Q 21. - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: 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:

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

Step 3: 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 $4 x-1 y+5 z=0$