Question

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

The plane contains the points $(1,0,0),(0,1,0)$ and $(0,0,1)$

Short answer

Plane that contains the points $(1,0,0),(0,1,0)$and $(0,0,1)$is $x+y+z=1$

Step-by-step solution

Given information

The plane that contains the points $(1,0,0),(0,1,0)$and $(0,0,1)$

Calculation

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

The general form of the equation of the plane is:

$ax+by+cz=d$

The equation of the plane satisfying the points $(1,0,0),(0,1,0)$and $(0,0,1)$is obtained by substituting the points in $ax+by+cz=d$and Solve for the constants $a,b,c$and $d$

The point $(1,0,0)$gives:

$a\left(1\right)+b\left(0\right)+c\left(0\right)=d\text{(Substitution)}$

$a=d\text{(Simplify)}$

The point $(0,1,0)$gives:

$$\begin{gathered} a\left(0\right)+b\left(1\right)+c\left(0\right)=d\text{(Substitution)} \\ b=d\text{(Simplify)} \end{gathered}$$

The point $(0,0,1)$gives

$a\left(0\right)+b\left(0\right)+c\left(1\right)=d\text{(Substitution)}$

$c=d(\text{Simplify)}$

Calculation

Substitute the values of constants $a,b$ and $c$ in the general form.

The equation of the plane is:

$dx+dy+dz=d\text{(Substitution)}$

$d(x+y+z)=d$ (Take common factor out)

$x+y+z=1$

Thus, the plane that contains the points $(1,0,0),(0,1,0)$and $(0,0,1)$is $x+y+z=1$