# Equation of the Line of Intersection of two Planes Help

• Jun 10th 2008, 07:33 AM
Morphayne
Equation of the Line of Intersection of two Planes Help
Problem:

Determine a vector equation of the line of intersection of the planes $3x-y+4z-2=0$ and $x+6y+10z+8=0$.

$[x,y,z]=[2,0,-1]+t[34,26,-19]
$

• Jun 10th 2008, 08:38 AM
o_O
Taking the cross product of their normals will give you a vector parallel to the line of intersection, i.e. your direction vector.

Find a position vector that lies on the line of intersection (i.e. that satisfies both equations of the plane). This can be pretty much anything and is why your answer will vary. What you can do is fix a value for x (or y or z) and solve for remaining simpler system of equations for the other 2 variables to get a point common to both planes. This point will form the position vector needed to define your line.
• Jun 10th 2008, 10:45 AM
Soroban
Hello, Morphayne!

o_O has the best method.

I'll show you another . . .

Quote:

Determine the equations of the line of intersection of the planes
. . $3x-y+4z-2\:=\:0$ and $x+6y+10z+8\:=\:0$.

Answers may vary. . . . . . .
Finally ... a book that gives fair warning!

. . $[x,y,z]\:=\:[2,0,-1]+t[34,26,-19]$

We have: . $\begin{array}{cccc}3x - y + 4z &=& 2 & {\color{blue}[1]} \\ x + 6y + 10z &=& \text{-}8 & {\color{blue}[2]} \end{array}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[1]} by 6:} & 18x - 6y + 24z &=& 12 \\ \text{Add {\color{blue}[2]}:} & x + 6y + 10z &=& -8 \\
\text{and we have:} & 19x + 34z &=& 4\end{array}$

. . Hence: . $x \:=\:\frac{4}{19} - \frac{34}{19}z$

Substitute into [2]: . $\left(\frac{4}{19} - \frac{34}{19}z\right) + 6y + 10z \:=\:\text{-}8 \quad\Rightarrow\quad y \:=\:\text{-}\frac{2}{19} - \frac{26}{19}z$

We have: . $\begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}z \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}z \\ \\[-3mm] z &=& z \end{Bmatrix}$

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

And we have: . $\begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}t \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix}$

Therefore: . $[x,y,z] \;=\;\left[\frac{4}{19},\:\text{-}\frac{2}{19},\:0\right] + t\left[\text{-}\frac{34}{19},\:\text{-}\frac{26}{19},\:1\right]$

• Jun 10th 2008, 10:49 AM
Morphayne
Quote:

Originally Posted by Soroban
Hello, Morphayne!

o_O has the best method.

I'll show you another . . .

We have: . $\begin{array}{cccc}3x - y + 4z &=& 2 & {\color{blue}[1]} \\ x + 6y + 10z &=& \text{-}8 & {\color{blue}[2]} \end{array}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[1]} by 6:} & 18x - 6y + 24z &=& 12 \\ \text{Add {\color{blue}[2]}:} & x + 6y + 10z &=& -8 \\
\text{and we have:} & 19x + 34z &=& 4\end{array}$

. . Hence: . $x \:=\:\frac{4}{19} - \frac{34}{19}z$

Substitute into [2]: . $\left(\frac{4}{19} - \frac{34}{19}z\right) + 6y + 10z \:=\:\text{-}8 \quad\Rightarrow\quad y \:=\:\text{-}\frac{2}{19} - \frac{26}{19}z$

We have: . $\begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}z \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}z \\ \\[-3mm] z &=& z \end{Bmatrix}$

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

And we have: . $\begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}t \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix}$

Therefore: . $[x,y,z] \;=\;\left[\frac{4}{19},\:\text{-}\frac{2}{19},\:0\right] + t\left[\text{-}\frac{34}{19},\:\text{-}\frac{26}{19},\:1\right]$

That is the answer I got!(Rofl) I know it says "Answers may vary", but I thought I had to find and equation without the fractions in the direction vector.(Headbang) Thanks anyway though.(Clapping)