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## 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 $\vec a$and $\vec b$. 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.