LinAlg4.2.22

From Exampleproblems

Find the vector equation of the line passing through three non-collinear points $-2\bar{i}+6\bar{j}-6\bar{k},-3\bar{i}+10\bar{j}-9\bar{k},-5\bar{i}-6\bar{k}\,$ .Also find its cartesian equation.

Let $P=\bar{r}\,$ be any point in the plane ABC.

Now the vector equation of the plane is $(\bar{r}-\bar{a})\cdot(\bar{b}-\bar{a})\times(\bar{c}-\bar{a})=0\,$

Now $\bar{b}-\bar{a}=-\bar{i}+4\bar{j}-3\bar{k},\bar{c}-\bar{a}=-3\bar{i}-6\bar{j}\,$

Therefore $(\bar{b}-\bar{a})\times(\bar{c}-\bar{a})=\begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ -1 & 4 & -3 \\ -3 & -6 & 0 \end{vmatrix} \,$ which is equal to