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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Find the line of intersection,
L, of the planes given by
Lets graph this data on paper. To plot plane P1, we can use its intercepts on the axes. How can we find its x-intercept?
Since
for all points on the
x-axis, we can substitute those zero values into the equation of the plane and solve for
.
Do that.
Find the y-intercept in a similar way.
Find the z-intercept in a similar way.
Show these points on a graph and connect them to show plane P1. Draw the x-axis out of the plane of the paper, as usual. Then check your graph by clicking on Next.
Determine the intercepts for plane P2.
We get
Show these points on a new graph and connect them to show plane P2. Then check your graph by clicking on Next.
How is L, the line of intersection of the planes, related geometrically to the planes?
It lies in each of them.
How is this line related to the normals to the plane?
Since the normals
are perpendicular to the planes, they must be perpendicular to every line in the planes.
What is the consequence for the line L?
It must be perpendicular to the normals.
Now consider the Cross Product of the normals. How must its direction be related to the directions of the normals?
The vector resulting from the Cross Product is always perpendicular to the vectors used in its calculation.
What can we conclude about the direction of L and the direction of the Cross Product of the normals?
Since each is perpendicular to both normals, they must be parallel. So here is our strategy. Determine the normals, get their Cross Product, and construct a line parallel to the Cross Product and contained in each plane. Can we determine the normals from the given information?
Yes.
How?
The equation of each plane is given in the form
,
where
a, b, c are the direction numbers of the normal.
State
in terms of its components.
State
in terms of its components.
Now we are ready to find the cross product of these two vectors. Set up
in determinant form.
Evaluate the determinant.
We get
Simplify.
We get
We have the vector, V, the Cross Product of the normals to the two planes. It is perpendicular to each normal and is parallel to the line of intersection.
Now we need a point A on the line. Set
x = 0 in the equation of each plane and solve for
. Apply this idea to plane
P1.
We get
Do the same for plane P2.
We get
Solve the equations simultaneously.
We get
State the coordinates of point A.
Point A: (0, 0, 4)
Use this point to set up the parametric equations of the line.
Using the components of V (which is parallel to the line L), we get
.
Find a second point, B, by substituting t = 3.
We get B: (6, -3, -2)
Plot the two points and show the line of intersection.
Add the two planes to this diagram.
How can we check these results?
Since these points are on both planes, we can check their coordinates in the equation of each plane.
Check point A (0,0,4).
For each plane, we get
, which is correct.
Check point B (6, – 3, – 2).
For plane
P1:
For plane
P2:
These are correct.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
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