Question

Find the vector equation of the plane passing through the point $R(2,5,-3)S(-2,-3,5)$ and $T(5,3,-3)$

Answer 1

$R(2,5,-3),S(-2,-3,5)$ and $T(5,3,-3)$

$\overrightarrow{n}$ be direction of normal to plane.

$\overrightarrow{n}=\overrightarrow{RS}\times \overrightarrow{TS}=(-4\hat{i}-8\hat{j}+8\hat{k})\times (7\hat{i}+6\hat{j}-8\hat{k})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -4& -8& 8\\ 7& 6& -8\end{array}\right|=16\hat{i}+24\hat{j}+32\hat{k}$

$\therefore$ equation $\Rightarrow \overrightarrow{r}.(16\hat{i}+24\hat{j}+32\hat{k})=p$

$\overrightarrow{r}=2\hat{i}+5\hat{j}-3\hat{k}$ satisfies equation.

$(2\hat{i}+5\hat{j}-3\hat{k}).(16\hat{i}+24\hat{j}+32\hat{k})=p\Rightarrow p=56\therefore \overrightarrow{r}.(16\hat{i}+24\hat{j}+32\hat{k})=56$