Prove that, if \(\bar{a}, \bar{b}, \bar{c}\) are three non coplanar vectors then any vector \(\bar{r}\) in the space can be uniquely expressed as a linear combination of \(\bar{a}, \bar{b}, \bar{c}\).

Answer: Proof (existence by a parallelepiped construction, uniqueness by non-coplanarity).

Existence. Take an origin O and let \(\overline{OA} = \bar{a}\), \(\overline{OB} = \bar{b}\), \(\overline{OC} = \bar{c}\), \(\overline{OP} = \bar{r}\). Since \(\bar{a}, \bar{b}, \bar{c}\) are non coplanar, OA, OB, OC are not in one plane. Through P draw planes parallel to the planes OBC, OCA and OAB, meeting the lines OA, OB, OC at L, M, N. This gives a parallelepiped with OP as a diagonal. [Figure: parallelepiped with edges OL, OM, ON along OA, OB, OC and diagonal OP.]

Then \(\overline{OP} = \overline{OL} + \overline{OM} + \overline{ON}\). As \(\overline{OL}\) is collinear with \(\bar{a}\), \(\overline{OL} = x\bar{a}\) for some scalar \(x\); similarly \(\overline{OM} = y\bar{b}\), \(\overline{ON} = z\bar{c}\). Hence \(\bar{r} = x\bar{a} + y\bar{b} + z\bar{c}\).

Uniqueness. Suppose also \(\bar{r} = x_1\bar{a} + y_1\bar{b} + z_1\bar{c}\). Subtracting, \((x - x_1)\bar{a} + (y - y_1)\bar{b} + (z - z_1)\bar{c} = \bar{0}\).

If \(x - x_1 \ne 0\), then \(\bar{a} = -\frac{y - y_1}{x - x_1}\bar{b} - \frac{z - z_1}{x - x_1}\bar{c}\), so \(\bar{a}\) would be a linear combination of \(\bar{b}\) and \(\bar{c}\) and the three vectors would be coplanar, a contradiction. So \(x = x_1\). In the same way \(y = y_1\) and \(z = z_1\).

Hence \(\bar{r}\) is expressed uniquely as \(x\bar{a} + y\bar{b} + z\bar{c}\).