VECTOR ALGEBRA | LINEAR COMBINATION LINEAR INDEPENDENCE AND LINEAR DEPENDENCE | Definition and physical interpretation: Linear Combination, Linear Combination: Linear Independence And Linear Dependence, Linearly Independent, Linearly Dependent, Theorem 1: If `veca` and `vecb` are two non collinear vectors; then every vector `vecr` coplanar with `veca` and `vecb` can be expressed in one and only one way as a linear combination: x`veca`+y`vecb`., Theorem 2: If `veca`, `vecb` and `vecc` are non coplanar vectors; then any vector `vecr` can be expressed as linear combination: x`veca`+y`vecb`+z`vecc`, Theorem 3:If vectors `veca`, `vecb` and `vecc` are coplanar then det(`veca` `vecb` `vecc`) = 0, Examples: Prove that the segment joining the middle points of two non parallel sides of a trapezium is parallel to the parallel sides and half of their sum., Components of a vector in terms of coordinates of its end points

Explanation

The vector `veca(vecaxxvecr)` is coplanar with `veca` and `vecr` is perpendicular to `veca`. Hence `vecr` can be expressed as a linear combination of `veca` and `veca(vecaxxvecr)`.
`:. Vecr=xveca+y(vecaxx(vecaxxvecr)),x,y` being scalars.
Taking scalar product with `veca`,
`vecr.veca=x(veca.veca)+yveca.{vecaxx(vecavecr)}`
`rArr vecr.veca=x|veca|^(2)rArr x=(vecr.veca)/(|veca|^(2))`
Again multiplying the expression for `vecr` with `veca` (vectorially)
`vecrxxveca=x(vecaxxveca)+y{vecaxx(vecaxxvecr)}xxveca`
`rArr vecrxxveca-y{(veca.vecr)veca-(veca.veca)vecr}xxveca`
`rArr vecrxxveca=-y(veca.veca)(vecrxxveca)`
`rArr vecrxxveca=y(veca.veca)(vecaxxvecr)`
`rArr (vecaxxvecr){y(veca.veca)+1}=0`
`vecaxxvecrne 0," so "y=-(1)/(veca.veca)`
So we have, `vecr=((vecr.veca)/(|veca|^(2)))veca+(-(1)/(veca.veca)){vecaxx(vecaxxvecr)}`
`=((vecr.veca)/(|veca|^(2)))veca+((vecaxxvecr)xxveca)/(|veca|^(2))`