The vector is the size characterized by the numerical value and the direction. In other words, the vector is the directed piece. The provision of a vector of AB in space is set by coordinates of a point of the beginning of a vector of A and point of the end of a vector of B. Let's consider how to determine coordinates of the middle of a vector.
Instruction
1. For a start we will decide on designations of the beginning and end of a vector. If the vector is written down as AB, then the point of A is the beginning of a vector, and B point – the end. To the contrary, for BA vector the point of B is the beginning of a vector, and A point – the end. Let to us AB vector with coordinates of the beginning of a vector of A = (a1, a2, a3) and the end of a vector of B = be set (b1, b2, b3). Then coordinates of a vector of AB will be following: AB = (b1 – a1, b2 – a2, b3 – a3), i.e. it is necessary to subtract the corresponding coordinate of the beginning of a vector from the coordinate of the end of a vector. AB vector length (or its module) is calculated as a root square of the sum of squares of its coordinates: |AB| = √ ((b1 – a1) ^2 + (b2 – a2) ^2 + (b3 – a3) ^2).
2. Let's find coordinates of the point which is the middle of a vector. Let's designate it by letter O = (o1, o2, o3). There are coordinates of the middle of a vector the same as coordinate of the middle of a usual piece, on the following formulas: o1 = (a1 + b1)/2, o2 = (a2 + b2)/2, o3 = (a3 + b3)/2. Let's find AO vector coordinates: AO = (o1 – a1, o2 – a2, o3 – a3) = ((b1 – a1)/2, (b2 – a2)/2, (b3 – a3)/2).
3. Let's review an example. Let AB vector with coordinates of the beginning of a vector of A = (1, 3, 5) and the end of a vector of B = be given (3, 5, 7). Then coordinates of a vector of AB can be written down as AB = (3 – 1, 5 – 3, 7 – 5) = (2, 2, 2). Let's find AB vector module: |AB| = √ (4 + 4 + 4) = 2 * √3. The value of length of the set vector will help us for further check of correctness of coordinates of the middle of a vector. Further we will find O point coordinates: O = ((1 + 3)/2, (3 + 5)/2, (5 + 7)/2) = (2, 4, 6). Then coordinates of a vector of AO are counted as AO = (2 – 1, 4 – 3, 6 – 5) = (1, 1, 1).
4. Let's execute check. AO vector length = √ (1 + 1 + 1) = √3. Let's remember that length of an initial vector is equal 2 * √3, i.e. a half of a vector is really equal to a half of length of an initial vector. Now we will calculate OB vector coordinates: OB = (3 – 2, 5 – 4, 7 – 6) = (1, 1, 1). Let's find the sum of vectors of AO and OB: AO + OB = (1 + 1, 1 + 1, 1 + 1) = (2, 2, 2) = AB. Therefore, coordinates of the middle of a vector were found truly.