• Apr 25th 2010, 08:27 AM
the prince
If we have to find the equation of the plane determined with two lines for example
$\displaystyle l_1 : \frac{x-5}{13}=\frac{y-6}{1}=\frac{z+3}{-4}$
$\displaystyle l_2 : \frac{x +3}{6}=\frac{y+2}{7}=\frac{z-4}{-3}$
and also we know that the lines intersect in a point K.
the normal vector of the plane we find using cross product of vectors from lines.
we find the eqution using this formula $\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0$
it is wrong to take $\displaystyle M(x_0,y_0,z_0)$ the point of $\displaystyle l_1$ or $\displaystyle l_2$?
should we take the point $\displaystyle K$ as $\displaystyle M(x_0,y_0,z_0)$??

and can someone explain if we take different points of the plane to find the equation we get different equations but it's still the same plane, should we write the equations in normal form?

• Apr 25th 2010, 01:04 PM
schmeling
1.the two equations you have given are equations of planes, not lines.

2.if you have two coplanar lines that are not parallel to each other taking their cross product should give you a normal vector to the plane.

3. I assume that by M(x,y,z) you mean any point on the plane, you are allowed to take any point on the plane so taking one that lies on one of the lines that are on the plane is ok.

4. you are also allowed to take the point of intersection of the coplanar lines. It really makes no difference which point you choose as long as it is on the plane.
• Apr 25th 2010, 11:30 PM
the prince
Quote:

Originally Posted by schmeling
1.the two equations you have given are equations of planes, not lines.

really thanks for help, but at this point i think you are wrong
the two equations are lines and these two together determined the plane.
• Apr 26th 2010, 01:52 AM
HallsofIvy
Quote:

Originally Posted by schmeling
1.the two equations you have given are equations of planes, not lines.

No, they are lines. A plane is given by a single linear equation. The given equations are two (or three dependent) equations. Since the three given fractions are equal, you can change to parametric equations by setting them all equal to a parameter, t:
$\displaystyle \frac{x-5}{13}=\frac{y-6}{1}=\frac{z+3}{-4}$
is equivalent to $\displaystyle \frac{x- 5}{13}= t$, so x- 5= 13t and x= 13t+ 5, $\displaystyle \frac{y-5}{1}= t$, so y- 5= t and y= t+ 5, and $\displaystyle \frac{z+3}{-4}= t$ so z+ 3= -4t and z= -4t- 3.

Those are parametric equations for the first line.

the prince, The equation of a plane with normal vector <A, B, C> and containing point $\displaystyle (x_0, y_0, z_0)$ is $\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$, as you clearly know. That is the same as $\displaystyle Ax+ By+ Cz= Ax_0+ By_0+ Cz_0$.

Now, suppose $\displaystyle (x_1, y_1, z_1)$ is another point on that plane. The the formula becomes $\displaystyle Ax+ By+ Cz= Ax_1+ By_1+ Cz_1$.

But the fact that $\displaystyle (x_1, y_1, z_1)$ lies on the plane means it must satisfy, by putting $\displaystyle x_1, y_1, z_1$ for x, y, z in the original formula, $\displaystyle Ax_1+ By_1+ Cz_1= Ax_0+ By_0+ Cz_0$ so the constants on the right side, and the equations, are exactly the same.
• Apr 26th 2010, 05:30 AM
schmeling
of course they are lines, i cant recall why i insisted on this crazy plane theory of mine...