# Find the vector and cartesian equations of the plane passing through the points $A(1 , -2 , 3 )$ and $B(-1 , 2 , -1 )$ and is parallel to the line $\large\frac{x-2}{2}=\frac{y+1}{3}=\frac{z-1}{4}.$

Toolbox:
• Equation of a plane through points $A(x_1, y_1, z_1)(pv \overrightarrow a)$ and $B(x_2, y_2, z_2)(pv\: \overrightarrow b)$ and parallel to a vector $\overrightarrow u$ (with D.r.s $l, m, n)$ $\overrightarrow r = (1-s) \overrightarrow a+s\overrightarrow b+t\overrightarrow v$ (vector form) $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n \end{vmatrix}=0$ ( cartesian form) $[ \overrightarrow r-\overrightarrow a\: \: \overrightarrow b-\overrightarrow a\: \: \overrightarrow u]=0$ ( Non parametre vector equation)
• The direction ratios of the normal to the plane $ax+by+cz=q$ (i.e., . $\overrightarrow r.(a\overrightarrow i+b\overrightarrow j+c\overrightarrow k)=q)$ are $(a, b, c)$
Step 1
The plane passes through the points $A(1, -2, 3) ( \overrightarrow a= \overrightarrow i-2 \overrightarrow j+3 \overrightarrow k)$ and
$B(-1, 2, -1) ( \overrightarrow b=- \overrightarrow i+2 \overrightarrow j- \overrightarrow k)$
It is parallel to the line $\large\frac{x-2}{2}=\large\frac{y+1}{3}=\large\frac{z-1}{4}$ i.e., to the vector $\overrightarrow v=2 \overrightarrow i+3 \overrightarrow j+4 \overrightarrow k$
Step 2
The vector equation of the plane is $\overrightarrow r=(1-s)( \overrightarrow i-2 \overrightarrow j+3 \overrightarrow k)+s(- \overrightarrow i+2 \overrightarrow j- \overrightarrow k)+t(2 \overrightarrow i+3 \overrightarrow j+4 \overrightarrow k)$
Step 3
The cartesian equation is $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n \end{vmatrix} = 0$
where $(x_1, y_1, z_1)=(1, -2, 3), (x_2, y_2, z_2) = (-1, 2, -1)$
$(l, m, n) = (2, 3, 4)$
$\therefore \begin{vmatrix} x-1 & y+2 & z-3 \\ -2 & 4 & -4 \\ 2 & 3 & 4 \end{vmatrix}=0$
$(x-1)(16+12)-(y+2)(-8+8)+(z-3)(-6-8)=0$
$28x-28-14z+42=0$
$28x-14z=-14$
$2x-z=-1$

edited Jun 25, 2013