Question

Find the equations of the planes determined by the given conditions.

The plane contains the point $(-2,3,-1)$and is normal to the vector $3i-2j$

Short answer

The equation of the plane that is determined by the given conditions is $3x-2y+12=0$

Step-by-step solution

Given information

The plane that contains the point $(-2,3,-1)$ and is normal to the vector $3\mathbf{i}-2\mathbf{j}$

Calculation

The goal is to determine the plane equation that is determined by the given conditions.

The equation of the plane with ${\mathbf{r}}_{\mathbf{0}}$ a point that lies in the plane and $\mathbf{n}$ a vector normal to the plane is given by:

$\mathbf{n}·\left(\mathbf{r}-{\mathbf{r}}_{\mathbf{0}}\right)=0$

The normal vector is:

$\mathbf{n}=⟨3,-2,0⟩$

The point ${\mathbf{r}}_{\mathbf{0}}$is:

${\mathbf{r}}_0=(-2,3,-1)$

Calculation

The equation of the plane that contains the point $(-2,3,-1)$ and is normal to the vector $3\mathbf{i}-2\mathbf{j}$ is:

$$\begin{gathered} ⟨3,-2,0⟩·(⟨x,y,z⟩-(-2,3,-1\left)\right)=0\text{(Substitution)} \\ 3x-2y-\left(3\right(-2)-2(3)+0(-1\left)\right)=0\text{(Simplify)} \\ 3x-2y-(-6-6+0)=0 \\ 3x-2y+12=0 \end{gathered}$$

Thus, the equation of the plane that is determined by the given conditions is $3x-2y+12=0$