Symmetric equation of the line of intersection of the two planes

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September 21st 2010, 10:45 AM
statman101

Symmetric equation of the line of intersection of the two planes

given the two planes

3x+4y-5z=12
4x+y+z=15

how can I find the symmetric equation of the line of intersection of the two planes.

i got the cross product=9i-23j-13k
but why is the point (3,2,1) a point on both the planes?
September 21st 2010, 10:49 AM
Unknown008

You have to get any point which is found on both planes. The point you take can be any point, as long as it satisfies both equations.

I could take for example,

If I take x = 1, I get z = 17/3 and y = 28/3

(Happy)
September 21st 2010, 12:17 PM
Soroban

Hello, statman101!

The cross-product approach is the best way.

Here's another (primitive and longer) approach.
(Use it when you're desperate.)

Quote:

$\text{Given the two planes: }\;\begin{Bmatrix}3x+4y-5z& =& 12 & [1] \\ 4x+y+z&=& 15 & [2] \end{Bmatrix}$

$\text{Find the symmetric equation of the line of intersection of the two planes.}$

$\begin{array}{cccccccc} \text{Multiply [2] by -4:} & \text{-}16x - 4y - 4z &=& \text{-}60 \\ \text{Add [1]:} & 3x + 4y - 5x &=& 12 \end{array}$

And we have: . $-13x - 9z \:=\:-48 \quad\Rightarrow\quad x \:=\:\dfrac{48-9z}{13}$

Substitute into [2]: . $4\left(\frac{48-9z}{13}\right) + y + z \:=\:15 \quad\Rightarrow\quad y \:=\:\dfrac{3+23z}{13}$

So we have: . $\begin{Bmatrix}x &=& \frac{48-9z}{13} &=& \frac{48}{13} - \frac{9}{13}z\\ \\[-3mm] y &=& \frac{3 + 23z}{13} &=& \frac{3}{13} + \frac{23}{13}z \\ \\[-3mm] z &=& z \end{Bmatrix}$

On the right side, replace $\,z$ with a parameter $\,t.$

. . . . . $\begin{Bmatrix} x &=& \frac{48}{13} - \frac{9}{13}t \\ \\[-3mm] y &=& \frac{3}{13} + \frac{23}{13}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix}$

These are the parametric equations of the line through $\left(\frac{48}{13},\:\frac{3}{13},\:0\right)$

. . with direction vector: . $\vec v \:=\:\langle \text{-}\frac{9}{13},\:\frac{23}{13},\:1\rangle \;=\;\langle \text{-}9,\:23,\:13\rangle$

The symmetric equation is: . $\displaystyle \frac{x-\frac{48}{13}}{-9} \;=\;\frac{y-\frac{3}{13}}{23} \;=\;\frac{z-0}{1}$

Note: There are fifteen brizillion different forms of the symmetric equation.
. . . . .This is just one of them.
September 22nd 2010, 02:46 AM
HallsofIvy

Quote:

Originally Posted by Soroban

Hello, statman101!

The cross-product approach is the best way.

Here's another (primitive and longer) approach.
(Use it when you're desperate.)

$\begin{array}{cccccccc} \text{Multiply [2] by -4:} & \text{-}16x - 4y - 4z &=& \text{-}60 \\ \text{Add [1]:} & 3x + 4y - 5x &=& 12 \end{array}$

And we have: . $-13x - 9z \:=\:-48 \quad\Rightarrow\quad x \:=\:\dfrac{48-9z}{13}$

Substitute into [2]: . $4\left(\frac{48-9z}{13}\right) + y + z \:=\:15 \quad\Rightarrow\quad y \:=\:\dfrac{3+23z}{13}$

So we have: . $\begin{Bmatrix}x &=& \frac{48-9z}{13} &=& \frac{48}{13} - \frac{9}{13}z\\ \\[-3mm] y &=& \frac{3 + 23z}{13} &=& \frac{3}{13} + \frac{23}{13}z \\ \\[-3mm] z &=& z \end{Bmatrix}$

On the right side, replace $\,z$ with a parameter $\,t.$

. . . . . $\begin{Bmatrix} x &=& \frac{48}{13} - \frac{9}{13}t \\ \\[-3mm] y &=& \frac{3}{13} + \frac{23}{13}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix}$

These are the parametric equations of the line through $\left(\frac{48}{13},\:\frac{3}{13},\:0\right)$

. . with direction vector: . $\vec v \:=\:\langle \text{-}\frac{9}{13},\:\frac{23}{13},\:1\rangle \;=\;\langle \text{-}9,\:23,\:13\rangle$

The symmetric equation is: . $\displaystyle \frac{x-\frac{48}{13}}{-9} \;=\;\frac{y-\frac{3}{13}}{23} \;=\;\frac{z-0}{1}$

Note: There are fifteen brizillion different forms of the symmetric equation.
. . . . .This is just one of them.

Only fifteen brizillion? I could have sworn there were more!(Rofl)