Decomposition of vectors.

Theorem 1. An arbitrary vector in the plane can be decomposed into two noncollinear vectors:

.

Theorem 2. An arbitrary vector in space can be decomposed into three noncoplanar vectors

.

Let be noncoplanar vectors.

The Cartesian system of coordinates. Consider the following coordinate system: take mutually perpendicular unit vectors and , draw coordinate axes x,y and z along them, and fix a unit on metric scale:

,

.

 

 

Definition. The triple of vectors is called right if, looking from the endpoint of the last vector, we see that the shorter rotation from the first vector to the second is anticlockwise.

From the triangle ОММ₁, we obtain

,

Since the vector is collinear to the unit vector , it follows that

.

From the triangle ОАМ_{1 ,} we obtain

,

because, by analogy, the vectors and are collinear to the unit vectors and . Substituting the vector thus obtained, we see that

. (2)

Thus, the radius vector is represented as the sum of and multiplied by the corresponding coordinates of the point М.

Consider the vectors

and

and their sum

.

Under addition the respective coordinates are added

Let us multiply the vector by a number l:

.

When a vector is multiplied by a number l, each coordinate of this vector is multiplied by this number.

Example. Find the vector if

; .

Let us find the required vector in vector notation:

.

To find the same vector in vector notation, we multiply the first vector by 4 and the second by –3 and sum their coordinates:

.

Given two points and in space, find the vector .

Thus, we have found the required vector in the coordinate notation:

 

. (2¢¢)

 

To find the coordinates of a vector, we must subtract the coordinates of its tail from the coordinates of the head.

For example, let us find vectors with given coordinates of heads and tails:

М ₁(7;4;–3); М ₂(1;–2;–2);

 

={–6; –6; 1}; ={6; 6; –1}.

 

Find the length of a vector :

.

From the right triangle ОМ ₁ М ₂, we find the hypotenuse

,

 

z

 

where .

M ₁

From the other right triangle ОАМ ₂, 0 z₁ y

we find the hypotenuse .

Substituting it into А x ₁

the first hypotenuse, we obtain x y ₁ M ₂

.

Thus, the length of a vector is defined by the formula

. (3)