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

Answer 1

Toolbox:

• Equation of the plane through a point $A(x_1, y_1, z_1)(pv \: is \: \overrightarrow a)$ and parallel to vectors $ \overrightarrow u$ and $\overrightarrow v$ $ \overrightarrow r=\overrightarrow a+s\overrightarrow u+t\overrightarrow v$ where $t,s$ are scalars (vector form) $ \begin{vmatrix}x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ (cartesian form) where $\overrightarrow u=l_1\overrightarrow i+m_1\overrightarrow j+n_1\overrightarrow k, \: \overrightarrow v=l_2\overrightarrow i+m_2\overrightarrow j+n_2\overrightarrow k$ Non-para meter vector equation : $ [ \overrightarrow r\: \overrightarrow u  \: \overrightarrow v]=[\overrightarrow a\: \overrightarrow u\: \overrightarrow v]$
• The line $ \large\frac{x-x_1}{l}=\large\frac{y-y_1}{m}=\large\frac{z-z_1}{n}$ contains the point $(x_1, y_1, z_1) $ and is parallel to the vector $ l\overrightarrow i+m\overrightarrow j + n\overrightarrow k$

Step 1

The required plane contains the line $ \large\frac{x-2}{2}=\large\frac{y-2}{3}=\large\frac{z-1}{3}$

$ \therefore $ it contains the point $A(p, v \: \overrightarrow a=2\overrightarrow i+2\overrightarrow j+\overrightarrow k)$ that lies on the line and it is parallel to the vector $ \overrightarrow u=2\overrightarrow i+3\overrightarrow j+3\overrightarrow k$

The plane is also parallel to the line $ \large\frac{x+1}{3}=\large\frac{y-1}{2}=\large\frac{z+1}{1}$ ans so it is parallel to $ \overrightarrow v=3\overrightarrow i+2\overrightarrow j+\overrightarrow k$

Step 2

The vector equation of the plane is of the form

$ \overrightarrow r=\overrightarrow a+s\overrightarrow u+t\overrightarrow v$

$\overrightarrow r=2\overrightarrow i+2\overrightarrow j+\overrightarrow k+s(2\overrightarrow i+2\overrightarrow j+3\overrightarrow k)+t(3\overrightarrow i+2\overrightarrow j+\overrightarrow k)$

Step 3

The cartesian equation is

$ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$

$ (x_1, y_1, z_1) = (2, 2, 1), (l_1, m_1, n_1)=(2, 3, 3), (l_2, m_2, n_2) = (3, 2, 1)$

$ \therefore \begin{vmatrix} x-2 & y-2 & z-1 \\ 2 & 3 & 3 \\ 3 & 2 & 1 \end{vmatrix}=0$

$(x-2)(3-6)-(y-2)(2-9)+(z-1)(4-9)=0$

$ -3x+6+7y-14-5z+5=0$

$ -3x+7y-5z-3=0$

or $ 3x-7y+5z=-3$