Finding the intersection of two planes

**brumby_3** · March 21st 2011, 04:30 PM

Find where the plane x-4y+3z=2 intersects the plane 9x+3y+z=5

**earboth** · March 22nd 2011, 12:21 AM

Originally Posted by brumby_3

Find where the plane x-4y+3z=2 intersects the plane 9x+3y+z=5

1. Solve the system of simultaneous equations for x, y:

$\left|\begin{array}{l}x-4y=2-3z \\ 9x+3y=5-z\end{array}\right.$

2. You should come out with the result:

$\left|\begin{array}{l}x=\frac23-\frac13 z \\ y= -\frac13 + \frac23 z \\ z = z\end{array}\right.$

3. Replace z = t and you'll get the equation of the line as:

$\langle x,y,z \rangle = \langle \frac23, -\frac13 , 0\rangle + t \cdot \langle -\frac13, \frac23, 1 \rangle$

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