Let \(\bar a\) and \(\bar b\) be non-collinear vectors. If vector \(\bar r\) is co-planar with \(\bar a\) and \(\bar b\) then prove that there exists unique scalars \(t_1\) and \(t_2\) such that \(\bar r=t_1\bar a+t_2\bar b\). Hence find \(t_1\) and…

Answer: Proved; \(t_1=1\), \(t_2=-1\).

Existence: Take \(O\) as origin and let \(\overline{OA}=\bar a\), \(\overline{OB}=\bar b\), \(\overline{OP}=\bar r\), all in one plane. Through \(P\) draw lines parallel to \(OB\) and \(OA\), meeting line \(OA\) at \(M\) and line \(OB\) at \(N\). Then \(OMPN\) is a parallelogram and \(\overline{OP}=\overline{OM}+\overline{MP}=\overline{OM}+\overline{ON}\).

\(\overline{OM}\) is collinear with \(\bar a\), so \(\overline{OM}=t_1\bar a\); \(\overline{ON}\) is collinear with \(\bar b\), so \(\overline{ON}=t_2\bar b\). Hence \(\bar r=t_1\bar a+t_2\bar b\).

Uniqueness: Suppose also \(\bar r=s_1\bar a+s_2\bar b\). Subtracting, \((t_1-s_1)\bar a+(t_2-s_2)\bar b=\bar0\). If \(t_1\ne s_1\), then \(\bar a=-\dfrac{t_2-s_2}{t_1-s_1}\bar b\), so \(\bar a,\bar b\) would be collinear, a contradiction. So \(t_1=s_1\), and then \((t_2-s_2)\bar b=\bar0\) gives \(t_2=s_2\) (\(\bar b\ne\bar0\)). Hence \(t_1,t_2\) are unique.

Hence part: \(\hat i+\hat j=t_1(2\hat i-\hat j)+t_2(\hat i-2\hat j)=(2t_1+t_2)\hat i+(-t_1-2t_2)\hat j\).

\(2t_1+t_2=1\) and \(-t_1-2t_2=1\). From the first, \(t_2=1-2t_1\); then \(-t_1-2+4t_1=1\Rightarrow t_1=1\), \(t_2=-1\).

Check: \((2\hat i-\hat j)-(\hat i-2\hat j)=\hat i+\hat j\).