If `bar a and bar b` are any two non-zero and non-collinear vectors then prove that any vector `bar r ` coplanar with `bar a and bar b` can be uniquely expressed as `bar r=t_1bara+t_2barb` , where ` t_1 and t_2` are scalars
Let ` bar r = bar(OR)` be any vector in a plane.
Let bara and barb be two non-zero non-collinear vectors in the same plane.
We have to prove that barr can be expressed as a linear combination of bara and barb and the linear combination is unique.
Passing through the point O, draw a line parallel to `bara`, and passing through the point R draw another line || to `barb` and let them intersect at the point P.
`therefore bar(OP)|| bara and bar(PR)|| barb`
`therefore bar(OP)=t_1bara and bar(PR)=t_2 barb`
(where ` t_1 and t_2` are any two scalars)
In triangle OPR by triangle law
`bar(OR)=bar(OP)+bar(PR) " i.e " bar r=t_1bara+t_2barb`
Thus barr is expressed as a linear combination of `bar a and bar b `
To prove the uniqueness of the linear combination.
If possible let ` alpha' and beta'` be two scalars
Such that: ` bar r =alpha' a +beta' b ,`
`where t_1 ne alpha' and t_2 ne beta'`
`therefore " we get " t_1bara +t_2 barb=alpha'bara+beta'barb`
Dividing throughout by `t_1-alpha' ` we get :
This is of the form `bar a = k barb` , which shows that `bar a & bar b` are collinear which is a contradiction.
Our assumption that `t_1 ne alpha'` is wrong.
`t_1=alpha'` . Similarly, we can prove that `t_2=beta'`
∴ The linear combination is unique.