Q 24. - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: 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:

$a x+b y+c z=d$

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 $a x+b y+c z=d$ 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

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

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

Step 3: 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 $x+y+z=1$