Coplanar vectors are the parallelepiped rule.

TEXT TRANSCRIPT OF THE LESSON:

The addition of several vectors in space is performed as follows: the first vector is added to the second, then their sum is added to the third vector, and so on. We know this rule as the polygon rule. The figure shows the addition of three vectors in space.

From point O the vector OA is equal to vector a, then from point A the vector AB is equal to be, from point B the next vector BC is equal to tse, and we connect the first and last points O with C we get the vector OS equal to the sum of vectors a, be and tse.

Let us formulate the polygon rule.

The figure shows the sum of six vectors.

If the beginning of the vector coincides with the end of the last one, then the sum is equal to the zero vector.

Consider the sum of vectors

After performing the addition according to the polygon rule, we obtain the vector AA or zero vector.

Let's solve problem No. 337 (c)

Simplify the expression

Solution: Let's replace the subtraction in the expression with the sum. To do this, replace the negative vectors with their opposite ones. The minus vector BC is equal to the vector SV, the minus vector RM is equal to the vector MR. The minus vector AP is equal to the vector RA. The AC vector added to the CB vector gives the AB vector. Vectors MR and RA give vector MA. Then, adding the vectors AB and BM, we obtain the vector AM. As a result, the sum of the vectors AM and MA gives a zero vector. The expression is simplified.

Let's solve proof problem No. 338.

Given parallelepiped ABCDA1B1C1D1. Prove that, where O is an arbitrary point in space.

Proof. Let's transform the left side of the equality. We represent vector OA as the sum of vectors OA1 and A1A according to the triangle rule. Vector A1A is equal to vector C1C as the opposite edges of the parallelepiped. Adding the vectors OC1 and C1C, we obtain OC. As a result of the transformations, we obtained the right side of the equality. The proof is over.

Parallelepiped rule. A vector lying on the diagonal of a parallelepiped is equal to the sum of vectors drawn from the same point and lying on three dimensions of the parallelepiped. B1. C1. A1. D1. B.C.A.D.

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