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equation of a line intersecting two planes

Hi,

I've got the question:

plane1:r.(5ij2k)=16 plane_1 : \textbf{r}.(5\textbf{i} - \textbf{j} - 2\textbf{k}) = 16
plane2:r.(16i5j4k)=53 plane_2 : \textbf{r}.(16\textbf{i} - 5\textbf{j} - 4\textbf{k}) = 53

Find the equation of the line intersecting these two planes.

I think these sorts of questions have multiple answers but my textbook only gives one (which is different to mine) does anyone know a way I can check my answer to see if it is correct?

By the way, my answer is:

(7j92k)+λ(6i12j9k)(-7\textbf{j} - \frac{9}{2}\textbf{k}) + \lambda(-6\textbf{i} - 12\textbf{j} - 9\textbf{k})

Thank you :smile:
Reply 1
sorry are you trying to find the equation of the line describing where the 2 planes intersect each other?
Reply 2
Original post by just george
sorry are you trying to find the equation of the line describing where the 2 planes intersect each other?


Yes, sorry if that wasn't clear :smile:
Reply 3
Its okay, I haven't done this stuff in absolutely ages, so I'm afraid I might not be much help..

is the answer λ(3i+9j+14k) \lambda(3i + 9j + 14k)

if its nothing like that I'll leave someone who knows what they're talking about to help.. :tongue:
Reply 4
Original post by just george
Its okay, I haven't done this stuff in absolutely ages, so I'm afraid I might not be much help..

is the answer λ(3i+9j+14k) \lambda(3i + 9j + 14k)

if its nothing like that I'll leave someone who knows what they're talking about to help.. :tongue:


Hi, thanks :biggrin:

I'm assuming that's a multiple of the direction vector of the line? If so I don't think it's quite right, but perhaps I'm misunderstanding something?
Reply 5
Original post by so it goes
Hi, thanks :biggrin:

I'm assuming that's a multiple of the direction vector of the line? If so I don't think it's quite right, but perhaps I'm misunderstanding something?


Sorry I'm pretty sure I'm wrong actually. Like I said its been absolutely ages since I did anything like this :L

Perhaps if you go through how you got your answer I (or someone else) could see if it seems reasonable or spot what you did wrong?

EDIT: Right, I've now done some revision for you.. Your direction vector is the same as what I've now got, and you can check whether your solution is right by picking a value for lambda (anything you like), and seeing if the point you get lies on both planes. if you pick a couple of different values and they all lie on both planes then your solution is correct.

Another thing, while it doesnt really matter, is that for your direction vector, you can write:
h(-6i - 12j - 9k)
as h(2i + 4j + 3k), as h = any number, you can effectively multiply through by -1 to get all of them positive and divide by 3 to make them the smallest integers - Its just the ratio that is important. This is just for neatness though, so if its in any way confusing, ignore me :smile:
(edited 10 years ago)
Reply 6
Original post by just george
Sorry I'm pretty sure I'm wrong actually. Like I said its been absolutely ages since I did anything like this :L

Perhaps if you go through how you got your answer I (or someone else) could see if it seems reasonable or spot what you did wrong?


That sounds great, thanks :biggrin:

Okay, so first I wrote my two planes in Cartesian form:

plane1:5xy2z=16plane_1: 5x - y - 2z = 16
plane2:16x5y4z=53plane_2: 16x - 5y - 4z = 53

I wanted to first find any point on the line. I know that at some point x=0 on the line created by the intersection of the two planes, so I let x=0 and solved the simultaneous equations:

y2z=16 -y -2z = 16
5y4z=53 -5y -4z = 53

When I solved this simultaneously, I found that:

y=7y=-7 and z=92 z = - \frac{9}{2}

I can therefore say a point on the line has coordinates:

(0,7,92)(0, -7, -\frac{9}{2})

Now I want to find the direction of the line. I have the two direction vectors of the normals to each plane, so I know their cross product will be perpendicular to each normal and therefore in the direction of the line we want.

So to find the direction of the line I did this:

(512)\begin{pmatrix} 5 \\ -1 \\ -2 \end{pmatrix} X (1654) \begin{pmatrix} 16 \\ -5 \\ -4 \end{pmatrix}

So working this out told me that the direction vector of the line will be:

(6i12j9k)(-6\textbf{i} - 12\textbf{j} - 9\textbf{k})

Combining my position and direction vectors I get that the equation of the line creatd by the intersection of the two planes is:

(7j92k)+λ(6i12j9k)(-7\textbf{j} - \frac{9}{2}\textbf{k}) + \lambda(-6\textbf{i} - 12\textbf{j} - 9\textbf{k})

Does this look correct?

Thank you again :smile:
Reply 7
Original post by so it goes
x


Just editted my last post - yeah that looks perfect, assuming no mistakes in the calculations :smile: you can try putting a couple of values of lambda in to check that the points on your line are on both of the planes to make sure :smile:
Reply 8
Original post by just george
Sorry I'm pretty sure I'm wrong actually. Like I said its been absolutely ages since I did anything like this :L

Perhaps if you go through how you got your answer I (or someone else) could see if it seems reasonable or spot what you did wrong?

EDIT: Right, I've now done some revision for you.. Your direction vector is the same as what I've now got, and you can check whether your solution is right by picking a value for lambda (anything you like), and seeing if the point you get lies on both planes. if you pick a couple of different values and they all lie on both planes then your solution is correct.

Another thing, while it doesnt really matter, is that for your direction vector, you can write:
h(-6i - 12j - 9k)
as h(2i + 4j + 3k), as h = any number, you can effectively multiply through by -1 to get all of them positive and divide by 3 to make them the smallest integers - Its just the ratio that is important. This is just for neatness though, so if its in any way confusing, ignore me :smile:



Original post by just george
Just editted my last post - yeah that looks perfect, assuming no mistakes in the calculations :smile: you can try putting a couple of values of lambda in to check that the points on your line are on both of the planes to make sure :smile:


Thank you so much! :biggrin: I'll try a few values for lambda to make sure they work out :smile:
Reply 9
Original post by so it goes
Hi,

I've got the question:

plane1:r.(5ij2k)=16 plane_1 : \textbf{r}.(5\textbf{i} - \textbf{j} - 2\textbf{k}) = 16
plane2:r.(16i5j4k)=53 plane_2 : \textbf{r}.(16\textbf{i} - 5\textbf{j} - 4\textbf{k}) = 53

Find the equation of the line intersecting these two planes.

I think these sorts of questions have multiple answers but my textbook only gives one (which is different to mine) does anyone know a way I can check my answer to see if it is correct?

By the way, my answer is:

(7j92k)+λ(6i12j9k)(-7\textbf{j} - \frac{9}{2}\textbf{k}) + \lambda(-6\textbf{i} - 12\textbf{j} - 9\textbf{k})

Thank you :smile:



Is this A2 Maths?
Reply 10
Original post by Vorsah
Is this A2 Maths?


Ye, it's Edexcel FP3 :biggrin:
Original post by so it goes
Hi,

I've got the question:

plane1:r.(5ij2k)=16 plane_1 : \textbf{r}.(5\textbf{i} - \textbf{j} - 2\textbf{k}) = 16
plane2:r.(16i5j4k)=53 plane_2 : \textbf{r}.(16\textbf{i} - 5\textbf{j} - 4\textbf{k}) = 53

Find the equation of the line intersecting these two planes.

I think these sorts of questions have multiple answers but my textbook only gives one (which is different to mine) does anyone know a way I can check my answer to see if it is correct?

By the way, my answer is:

(7j92k)+λ(6i12j9k)(-7\textbf{j} - \frac{9}{2}\textbf{k}) + \lambda(-6\textbf{i} - 12\textbf{j} - 9\textbf{k})

Thank you :smile:


There will be an infinite number of solutions depending on your choice of a point on the line and the way you write the direction vector.
(edited 10 years ago)
Reply 12
Original post by brianeverit
There will be an infinite number of solutions depending on your choice of a point on the line and the way you write the direction vector.


Thank you :biggrin: I was thinking along those lines but wasn't too sure :tongue:

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