# LinAlg4.1.16

Obtain the point of intersection of the line joining the points $\bar{i}-2\bar{j}-\bar{k},2\bar{i}+3\bar{j}+\bar{k}\,$ with the plane through the points $2\bar{i}+\bar{j}-3\bar{k},4\bar{i}-\bar{j}+2\bar{k}\,$ and $3\bar{i}+\bar{k}\,$

The vector equation of the line joining the points $\bar{i}-2\bar{j}-\bar{k},2\bar{i}+3\bar{j}+\bar{k}\,$ is $\bar{r}=(1-t)(\bar{i}-2\bar{j}-\bar{k})+t(2\bar{i}+3\bar{j}+\bar{k})=(1+t)\bar{i}+(5t-2)\bar{j}+(2t-1)\bar{k}\,$

Vector equation of the plane through the given points is $\bar{r}=(1-s-p)(2\bar{i}+\bar{j}-3\bar{k})+s(4\bar{i}-\bar{j}+2\bar{k})+p(3\bar{i}+\bar{k})=(2+2s+p)\bar{i}+(1-2s-p)\bar{j}+(5s+4p-3)\bar{k}\,$

Solving the above two equations, $2+2s+p=1+t,2s+p+5t=3,5s+4p-2t=2\,$

Solving the first two of the above three equations,we get $t=\frac{2}{3}\,$

Hence the point of intersection is $\bar{r}=\frac{1}{3}(5\bar{i}+4\bar{j}+\bar{k})\,$

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