Question

If `bara and barb` are any two non-zero and non-collinear vectors then prove that any vector `barr` coplanar with  `bara` and `barb` can be uniquely expressed as `barr = t_1bara + t_2barb` , where t₁ and t₂ are scalars.

Answer 1

Let `bara, barb, barr` be coplanar.

Take any point O in the plane of `bara, barb` and `barr`.

Represents the vectors `bara, barb` and `barr` by `bar(OA), bar(OB)` and `bar(OR)`.

Take the point P on `bara` and Q on `barb` such that OPRQ is a parallelogram.

Now, `bar(OP)` and `bar(OA)` are collinear vectors.

∴ There exists a non-zero scalar t₁ such that

`bar(OP) = t_1 * bar(OA) = t_1 * bara`

Also, `bar(OQ)` and `bar(OB)` are collinear vectors.

∴ There exists a non-zero scalar t₂ such that

`bar(OP) = t_2 * bar(OB) = t_2 * barb`

Now, by parallelogram law of addition of vectors,

`bar(OR) = bar(OP) + bar(OQ)`

∴ `barr = t_1bara + t_2barb`

Thus, `barr` is expressed as a linear combination `t_1bara + t_2barb`.

Uniqueness: Let, if possible, 

`barr = t_1^'bara + t_2^'barb`, where t₁', t₂' are non-zero scalars.

Then, `t_1bara + t_2barb = t_1^'bara + t_2^'barb`

∴ `(t_1 - t_1^')bara = -(t_2 - t_2^')barb`   .....(1)

We want to show that t₁ = t₁' and t₂ = t₂'

Suppose t₁ ≠ t₁', i.e. t₁ – t₁' ≠ 0 and t₂ ≠ t₂, i.e. t₂ – t₂ ≠ 0

Then dividing both sides of (1) by t₁ – t₁', we get

`bara = -((t_2 - t_2^')/(t_1 - t_1^'))barb`

This shows that the vector `bara` is a non-zero scalar multiple of `barb`.

∴ `bara` and `barb` are collinear vectors.

This is a contradiction, since `bara, barb` are given to be non-collinear.

∴ t₁ = t₁'

Similarly, we can show that t₂ = t₂'

This shows that `barr` is uniquely expressed as a linear combination `t_1bara + t_2barb`.

Conversely: Let `barr = t_1bara + t_2barb`, where t₁, t₂ are scalars.

Since `bara, barb` are coplanar, `t_1bara, t_2barb` are also coplanar.

∴ `barr = t_1bara + t_2barb` is coplanar with `bara` and `barb`.