# Find the equation of the plane through the point whose $p.v.$ is $\overrightarrow{2i}-\overrightarrow{j}+\overrightarrow{k}$ and perpendicular to the vector $\overrightarrow{4i}+\overrightarrow{2j}-\overrightarrow{3k}$

Toolbox:
• Equation of the plane through a point $A(x_1, y_1, z_1) [pv=\overrightarrow a]$ and perpendicular to vector $\overrightarrow n = a \overrightarrow i+b \overrightarrow j+c \overrightarrow k$
• $\overrightarrow r. \overrightarrow n = \overrightarrow a. \overrightarrow n$ (vector equation )
• $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ (cartesian equation)
Step 1
The point $A$ with $PV$ $\overrightarrow a=2\overrightarrow i-\overrightarrow j+\overrightarrow k$ lies in the plane and $\overrightarrow n=4\overrightarrow i+2\overrightarrow j-3\overrightarrow k$ is normal the plane.
Step 2
The vector equation of the plane is
$\overrightarrow r.\overrightarrow n=\overrightarrow a.\overrightarrow n$
$\overrightarrow r.(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)=(2\overrightarrow i-\overrightarrow j+\overrightarrow k).(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)$
$\overrightarrow r.(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)=8-2-3$
$\overrightarrow r.(4\overrightarrow i+2\overrightarrow j-3\overrightarrow k)=3$
$\overrightarrow r=x\overrightarrow i+y\overrightarrow j+z\overrightarrow k$
Step 3
The cartesian equation is
$4x+2y-3y=3$