If `bar(a)` and `bar(b)` any two non-collinear vectors lying in the same plane, then prove that any vector `bar(r)` coplanar with them can be uniquely expressed as `bar(r)=t_(1)bar(a)+t_(2)bar(b)`, where `t_(1)andt_(2)` are scalars.

Explanation

Take any points O in the plane of `bar(a),bar(b)andbar(r)`. Represents the vectors `bar(a),bar(b)andbar(r)` by `bar(OA),bar(OB)andbar(OR)`. Take the points P on `bar(a)` and Q `bar(b)` such that OPRQ is a parallelogram.
Now `bar(OP)andbar(OA)` are collinear vectors.
`:.` there exists a non - zero scalar `t_(1)` such that
`bar(OP)=t_(a)*bar(OA)=t_(1)*bar(a)`.
Also `bar(OQ)andbar(OB)` are collinear vectors.
`:.` there exixts a non-zero scalar `t_(2)` such that `bar(OQ)=t_(2)*bar(OB)=t_(2)*bar(b)`.
Now, by parallelogram law of addition of vectors,
`bar(OR)=bar(OP)+bar(OQ)" ":.bar(r)=t_(1)bar(a)+t_(2)bar(b)`
Thus `bar(r)` expressed as linear combination `t_(1)bar(a)+t_(2)bar(b)`
Uniqueness:
Let, if possible, `bar(r)=t_(1)^(')bar(a)+t_(2)^(')bar(b)`, where `t_(1)^('),t_(2)^(')` are non-zero scalars. Then
`t_(1)bar(a)+t_(2)bar(b)=t_(1)^(')bar(a)+t_(2)^(')bar(b)`
`:.(t_(1)-t_(1)^('))bar(a)=-(t_(2)-t_(2)^('))bar(b)` . . . . (1)
WE want to show that `t_(1)=t_(1)^(')andt_(2)=t_(2)^(')`.
Suppose `t_(1)!=t_(1)^('),i.e.,t_(1)-t_(1)^(')!=0andt_(2)!=t_(2)^(')!=0`.
Then dividing both sides of (1) by `t_(1)-t_(1)^(')`, we get,
`bar(a)=-((t_(2)-t_(2)^('))/(t_(1)-t_(1)^(')))bar(b)`
This shows that the vector `bar(a)` is a non-zero scalar multiple of `bar(b)`.
`:.bar(a)andbar(b)` are collinear vectors.
This is a contradiction, since `bar(a),bar(b)` are given to be non-collinear.
`:.t_(1)=t_(1)^(')`
Similarly, we can show tath `t_(2)=t_(2)^(')`
This shows that `bar(r)` is uniquwly expressed as a linear combination `t_(1)bar(a)+t_(2)bar(b)`.