CHAPTER 6 - Multiplication of a Vector by a Scalar

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

Collinear vectors

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.