Answer: Proof below; \(t_1 = 2\), \(t_2 = 3\).
Existence: Let \(\overline{OA} = \bar a\), \(\overline{OB} = \bar b\), \(\overline{OR} = \bar r\) with O, A, B, R in one plane. Since \(\bar a\), \(\bar b\) are non-collinear, the lines OA and OB are not the same. Through R draw lines parallel to OB and OA meeting line OA at P and line OB at Q. Then OPRQ is a parallelogram, so
\(\overline{OR} = \overline{OP} + \overline{PR} = \overline{OP} + \overline{OQ}\).
\(\overline{OP}\) is collinear with \(\bar a\), so \(\overline{OP} = t_1\bar a\) for some scalar \(t_1\); \(\overline{OQ}\) is collinear with \(\bar b\), so \(\overline{OQ} = t_2\bar b\). Hence \(\bar r = t_1\bar a + t_2\bar b\). (If R lies on OA or OB, one scalar is zero.)
Uniqueness: Suppose also \(\bar r = s_1\bar a + s_2\bar b\). Subtracting, \((t_1 - s_1)\bar a + (t_2 - s_2)\bar b = \bar 0\). If \(t_1 \ne s_1\), then \(\bar a = -\dfrac{t_2 - s_2}{t_1 - s_1}\bar b\), making \(\bar a\), \(\bar b\) collinear, a contradiction. So \(t_1 = s_1\), and then \((t_2 - s_2)\bar b = \bar 0\) gives \(t_2 = s_2\) (\(\bar b \ne \bar 0\)). Hence the scalars are unique.
Finding \(t_1, t_2\): \(2\hat i + 7\hat j + 9\hat k = t_1(\hat i + 2\hat j) + t_2(\hat j + 3\hat k) = t_1\hat i + (2t_1 + t_2)\hat j + 3t_2\hat k\).
Comparing: \(t_1 = 2\); \(3t_2 = 9 \Rightarrow t_2 = 3\); check: \(2t_1 + t_2 = 4 + 3 = 7\).
So \(t_1 = 2\), \(t_2 = 3\).
