Line of intersection of two planes

**cubs3205** · April 4th 2009, 02:23 PM

Hi, I am not sure of the best strategy to approach finding the parametric description of an intersection line in 3-D where the two planes are also described parametrically.
The planes are described {t - u, 3t + u, u} and {4r - s, 0, s + r}

I know how to do this if the equations of the planes are Cartesian. Should I somehow convert them and eliminate the parameters?

Thanks!

**Jhevon** · April 4th 2009, 02:30 PM

Originally Posted by cubs3205

Hi, I am not sure of the best strategy to approach finding the parametric description of an intersection line in 3-D where the two planes are also described parametrically.
The planes are described {t - u, 3t + u, u} and {4r - s, 0, s + r}

I know how to do this if the equations of the planes are Cartesian. Should I somehow convert them and eliminate the parameters?

Thanks!

i don't think you can describe planes this way. these look more like lines. and you will need to know what to think of as parameters and what as constants. you should type the problem in its entirety so that we can have a clue about what this all means

**cubs3205** · April 4th 2009, 03:22 PM

The question is:
Write the parametric description of the line that is the intersection of planes described parametrically by {(t - u, 3t + u, u | t, u $\in$ R} and {(4r - s, 0, s + r) | r, s $\in$ R}

**Soroban** · April 4th 2009, 03:29 PM

Hello, cubs3205!

I've never had a plane described with two parameters.
But I followed my instincts and came up with a solution.

Hi, I am not sure of the best strategy to approach finding the parametric description
of an intersection line in 3-D where the two planes are also described parametrically.

The planes are described: {t - u, 3t + u, u} and {4r - s, 0, s + r}

We have these two planes: . $\begin{Bmatrix}x &=& t-u \\ y &=& 3t + u \\ z &=& u\end{Bmatrix}\;\;\text{ and }\;\;\begin{Bmatrix}x &=& 4r-s \\ y &=& 0 \\ z &=& s+r \end{Bmatrix}$ .(a)

Where they intersect, the $x$ 's, $y$ 's and $z$ 's are equal.

So we have: . $\begin{array}{ccc}t - u &=& 4r-s \\ 3t + u &=& 0 \\ u &=& s+t \end{array}\quad\Rightarrow\quad \begin{array}{cccc}4r-s-t+u &=&0 & [1] \\<br /> \qquad\quad 3t + u &=& 0 & [2] \\<br /> \qquad s + t - u &=&0 & [3] \end{array}$

. . Add [1] and [3]: . $4r \:=\:0\quad\Rightarrow\quad\boxed{ r \:=\:0}$

Then we have: . $\begin{array}{cccc}\text{-}s -t + u &=& 0 & [1] \\ \quad 3t + u &=& 0 & [2] \end{array}$

. . Subtract [2] - [1]: . $4t + s \:=\:0 \quad\Rightarrow\quad \boxed{s \:=\:-4t}$

Substitute into [3]: . $-4t + t - u \:=\:0 \quad\Rightarrow\quad \boxed{u \:=\:-3t}$

So we have: . $\begin{array}{ccc}r &=& 0 \\ s &=& \text{-}4t \\ t &=& t \\ u &=& \text{-}3t \end{array}$
On the right, replace $t$ with a parameter $k\!:\;\;\begin{array}{ccc}r &=& 0 \\ s &=& \text{-}4k \\ t &=& k \\ u &=& \text{-}3k \end{array}$

Substitute these into (a) and we have: . $\begin{Bmatrix}x &=& 4k \\ y &=& 0 \\ z &=& \text{-}3k \end{Bmatrix}$

And these are the parametric equations of the line of intersection.

**mr fantastic** · April 4th 2009, 03:39 PM

Originally Posted by cubs3205

Hi, I am not sure of the best strategy to approach finding the parametric description of an intersection line in 3-D where the two planes are also described parametrically.
The planes are described {t - u, 3t + u, u} and {4r - s, 0, s + r}

I know how to do this if the equations of the planes are Cartesian. Should I somehow convert them and eliminate the parameters?

Thanks!

For the first plane you have:

x = t - u ... (1)
y = 3t + u ... (2)
z = u ... (3)

Substitute (3) into (1) and (2):

x = t - z => t = x + z .... (A)
y = 3t + z .... (B)

Substitute (A) into (B):

y = 3(x + z) + z => 3x - y + 4z = 0.

This is the cartesian equation of the first plane.

The cartesian equation equation of the second plane is just y = 0.

So the line of intersection is found by solving 3x - y + 4z = 0 and y = 0 simultaneously: 3x + 4z = 0.

I'm sure you can express this line parametrically.

Edit: This is an alternative approach to Soroban (hey, you beat me this time! You must have started typing an hour ago lol!)

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