Question

Find the Vector and Cartesian equations of the plane containing the line $\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}$ and passing through the point $(-1,1,-1)$

Answer 1

The required plane contains the line $\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}$ i.e., it passes through $(2,2,1)$ and is parallel to the vector $2\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k}$.
Also it is passing through $(-1,1,-1)$
$(x_1,y_1,z_1)=(2,2,1);(x_2,y_2,z_2)=(-1,1,-1);(l_1,m_1,n_1)$
$(2,3,-2)$
Vector Equation
The vector equation of the required plane is
$\overrightarrow{r}=(1-s)(-\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k})+s(2\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k})+t(2\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})$
$\overrightarrow{r}=-\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}+s(3\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k})+(2\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})$
Cartesian Equation:
The required cartesian equation is
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x-2& y-2& z-1\\ -3& -1& -2\\ 2& 3& -2\end{array}\right|=0$
$\Rightarrow (x-2)(2+6)-(y-2)(6+4)+(z-1)(-9+2)=0$
$\Rightarrow 8(x-2)-10(y-2)-7(z-1)=0$
$\Rightarrow 8x-16-10y+20-7z+7=0$
$\Rightarrow 8x-10y-7z+11=0$