Finding an equation for the plane....

• Mar 7th 2010, 09:49 AM
chrisks
Finding an equation for the plane....
I am unsure of how to solve this linear algebra problem...

Find an equation for the plane that passes through (2,4,-1) and contains the line of intersection of the planes x-y-4z=2 and -2x+y+2z=3

Help would be greatly appreciated!
• Mar 7th 2010, 10:06 AM
icemanfan
The normal vectors to the planes \$\displaystyle x - y - 4z = 2\$ and \$\displaystyle -2x + y + 2z = 3\$ are \$\displaystyle <1, -1, -4>\$ and \$\displaystyle <-2, 1, 2>\$. If you take the cross-product of these two vectors, you will get a vector that points in the same direction of the line. Then, all you have to do is find two points A and B on that line, and let C = \$\displaystyle (2, 4, -1)\$. Then the cross-product of the vectors AC and BC will be a normal vector \$\displaystyle <p, q, r>\$ to the desired plane, whose equation will be \$\displaystyle px + qy + rz = s\$, and s can be determined by putting point C into the equation: \$\displaystyle 2p + 4q - r = s\$.
• Mar 19th 2010, 09:44 AM
eleahy
how can you find the equation of the plane in ax+by+cz=d format through the point (1,2,3) for example with normal vector (2,-1,4)
thanks for any help
• Mar 19th 2010, 09:48 AM
eleahy
sorry just dont understand wher yo use the normal vector after gettin x+y+z= s