Q.Let $\bar a$ and $\bar b$ be non-collinear vectors. If vector $\bar r$ is coplanar with $\bar a$ and $\bar b$ then show that there exist unique scalars $t_1$ and $t_2$ such that $\bar r=t_1\bar a+t_2\bar b$. For $\bar r=2\hat i+7\hat j+9\hat k$, $\bar a=\hat i+2\hat j$, $\bar b=\hat j+3\hat k$, find $t_1, t_2$.

Existence/uniqueness follows from $\bar a,\bar b$ being a basis for the plane they span; then match components for the specific vectors given.

Proof of existence and uniqueness. Since $\bar a,\bar b$ are non-collinear, they are linearly independent and span a plane $\pi$. Since $\bar r$ is coplanar with $\bar a,\bar b$, $\bar r$ lies in $\pi$, so it can be expressed as $\bar r=t_1\bar a+t_2\bar b$ for some scalars $t_1,t_2$ (existence, by the parallelogram/plane-spanning argument).

Uniqueness: Suppose also $\bar r=t_1'\bar a+t_2'\bar b$. Then

$t_1\bar a+t_2\bar b=t_1'\bar a+t_2'\bar b \implies (t_1-t_1')\bar a=(t_2'-t_2)\bar b$

If $t_1\ne t_1'$, then $\bar a=\dfrac{t_2'-t_2}{t_1-t_1'}\bar b$, i.e. $\bar a$ is a scalar multiple of $\bar b$ — meaning $\bar a,\bar b$ are collinear, contradicting the hypothesis. Hence $t_1=t_1'$, and then $(t_2'-t_2)\bar b=\bar 0$ with $\bar b\ne\bar 0$ forces $t_2=t_2'$. So $t_1,t_2$ are unique. $\blacksquare$

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