Chapter 7.3, Problem 52E

To determine

To sketch the plane represented by the linear equation and to list the points that lie in the plane

Explanation of Solution

Given information:

The linear equation is $5x+y+3z=15$

Graph:

When, $y=0,z=0$ then,

  $\begin{array}{l}5x+0+0=15\\ 5x=15\\ x=3\end{array}$

  $x$ -intercept: $(3,0,0)$

When, $x=0,z=0$ then,

  $\begin{array}{l}0+y+0=15\\ y=15\\ y=15\end{array}$

  $y$ -intercept: $(0,15,0)$

When, $x=0,y=0$ then,

  $\begin{array}{l}0+0+3z=15\\ 3z=15\\ z=3\end{array}$

  $z$ -intercept: $(0,0,3)$

Hence, the plane of the given equation is obtained by connecting all the three points and the plane thus obtained is the required graph of the equation.

The graph is obtained as:

  

Let, $y=3,z=2$ then,

  $\begin{array}{l}5x+3+2=15\\ 5x=15-5\\ 5x=10\\ x=2\end{array}$

Let, $x=3,z=2$ then,

  $\begin{array}{l}5(3)+y+3(2)=15\\ 15+y+6=15\\ y=15-21\\ y=-6\end{array}$

Let, $x=1,y=4$ then,

  $\begin{array}{l}5(1)+4+3z=15\\ 3z=15-9\\ 3z=6\\ z=2\end{array}$

The points $(2,-6,2)$ , $(2,3,2)$ , $(3,-6,2)$ and $(1,4,2)$ lie on the plane.

Interpretation:

From the above graph, it can be observed that the points $(2,-6,2)$ , $(2,3,2)$ , $(3,-6,2)$ and $(1,4,2)$ lie on the plane.