**Intuitively, we can expect that if we multiply a vector by some scalar , the support of the vector will not change; only its magnitude and / or its sense will. Specifically, if is positive, the vector will have the same direction; only its length will get scaled according to the magnitude of . If is negative, the direction of the product vector will be opposite to that of the original vector; the length of the product vector will depend on the magnitude of .**

Note that for any vector , if we denote the unit vector along by , we have

Put in words, if we multiply the unit vector along a vector by its magnitude, we obtain that vector itself. Put in a slightly different way, we have

i.e, if we divide a vector by its magnitude, we obtain the unit vector along that vector’s direction.

Another very important result that follows from this discussion is that two vectors and are collinear if and only if there exists some such that

i.e, two vectors are collinear if one can be obtained from the other simply by multiplying the latter with a scalar.

This fact can be stated in another way : consider two non-collinear vectors and . If for some , the relation

is satisfied, then and must be zero. This is because can be written as

which would imply that is a scalar multiple of i.e. and , are collinear, contradicting our initial supposition that and are non-collinear.

In subsequent discussions, we’ll be talking a lot about linear combinations of vectors. Let us see what we mean by this. Consider arbitrary vectors A linear combination of these vectors is a vector such that

where are arbitrary scalars. Any sort of combination of the form in will be termed a linear combination.

Thus, using the terminology of linear combinations, we can restate the result we obtained earlier: for any two non-zero and non-collinear vectors and , if their linear combination is zero, then both the scalars in the linear combination must be zero.

We now come to a very important concept.

#### THE BASIS OF A VECTOR SPACE

Consider a two-dimensional plane, and any two arbitrary non-collinear vectors and in this plane. We make the two vectors co-initial and use their supports as our reference axes:

Observe carefully that any vector in the plane can be represented in terms of and . We find the components of along the directions of and ; those components must be some scalar multiples of and .

Thus, any vector in the plane can be written as

for some |

i.e, any vector in the plane can be expressed as a linear combination of and.

We state this fact in mathematical terms as follows: the vectors and form a basis of our vector space (which is a plane in this case). The term “basis” means that using only and , we can construct any vector lying in the plane of and .

Note that there’s nothing special about and ; any two non-collinear vectors can form a basis for the plane.

You must be very clear on the point that two collinear vectors cannot form the basis for a plane while any two non-collinear vectors can. Understanding this fact is very crucial to later discussions.

Try proving this: let and form the basis of a plane. For any vector in the plane of and , we can find scalars such that

Prove that this representation is unique

The basic principle that we’ve learnt in this discussion can be expressed in a very useful way as follows:

__Three vectors are coplanar if and only if one of them can be expressed as a linear combination of the other two. i.e., three vectors are coplanar if there exist scalars such that__

__We can write this as__

__This form equivalently tells us that three vectors are coplanar if we can find three scalars for which their linear combination is zero.__