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Friday, 8 August 2014

CHAPTER 8 - Coplanar Vectors

In the preceeding discussion, we talked about the basis of a plane. We can easily extend that discussion to observe that any three non-coplanar vectors can form a basis of three dimensional space:
In other words, any vector \vec r in 3-D  space can be expressed as a linear combination of three arbitrary non-coplanar vectors. From this, it also follows that for three non-coplanar vectors \vec a,\,\,\vec b,\,\,\vec c, if their linear combination is zero, i.e, if
\lambda \vec a + \mu \vec b + \gamma \vec c = \vec 0where {\lambda ,\mu ,\gamma  \in\mathbb{R} }
then \lambda ,\mu  and \gamma  must all be zero. To prove this, assume the contrary. Then, we have
\vec a = \left( { - \dfrac{\mu }{\lambda }} \right)\vec b + \left( { - \dfrac{\gamma }{\lambda }} \right)\vec c
which means that \vec a can be written as the linear combination of \vec b and \vec c. However, this would make \vec a,\,\,\vec b and \vec c coplanar, contradicting our initial supposition. Thus, \lambda ,\mu  and \gamma  must be zero.
We finally come to what we mean by linearly independent and linearly dependent vectors.
Linearly independent vectors:
A set of non-zero vectors {\vec a_1},{\vec a_2},{\vec a_3}\ldots ,{\vec a_n} is said to be linearly independent if
{\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + \ldots + {\lambda _n}{\vec a_n} = \vec 0
implies {\lambda _1} = {\lambda _2} = \ldots = {\lambda _n} = 0
Thus, a linear combination of linearly independent vectors cannot be zero unless all the scalars used to form the linear combination are zero.
Linear dependent vectors:
A set of non-zero vectors {\vec a_1},{\vec a_2},{\vec a_3},\ldots,{\vec a_n} is said to be linearly dependent if there exist scalars {\lambda _1},{\lambda _2}\ldots {\lambda _n}, not all zero such that,
{\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} +\ldots + {\lambda _n}{\vec a_n} = \vec 0
For example, based on our previous discussions, we see that
(i) Two non-zero, non-collinear vectors are linearly independent.
(ii) Two collinear vectors are linearly dependent
(iii) Three non-zero, non-coplanar vectors are linearly independent.
(iv) Three coplanar vectors are linearly dependent
(v) Any four vectors in 3-D  space are linearly dependent.
You are urged to prove for yourself all these assertions.
     Example: 5      

Let \vec a,\vec b and \vec c be non-coplanar vectors. Are the vectors 2\vec a - \vec b + 3\vec c,\,\,\vec a + \vec b - 2\vec c and \vec a + \vec b - 3\vec c coplanar or non-coplanar?
Solution: 5

Three vectors are coplanar if there exist scalars \lambda ,\mu  \in \mathbb{R} using which one vector can be expressed as the linear combination of the other two.
Let us try to find such scalars:
2\vec a - \vec b + 3\vec c = \lambda \left( {\vec a + \vec b - 2\vec c} \right) + \mu \left( {\vec a + \vec b - 3\vec c} \right)
 \Rightarrow  \,\,\,\, \left( {2 - \lambda  - \mu } \right)\vec a + \left( { - 1 - \lambda  - \mu } \right)\vec b + \left( {3 + 2\lambda  + 3\mu } \right)\vec c = \vec 0
Since \vec a,\,\,\vec b,\,\,\vec c are non-coplanar, we must have
2 - \lambda  - \mu  = 0
 - 1 - \lambda  - \mu  = 0
3 + 2\lambda  + 3\mu  = 0
This system, as can be easily verified , does not have a solution for \lambda  and \mu .
Thus, we cannot find scalars for which one vector can be expressed as the linear combination of the other two, implying the three vectors must be non-coplanar.
As an additional exercise, show that for three non-coplanar vectors \vec a,\,\,\vec band \vec c, the vectors \vec a - \,2\vec b\, + 3\vec c,\,\,\vec a - 3\vec b + 5\vec c and  - 2\vec a - \,3\vec b\, - 4\vec c are coplanar.
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