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?

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)

Hello, statman101!

The cross-product approach is the best way.

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

Quote:

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

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

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

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

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

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


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

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


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

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


The symmetric equation is: .$\displaystyle \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.

Quote:

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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.)


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

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

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

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


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

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


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

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


The symmetric equation is: .$\displaystyle \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.

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