Given that M divides AD in the ratio 1:2, we can say that AM is one-third of AD, and MD is two-thirds of AD. 2. Since ABCD is a parallelogram, we know that opposite sides are parallel and equal in length. Therefore, AB is parallel and equal in length to CD, and AD is parallel and equal in length to BC. 3. Since AB is parallel to CD, we can say that vector AB is equal to vector CD. 4. Similarly, since AD is parallel to BC, we can say that vector AD is equal to vector BC. 5. Now, let's express vector MN in terms of vectors AB and AD. We can write vector MN as the sum of vector MA and vector AN. 6. Vector MA can be expressed as vector AB minus vector BM. Similarly, vector AN can be expressed as vector AD minus vector DN. 7. Substituting the values of vector AB, vector AD, vector BM, and vector DN, we can express vector MN in terms of vectors AB and AD.

Therefore, the expression for vector MN in terms of vectors AB and AD is:

MN = MA + AN

MN = (AB - BM) + (AD - DN)

Substituting the values of vector AB, vector AD, vector BM, and vector DN, we get:

MN = (AB - (2/3)AD) + (AD - (1/3)AD)

Simplifying the expression, we get:

MN = AB - (2/3)AD + AD - (1/3)AD

Combining like terms, we get:

MN = AB - (2/3)AD + (2/3)AD

Simplifying further, we get:

MN = AB

Therefore, vector MN can be expressed solely in terms of vector AB.

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