The line intercept of two planes

[ThomasRiley]

I am trying to solve for the line that intersects any planes that fall under the arithmetic pattern of ax+(a+n)y+(a+2n)z=a+3n. The sample equations I've been using is (A) x+2y+3z=3, (B) 2x-1y-4y=-7 and (C) 6x-6y-18z=-30. Using a graphing program and Gauss' elimination I know it is a line where the planes intersect rather then a point (in the final steps of the Gauss' elimination I had a bottom row of all zeros which gives unlimited answers).

I did find the cross product of vectors A and B to be -18i+18j-18k but I am not sure what to do from here to get the equation of the line. If anyone could point me in the direction or even give me a step by step of what to do or where i went wrong that would be great thanks!

 

[Mark44]

I am trying to solve for the line that intersects any planes that fall under the arithmetic pattern of ax+(a+n)y+(a+2n)z=a+3n. The sample equations I've been using is (A) x+2y+3z=3, (B) 2x-1y-4y=-7 and (C) 6x-6y-18z=-30. Using a graphing program and Gauss' elimination I know it is a line where the planes intersect rather then a point (in the final steps of the Gauss' elimination I had a bottom row of all zeros which gives unlimited answers).

I did find the cross product of vectors A and B to be -18i+18j-18k but I am not sure what to do from here to get the equation of the line. If anyone could point me in the direction or even give me a step by step of what to do or where i went wrong that would be great thanks!

What are the vectors A and B? Are they the normals of the planes you identified as (A) and (B)? If they are, you should use different letters to avoid confusion. Assuming that a and b are the normals to the first two planes, a = <1, 2, 3> and b = <2, -1, -4>, since the normals aren't parallel, the planes have to intersect in a line.

You can use Gauss elimination, using an augmented matrix whose rows are the coefficients in the two plane equations, to find the equation of the line of intersection. If I understand what you're trying to do, start with this augmented matrix, and get it into reduced, row-echelon form.

[1 2 3 | 3]
[2 -1 -4| -7]

When it is reduced, row-echelon form, it will give you x in terms of z plus a constant, y in terms of z plus a constant, and (implied) z equal to itself. You will have the parametric equations for the line of intersection.

 

[ThomasRiley]

Using the augmented matrix which you suggested i could not get it to get into a complete row-echelon form, I could only reduce to

[1 2 3 | 3]
[0 -5 -2| -13]

And I do not know where to go after this.

 

[Mark44]

You made a mistake. When you add (-2) times the first row to the second row, the second row becomes
0 -5 -10 | -13

Now, multiply the second row by (-1/5) to get a leading entry of 1. Then use that 1 to eliminate the entry above it.

Then do what I said at the end of post #2.

 

[ThomasRiley]

I did get the parametric form of
x=5z +17
y=-2z+13/5
z=z

How would I put this into a 'ax+by+cz=d' form?
And thank you for all your help!

 

[Mark44]

I did get the parametric form of
x=5z +17
y=-2z+13/5
z=z

Your value for x is incorrect.

How would I put this into a 'ax+by+cz=d' form?

Why would you want to? That is the equation of a plane in space. What you're after are the parametric equations for the line of intersection of two planes. What you have above is the right form, but with the mistake as noted.

When you get the right equations, you can check them by substituting the values for x, y, and z into the equations of the two given planes. If your parametric equations are correct, you will get two true statements.

And thank you for all your help!

Sure, happy to do it!

 

[ThomasRiley]

Just checking my work is the correct x value x=17-5z ?
And again thanks so much you saved my final grade lol

 

[Mark44]

No. My first row looks like this: [1 0 -1 | -11/5]. Can you read off x from this?