Question

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

The plane contains the point $(2,-1,6)$and is normal to the vector$⟨2,-1,6⟩$

Short answer

The equation of the plane that is determined by the given conditions is

$2x-y+6z-41=0\text{.}$

Step-by-step solution

Given information

The plane that contains the point $(2,-1,6)$and is normal to the vector $⟨2,-1,6⟩$

Calculation

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

The equation of the plane with ${\mathbf{r}}_0$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:

$\mathbf{n}=⟨2,-1,6⟩$

The point ${\mathbf{r}}_0$is:

${\mathbf{r}}_0=(2,-1,6)$

Calculation

The equation of the plane that contains the point $(2,-1,6)$and is normal to the vector $⟨2,-1,6⟩$is:

$⟨2,-1,6⟩·(⟨x,y,z⟩-(2,-1,6\left)\right)=0$(Substitution)

$$\begin{gathered} 2x-y+6z-\left(2\right(2)-1(-1)+6(6\left)\right)=0\text{(Simplify)} \\ 2x-y+6z-(4+1+36)=0 \\ 2x-y+6z-41=0 \end{gathered}$$

Thus, the equation of the plane that is determined by the given conditions is

$2x-y+6z-41=0\text{.}$