Question

Find the vector equation of the plane passing through the points $4\overline{i}-3\overline{j}-\overline{k},3\overline{i}+7\overline{j}-10\overline{k}$ and $2\overline{i}+5\overline{5}-7\overline{k}$ and find if the point $\overline{i}+2\overline{j}-3\overline{k}$ lies on the plane.

Answer 1

The vector equation of plane through point with position vector

$\overline{a}=4\hat{i}-3\hat{j}-\hat{k}$

$\overline{b}=3\hat{i}+7\hat{j}-10\hat{k}$

$\overline{c}=2\hat{i}+5\hat{j}-7\hat{k}$

$=\left(1-s-t\right)\overline{a}+s\overline{b}+t\overline{c}$ s and t is scalar

$\overline{r}=\left(1-s-t\right)\left(4\hat{i}-3\hat{j}-\hat{k}\right)+s\left(3\hat{i}+t\hat{j}-10\hat{k}\right)+t\left(2i+5j-tk\right)$

$\overline{r}=4\hat{i}-3\hat{j}-\hat{k}+s\left(-\hat{i}+10\hat{j}-9\hat{k}\right)+t\left(-2\hat{i}+58-\widehat{6k}\right)$

to find if point $\hat{i}+2\hat{j}-3\hat{k}$ lies on the above plane

$\hat{i}+2\hat{j}-3\hat{k}=s\left(-\hat{i}+10\hat{j}-9\hat{k}\right)+t\left(-2\hat{i}+58-\widehat{6k}\right)$

$-3\hat{i}+5\hat{j}-2\hat{k}=s\left(-\hat{i}+10\hat{j}-9\hat{k}\right)+t\left(-2\hat{i}+58-\widehat{6k}\right)$

i , j ,k coefficient

-3 = -5 -2t is equation (1)

5 = 10s+8t is equation (2)

-2 = -9s -6t is equation (3)

solve equation (1) and (2)

$s=\frac{-7}{6}t=\frac{14}{6}$

RHS of equation $\left(3\right)-9s-6t$

$\begin{array}{c}=-9\left(\frac{-7}{6}\right)-6\left(\frac{14}{6}\right)\\ \frac{63-84}{6}=-\frac{-21}{6}=\frac{-7}{2}\\ \frac{-7}{2}\ne -2\end{array}$

$\left(\hat{i}+2\hat{j}-3\hat{k}\right)$ does not lie on above plane.