ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 8

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

$\lambda \vec a + \mu \vec b + \gamma \vec c = \vec 0$

where ${\lambda ,\mu ,\gamma \in\mathbb{R} }$

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 $\vec a,\,\,\vec b$ and $\vec c$ , 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.

Friday, 8 August 2014

CHAPTER 8 - Coplanar Vectors

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