Basis Of A Two Dimensional Plane

THE BASIS OF A VECTOR SPACE

Consider a two-dimensional plane, and any two arbitrary non-collinear vectors \(\vec a\,\,{\text{and}}\,\,\vec b\)  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  \(\vec a\,\,{\text{and}}\,\,\vec b\) . We find the components of  \(\vec r\) along the directions of \(\vec a\,\,{\text{and}}\,\,\vec b\); those components must be some scalar multiples of \(\vec a\,\,{\text{and}}\,\,\vec b\) .

Thus, any vector \(\vec r\)  in the plane can be written as

\[\vec r = \lambda \vec a + \mu \vec b\;\qquad\qquad\qquad for{\text{ }}some\;\lambda ,\mu  \in \mathbb{R}{\text{ }}\quad\qquad\qquad...{\text{ }}\left( 1 \right)\]

i.e, any vector \(\vec r\)  in the plane can be expressed as a linear combination of  \(\vec a\,\,{\text{and}}\,\,\vec b\).

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

Note that there’s nothing special about  \(\vec a\,\,{\text{and}}\,\,\vec b\) ; 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 \(\vec a\,\,{\text{and}}\,\,\vec b\)  form the basis of a plane. For any vector \(\vec r\) in the plane of \(\vec a\,\,{\text{and}}\,\,\vec b\) , 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:

Example – 4

Suppose that for three non-zero vectors  \(\vec a,\vec b,\vec c,\)  any two of them are non-collinear. If the vectors \(\left( {\vec a + 2\vec b} \right)\) and  \(\vec c\) are collinear and the vectors \(\left( {\vec b + 3\vec c} \right)\) and \(\vec a\) are collinear, prove that

\[\vec a + 2\vec b + 6\vec c = \vec 0\]

Solution: We must have some \(\lambda ,\mu  \in \mathbb{R}\)such that

\[\begin{align}&\vec a + 2\vec b = \lambda \vec c\quad\quad\quad...{\text{ }}\left( 1 \right) \hfill \\& \vec b + 3\vec c = \mu \vec a\quad\quad\quad...{\text{ }}\left( 2 \right) \hfill \\ \end{align} \]

From (1), we have

\[\vec c = \frac{1}{\lambda }\left( {\vec a + 2\vec b} \right)\quad\quad\quad...{\text{ }}\left( 3 \right)\]

We use this in (2) :

\[\vec b + \frac{3}{\lambda }\left( {\vec a + 2\vec b} \right) = \mu \vec a\]

\[ \Rightarrow  \quad  \left( {\frac{3}{\lambda } - \mu } \right)\vec a + \left( {1 + \frac{6}{\lambda }} \right)\vec b = \vec 0\]

Since \(\vec a\,\,{\text{and}}\,\,\vec b\) are non-collinear, their linear combination can be zero if and only if the two scalars are zero.

This gives

\[\frac{3}{\lambda } - \mu  = 0\]

\[1 + \frac{6}{\lambda } = 0\]

\[ \Rightarrow  \quad \lambda  =  - 6,\,\,\mu  =  - \frac{1}{2}\]

Using the value \(\lambda \) of in (3), we have

\[\vec a + 2\vec b + 6\vec c = 0\]