Differential Geometry - Line of Intersection

bucjen · **Joined:** Tue, 17 Jul 2007 04:34:16 UTC **Posts:** 8

Hi,

I need help with the line of intersection between 2 planes. I am quite embarassed that I don't remember how to do this :oops:

I need to find the equation of the plane through (-2,1,-4) which is perpendicular to the line of intersection of the planes 4x + 2y + 2z = -1 and 3x + 6y + 3z = 7.

I am stuck on finding the line of intersection. I have let n1=(4,2,2) and n2=(3,6,3) and have n1xn2=[-6,-6,18] and I now think that I have to let n1=n2 which will give me a point on the line of intersection but I am not too sure how to do it.

Any help appreciated - thanks in advance.
bucjen

Bilbo · **Posted:** Tue, 17 Jul 2007 05:52:26 UTC

bucjen wrote:

Hi,

I need help with the line of intersection between 2 planes. I am quite embarassed that I don't remember how to do this :oops:

I need to find the equation of the plane through (-2,1,-4) which is perpendicular to the line of intersection of the planes 4x + 2y + 2z = -1 and 3x + 6y + 3z = 7.

I am stuck on finding the line of intersection. I have let n1=(4,2,2) and n2=(3,6,3) and have n1xn2=[-6,-6,18] and I now think that I have to let n1=n2 which will give me a point on the line of intersection but I am not too sure how to do it.

Any help appreciated - thanks in advance.
bucjen

Let $\bold{m} = \bold{n_1} \times \bold{n_2}$ and $\bold{c}$ be any point on the intersection of the two planes (i.e., it satisfies the two equations). Then the line of intersection is the set of points of the form $t\bold{m} + \bold{c}$ for $t \in \mathbb{R}$ . The plane you want has normal $\bold{m}$ and its equation is $\bold{m(x-d)} = 0$ where $\bold{d} = (-2,1,-4)$ and $\bold{x} = (x,y,z)$ .

bucjen · **Joined:** Tue, 17 Jul 2007 04:34:16 UTC **Posts:** 8

Hey Bilbo

Thanks for the reply

I guess the trouble I am having is finding points to satisfy both equations. Can you help with that?

Thanks
bucjen

red_dog · **Joined:** Wed, 27 Jun 2007 09:10:55 UTC **Posts:** 66

To find the equation of the line, solve the system $\displaystyle \left\{\begin{array}{ll}4x+2y=-2z-1\\3x+6y=-3z+7\end{array}\right.$ with x,y

unknown and

as parameter.
We get $\displaystyle x=-\frac{1}{3}z-\frac{10}{9},y=-\frac{1}{3}z+\frac{31}{18}$ .
Then we can write $\displaystyle \frac{x+\frac{10}{9}}{-\frac{1}{3}}=\frac{y-\frac{31}{18}}{-\frac{1}{3}}=\frac{z}{1}$ or
$\displaystyle \frac{x+\frac{10}{9}}{1}=\frac{y-\frac{31}{18}}{1}=\frac{z}{-3}$ .
So the director vector of the line is (1,1,-3)

which it is also the normal vector of the plane.
Then the equation of the plane passing through the point (-2,1,4)

is
$(x+2)+(y-1)-3(z-4)=0\Leftrightarrow x+y-3z+13=0$

Bilbo · **Posted:** Tue, 17 Jul 2007 07:30:36 UTC

bucjen wrote:

Hey Bilbo

Thanks for the reply

I guess the trouble I am having is finding points to satisfy both equations. Can you help with that?

Thanks
bucjen

Notice that the equation of the plane $\bold{m}(\bold{x} - \bold{d}) = 0$ doesn't make any reference to a point on the line. All you need is the normal $\bold{m}$ , which you calculated from the cross product, and the point $\bold{d}$ you want the plane to go through. Also notice that the only calculation you need is the cross product.

If you need to find points on the line in addition to the equation of the plane, red_dog shows you how to do that. red_dog's calculation yields an equation for the plane that is a multiple (-6) of the one I give.

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Differential Geometry - Line of Intersection

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