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

#### Solution

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`

`(t_1alpha')bara=(t_2-beta')barb`

Dividing throughout by `t_1-alpha' ` we get :

`bar a=((beta'-t_2)/(t_1-alpha'))barb`

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.