Question

Prove that three vectors `bara, barb and barc ` are coplanar, if and only if, there exists a non-zero linear combination `xbara + ybarb + z barc = bar0`.

Answer 1

Let `bara, barb, barc` be coplanar vectors.

Case I: Suppose that any two of `bara, barb` and `barc` are collinear vectors, say `bara` and `barb`.

∴ There exist scalars x and y at least one of which is non-zero such that `xbara + ybarb = bar0`.

∴ `xbara + ybarb  + zbarc = bar0` is required non-zero linear combination where z = 0.

Case II: None of the two vectors `bara, barb` and `barc` are collinear.

Then any one of them, say `bara`, will be the linear combination of `barb and barc`.

∴ There exist scalars α and β such that `bara = α barb + β barc`

∴ `(-1)bara + α barb + β barc = bar0 `, i.e `xbara + ybarb + zbarc = bar0`

where x = –1, y = α, z = β which are not all zero simultaneously.

Conversely: Let there exist scalars x, y, z not all zero such that

`xbara + ybarb + zbarc`  ...(1)

Let x ≠ 0, then divide (1) by x, we get

i.e `bara + (y/x)barb + (z/x)barc = bar0`

∴ `bara = (-y/x)barb + (-z/x)barc`

 i.e. `bara = αbarb + βbarc, "where"  α = -y/x and β = -z/x` are scalars.

∴ `bara` is the linear combination of ` barb and barc.`

Hence, `bara, barb, barc` are coplanar.