Question

Find the vector equation of the plane passing through the points $(2,5,-3),(-2,-3,5)$ and $(5,3,-3)$.

Answer 1

Let the given points be $A(2,5,-3),B(-2,-3,5),C(5,3,-3)$

Let $\overrightarrow{a}=2\hat{i}+5\hat{j}-3\hat{k},\overrightarrow{b}=-2\hat{i}-3\hat{j}+5\hat{k},\overrightarrow{c}=5\hat{i}+3\hat{j}-3\hat{k}$

Hence the vector equation of the plane passing through the points is given by $(\overrightarrow{r}-\overrightarrow{a})\cdot (\overrightarrow{AB}\times \overrightarrow{AC})=0$

$\Rightarrow (\overrightarrow{r}-\overrightarrow{a})\cdot \left(\right(\overrightarrow{b}-\overrightarrow{a})\times (\overrightarrow{c}-\overrightarrow{a}\left)\right)=0$

$\Rightarrow (\overrightarrow{r}-\overrightarrow{a})\cdot \left(\right(-4\hat{i}-8\hat{j}+8\hat{k})\times (3\hat{i}-2\hat{j}\left)\right)=0$

$\Rightarrow (\overrightarrow{r}-\overrightarrow{a})\cdot \left|\begin{array}{ccc}i& j& k\\ -4& -8& 8\\ 3& -2& 0\end{array}\right|=0$

$\Rightarrow (\overrightarrow{r}-\overrightarrow{a})\cdot (16\hat{i}+24\hat{j}+32\hat{k})=0$

$\therefore (\overrightarrow{r}-(2\hat{i}+5\hat{j}-3\hat{k}\left)\right)\cdot (16\hat{i}+24\hat{j}+32\hat{k})=0$ is the required equation of the plane.