tg

tg
tgt

Friday, 8 August 2014

CHAPTER 6 - Multiplication of a Vector by a Scalar

Intuitively, we can expect that if we multiply a vector \vec a by some scalar \lambda , the support of the vector will not change; only its magnitude and / or its sense will. Specifically, if \lambda  is positive, the vector will have the same direction; only its length will get scaled according to the magnitude of \lambda . If \lambda  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 \lambda .
Note that for any vector \vec a, if we denote the unit vector along \vec a by \hat a, we have
{\vec a = \left| {\,\vec a\,} \right|\hat a}
Put in words, if we multiply the unit vector along a vector \vec a by its magnitude, we obtain that vector itself. Put in a slightly different way, we have
\hat a = \dfrac{{\vec a}}{{\left| {\vec a} \right|}}
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 \vec a and \vec b are collinear if and only if there exists some \lambda  \in \mathbb{R} such that
{\vec a = \lambda \vec b}
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 \vec aand \vec b. If for some \lambda ,\mu  \in \mathbb{R}, the relation
\lambda \vec a + \mu \vec b = \vec 0\ldots(1)
is satisfied, then \lambda  and \mu  must be zero. This is because (1) can be written as
\vec a = \left( { - \dfrac{\mu }{\lambda }} \right)\vec b
which would imply that \vec a is a scalar multiple of \vec b, i.e.\vec a and \vec b, are collinear, contradicting our initial supposition that \vec a and \vec b 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 n arbitrary vectors {\vec a_1},{\vec a_2}\ldots {\vec a_n}. A linear combination of these n vectors is a vector \vec r such that
\vec r = {\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + \ldots + {\lambda _n}{\vec a_n}\ldots(2)
where {\lambda _1},{\lambda _2}\ldots {\lambda _n} \in\mathbb{R}  are arbitrary scalars. Any sort of combination of the form in (2)  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 \vec a and \vec b, 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 \vec a and \vec b in this plane. We make the two vectors co-initial and use their supports as our reference axes:
Observe carefully that any vector \vec r in the plane can be represented in terms of \vec a and \vec b. We find the components of \vec r along the directions of \vec a and \vec b; those components must be some scalar multiples of \vec a and \vec b.
Thus, any vector \vec r in the plane can be written as
\vec r = \lambda \vec a + \mu \vec b for some \lambda ,\mu  \in\mathbb{R} \ldots(1)
i.e, any vector \vec r in the plane can be expressed as a linear combination of \vec a and\vec b.
We state this fact in mathematical terms as follows: the vectors \vec a and \vec b form a basis of our vector space (which is a plane in this case). The term “basis” means that using only \vec a and \vec b, we can construct any vector lying in the plane of \vec a and \vec b.
Note that there’s nothing special about \vec a and \vec b; 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 \vec a and \vec b form the basis of a plane. For any vector \vec r in the plane of \vec a and \vec b, we can find scalars \lambda ,\mu  \in\mathbb{R}  such that
\vec r = \lambda \vec a + \mu \vec b
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 \vec a,\,\,\vec b,\,\,\vec c are coplanar if there exist scalars {l_1},{l_2} \in\mathbb{R}  such that
\vec a = {l_1}\vec b + {l_2}\vec c
We can write this as
(1) \vec a + \left( { - {l_1}} \right)\vec b + \left( { - {l_2}} \right)\vec c = \vec 0
 \Rightarrow  \,\,\,\,\, \lambda \vec a + \mu \vec b + \gamma \vec c = 0
This form equivalently tells us that three vectors are coplanar if we can find three scalars \lambda ,\mu ,\gamma  \in \mathbb{R} for which their linear combination is zero.

Post a Comment