Theorem 2: If a; b and c are non coplanar vectors; then any vector r can be expressed as linear combination: xa+yb+zc

Explanation

Let `bar(OB)` represent the vector `bar(r)` and `bar(OA),bar(OB),bar(OC)` represement the three non-coplanar vectors `bar(a),bar(b)andbar(c)` respectively.
Through P draw planes parallel to the planes BOC, COA and AOB intersecting the lines OA,OB and OC in L,M and N respectively.
Now `bar(OL)andbar(a)` are collinear vectors. Hence there exists a non - zero scalar x such that `bar(OL)=xbar(a)`.
Similarlt, `bar(OM)andbar(b)` are collinear and `bar(ON)andbar(c)` are collinear. Hence there exist bob-zero scalars y and z such that `bar(OM)-ybar(b)andbar(ON)=zbar(c)`.
Now, `bar(OP)=bar(OL)+bar(LP)`
`:.bar(r)=bar(OL)+bar(LQ)+bar(QP)`
`=bar(OL)+bar(OM)+bar(ON)`
`=xbar(a)+ybar(b)+zbar(c)`.
Thus, `bar(r)` is expressed as a linear combination `xbar(a)+ybar(b)+zbar(c)`.
Uniqueness :
Let, if possible, `bar(r)=x'bar(a)+y'bar(b)+z'bar(c)`, where x',y',z' are scalars. Then
`xbar(a)+ybar(b)+zbar(c)=x'bar(a)+y'bar(b)+z'bar(c)`
`:.(x-x')bar(a)+(y-y')bar(b)=(z'-z)bar(c)`
We note that uniqueness of the linear combination for `bar(r)` will be established if we shoe that `x=x',y=y'andz=z'`
Suppose on the contrary that `z!=z',i.e.,z'-z!=0`.
Then dividing both sides of (1) by `z'-z(!=0)`, we get,
`bar(c)=((x-x')/(z'-z))bar(a)+((y-y')/(z'-z))bar(b)`
This shows that `bar(c)` is expressed as a linear combination of `bar(a)andbar(b)`.
`:.bar(a),bar(b)andbar(c)` are coplanar. This is a contradiction, since `bar(a),bar(b)andbar(c)` are given to be non - coplanar.
`:.z=z'`
Similarly, we can show that `x=x'and y=y'`.
This proves the uniqueness of the linear combination `xbar(a)+ybar(b)+zbar(c)`.