Question

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

The plane contains the origin and is normal to the vector$\left(4,-1,5\right)$

Short answer

The equation of the plane that is determined by the given conditions is $4x-1y+5z=0$

Step-by-step solution

Given information

The plane that is normal to the vector$\left(4,-1,5\right)$ and contains the origin.

Calculation

Consider the plane that is normal to the vector$⟨4,-1,5⟩$ and contains the origin.

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:

$\mathbf{n}·\left(\mathbf{r}-{\mathbf{r}}_0\right)=0$

The normal vector is:

$\mathbf{n}=⟨4,-1,5⟩$

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

${\mathbf{r}}_{\mathbf{0}}=(0,0,0)$

Calculation

The plane that contains the origin and is normal to the vector $⟨4,-1,5⟩$ has the equation:

$$\begin{gathered} ⟨4,-1,5⟩·(⟨x,y,z⟩-⟨0,0,0⟩)=0\text{(Substitution)} \\ ⟨4,-1,5⟩·⟨x,y,z⟩=0\text{(Simplify)} \\ 4x-1y+5z=0 \end{gathered}$$

Thus, the equation of the plane that is determined by the given conditions is $4x-1y+5z=0$