Question

x + 3y -2z = 4 and 2x + y = -1

what is the line of intersection?

Answer 1

Is this two planes?

Answer 2

yes

Answer 3

Ok, I'll give it a try (long time since I solved a problem like this)

Answer 4

The line of intersection has to be normal to both of the planes => find the cross product of the two planes' normal vectors.

Answer 5

Then we need a point.

Answer 6

I get cross product = 2x-4y-7z but not sure what to do next.

Answer 7

Please check your cross product again. It should be [2, -4, -5]

Answer 8

Well, in order to parameterize a line, we need a vector and a point.

Answer 9

sorry my bad i find a mistake in question line is 2x - y = -1

Answer 10

Oh, ok :)

Answer 11

we have vector ( 2,-4,-7)

Answer 12

can we peak a random point in this new plane?

Answer 13

The point can be found by trying to set z = 0 or y = 0 in one of the equations and then solve for the two variables.

Answer 14

I will calculate for a moment and give you my results.

Answer 15

Sorry, but again, you have the wrong cross product. It should be [2, 4, 7]. Since this is a directional vector, we can rewrite it as [2, 4, 7]

Answer 16

I mean, the cross product is [-2, -4, -7]

Answer 17

ohk i got that

Answer 18

but next we need to put y or z =0 and it will be our line?

Answer 19

Well, man should just try. Here, if we pick y = 0, we get our second equation (the equation for plane number 2) in terms of x and can be easily solved.

Answer 20

By setting y = 0 and solving for x in the second equation and solving for z in the first one, I get the point (x_1, y_1, z_1) = (-1/2, 0, -9/4 )

Answer 21

Hence, the line of intersection, l(t), can be given as l(t) = [2t, 4t, 7t ] + (-1/2, 0, -9/4).

Answer 22

Now, how do I know that this point is " valid "? Like why is it valid with y = 0? Well, that's something you have to test, we do make an assumption, right? So, when I found the x- and z-coordinate, I put them into the first equation and calculated the left side (using y = 0 of course). Then my left side turned out to be 4, which is equal to the right side, so my assumption with y = 0 was valid.

Answer 23

*we only made an assumption, right?

Answer 24

I do really bad in theory and such stuff, and i can get where everything come from, sorry

Answer 25

Oh, sorry.. What can I do to help?

Answer 26

I was confused with finding this line since i dont get why setting us y or z =0 give that line.
I got everything up to and include cross-product and the orthogonal plane that we found but nothing more.

Answer 27

Ok, I see... First, I did a mistake by saying that the line is ortogonal to the planes; it is of course parallell to the planes, and therefore ortogonal to the planes' normal vectors.

Answer 28

I am not able to explain why we set x = 0, y = 0, or z = 0... That's just a way to solve the equations. Cause we have two equations with three unknown, right, and that cannot be solved unless one of them is 0.

Answer 29

*unknowns

Answer 30

ohk, write will try to get it this way thank you.

Answer 31

Ok, hope you'll understand it :)