Q 23. - Calculus

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 $⟨2,-1,6⟩$

Step-by-Step Solution

Answer

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

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

Step 1: Given information

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

Step 2: 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:

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

Step 3: 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

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