Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Concept Byte: Linear Programming
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Exercise 23 Page 165

To graph a plane, you will need at least two lines contained in the plane. Begin by determining the lines for each plane.

Graph:

Intersection: Line

Practice makes perfect

Each of the equations represents a plane. Before we begin, remember that two lines are enough to define a plane. Therefore, we will need two lines for each plane in order to graph them. -2x-3y+5z=7 & (I) 2x-3y-4z=-4 & (II) Let's graph each plane separately.

Plane (I)

The lines contained in Plane (I) can be found by substituting 0 for one of its variables. The resulting equation represents a line. Let's first find the equation of the line that passes through x= 0.

-2x-3y+5z=7
-2( 0)-3y+5z=7
0-3y+5z=7
-3y+5z=7

Now that we know the equation of the line, we can use its intercepts to graph it. For the y-intercept, we will substitute z= 0. Similarly, for the z-intercept, we will substitute y= 0.

y-intercept z-intercept
Substitute -3y+5( 0)=7 -3( 0)+5z=7
Calculate y= - 7/3 z= 7/5
Point ( 0, - 7/3, 0) ( 0, 0, 7/5)
Next, we can plot the intercepts on a coordinate space and draw the line through them.

Now we will find the equation of another line on the plane. This time, let's look for the line that passes through y= 0.

-2x-3y+5z=7
-2x-3( 0)+5z=7
- 2x-0+5z=7
-2x+5z=7

To graph the line, we will find its intercepts by substituting z= 0 for the x-intercept and x= 0 for the z-intercept.

x-intercept z-intercept
Substitute -2x+5( 0)=7 -2( 0)+5z=7
Calculate x= - 3.5 z= 1.4
Point ( - 3.5, 0, 0) ( 0, 0, 1.4)

Now that we know the intercepts, let's graph the second line!

Finally, we have two lines to graph a plane.

Plane (II)

We will graph Plane (II) in the same way as we graphed Plane (I). Let's first find two lines such that one line passes through x= 0 and the other through y= 0.

Substitution Resulting Equation
x=0 2( 0)-3y-4z=-4 -3y-4z=-4
y=0 2x-3( 0)-4z=-4 2x-4z=-4

Next, we will find the intercepts of each line.

-3y-4z=-4 2x-4z=-4
Intercept y-intercept z-intercept x-intercept z-intercept
Substitution -3y-4( 0)=-4 -3( 0)-4z=-4 2x-4( 0)=-4 2( 0)-4z=-4
Calculation y= 4/3 z= 1 x= - 2 z= 1
Point ( 0, 4/3, 0) ( 0, 0, 1) ( - 2, 0, 0) ( 0, 0, 1)

Now that we know the intercepts of both lines, we can graph them.

We can graph the plane that contains the lines.

Combining the Graphs

In the final step, we will graph the planes in the same coordinate system.

Two planes can be parallel or not parallel. If the planes are parallel, then they do not intersect. Conversely, if the planes are not parallel, then their intersection is a line. Looking at the graph, we can see that these two planes are not parallel. Therefore, we can conclude that their intersection is a line.