# HELP Vectors- Finding equation of line of intersection between 2 planes

• October 31st 2008, 12:42 AM
NiCeBoY
HELP Vectors- Finding equation of line of intersection between 2 planes
Hello,
I got a question to do as assignment. I have been able to do the first part which is finding the angle between 2 planes but am not able to find the equation of line of intersection of the 2 planes..

Can anyone check the question and help me out please?

Also check if the fist part is right...

Thanks again...
• October 31st 2008, 04:14 AM
mr fantastic
Quote:

Originally Posted by NiCeBoY
Hello,
I got a question to do as assignment. I have been able to do the first part which is finding the angle between 2 planes but am not able to find the equation of line of intersection of the 2 planes..

Can anyone check the question and help me out please?

Also check if the fist part is right...

Thanks again...

First part looks OK.

Second part: Solve simultaneously:

3x - y + 4z = 3 .... (1)

4x - 2y + 7z = 8 .... (2)

There are an infinite number of solutions. One possible form for these solutions is go by letting z = t, say. Then

3x - y = 3 - 4t .... (1')

4x - 2y = 8 - 7t .... (2')

Solve (1') and (2') simultaneously to get x and y in terms of t. This soluion is the line of intersection of the two planes (written in parametric form).
• October 31st 2008, 04:14 AM
earboth
Quote:

Originally Posted by NiCeBoY
Hello,
I got a question to do as assignment. I have been able to do the first part which is finding the angle between 2 planes but am not able to find the equation of line of intersection of the 2 planes..

Can anyone check the question and help me out please?

Also check if the fist part is right...

Thanks again...

1. Your calculation of the angle is OK. (Clapping)

2. Line of intersection between 2 planes.

a) Check if the normal vectors (3, -1, 4) and (4, -2, 7) are collinear. Since $(3,-1,4) \ne k\cdot (4,-2,7)$ the planes are not parallel and therefore there exists a set of common points forming a line.

b) calculate x and y with respect of z from the given system of equations:

$\begin{array}{l}3x-y+4z=3 \\ 4x-2y+7z=8\end{array}$

I've got

$\begin {array}{l}-2y+5z=12 \\ 6x+3z=12\end{array}$

c) Now set z = t and solve both equations for x respectively y and substitute t instaed of z.

$l:\left\{\begin{array}{l}x=-\dfrac12t-1 \\y=\dfrac52t-6 \\z=t\end{array}\right.$

d) This is the parametric equation of the line:

$\vec r = (-1, -6, 0)+t\left(-\dfrac12\ ,\ \dfrac52\ ,\ 1\right)$

e) I've attached a drawing of the 2 planes and the line of intersection.

EDIT: Since the intersection line belongs to both planes the direction vector of the line must be perpendicular to both normal vectors:

$(3,-1,4) \times (4,-2,7) = (1,-5,-2) = (-2)\cdot \left(-\dfrac12\ ,\ \dfrac52\ ,\ 1\right)$
• October 31st 2008, 06:58 AM
NiCeBoY
• October 31st 2008, 07:05 AM
earboth
Quote:

Originally Posted by NiCeBoY

Take the equation of the line in parametric form, solve each equation for t and set equal the results:

$\begin{array}{l}x=-\dfrac12t-1 \\y=\dfrac52t-6 \\z=t\end{array}~~\implies~~\begin{array}{l}-2x-2=t \\\dfrac{2y+12}5=t \\z=t\end{array}$ will yield:

$-2x-2=\dfrac{2y+12}5=z$
• November 1st 2008, 05:06 PM
NiCeBoY
Hello,
I am getting a diferent answer.. a different directional vector..

Can anyone please check my work and let me know where is my mistake please?...

Thanks..
• November 1st 2008, 06:05 PM
mr fantastic
Quote:

Originally Posted by NiCeBoY
Hello,
I am getting a diferent answer.. a different directional vector..

Can anyone please check my work and let me know where is my mistake please?...

Thanks..

There's no mistake. You should have noticed that <1, -5, -2> and <-1/2, 5/2, 1> both have the same direction:

<1, -5, -2> = -2 <-1/2, 5/2, 1>

It's up to you which one you use.
• November 1st 2008, 06:57 PM
NiCeBoY
Oh yes....
Thank you very much man... I succeded in completing it...

Thanks ya again...