Question

Find the vector and cartesian equations of the plane containing the line $\frac{x-2}{2}=\frac{y-2}{2}=\frac{z-1}{3}$ and parallel to the line $\frac{x+1}{3}=\frac{y-1}{2}=\frac{z+1}{1}$.

Answer 1

Since the required plane containing the line
$\frac{x-2}{2}=\frac{y-2}{2}=\frac{z-1}{3}$, it contains the point $(2,2,1)$ and parallel to a vector $2\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}$
Again, the plane is parallel to the line $\frac{x+1}{3}=\frac{y-1}{2}=\frac{z+1}{1}$
$\therefore$ the plane is parallel to $3\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}$
The equation of a plane passing through point whose $p.v$ is $\overrightarrow{a}$ and parallel to the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ is
$\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{u}+t\overrightarrow{v}$
Where $\overrightarrow{a}=2\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}$
$\overrightarrow{u}=2\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}$
$\overrightarrow{v}=3\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}$
$\therefore \overrightarrow{r}=(2\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k})+s(2\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k})+t(3\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k})$
Cartesian Form
$(x_1,y_1,z_1)$
$(l_1,m_1,n_1)$
$(l_2,m_2,n_2)$
The equation of the plane is $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x-2& y-2& z-1\\ 2& 3& 3\\ 3& 2& 1\end{array}\right|=0$
$\Rightarrow (x-2)(3-6)-(y-2)(2-9)+(z-1)(4-9)=0$
$\Rightarrow 3x+6+7y-14-5z-3=0$
$\Rightarrow$ $3x-7y+5z+3=0$