# planes, lines and points type of question.

• Aug 14th 2006, 11:18 PM
sterps
planes, lines and points type of question.
I need help with this, do i use gaussian elimination or something else.

Three planes have eqn's given by:

x+3y-z = 4
2x+8y = 18
x-y-3z = -6

a) have no common point of intersection
b) intersect in a point
c) intersect in a line
d) intersect in a plane

cheers guys
• Aug 15th 2006, 12:22 AM
Glaysher
Just noticed that you're from Australia. This isn't part of the Australian Maths Challenge is it?
• Aug 15th 2006, 12:31 AM
sterps
when i used gaussian elimination i found that on the 3rd equation near the bottom that 0= 8, which shows its inconsistant.
• Aug 15th 2006, 12:35 AM
Glaysher
My graphical calculator has checked my previously posted answer but I won't repost it until I know for sure that this isn't part of the Challenge
• Aug 15th 2006, 04:04 AM
CaptainBlack
Quote:

Originally Posted by Glaysher
Just noticed that you're from Australia. This isn't part of the Australian Maths Challenge is it?

His profile says he a 2nd year undergrad, so not eligable, so you can

RonL
• Aug 15th 2006, 05:31 AM
Glaysher
See attached
• Aug 15th 2006, 06:54 AM
Soroban
Hello, sterps!

Yes, Gaussian elimination is a good approach.
(You must have made some errors, though.)

Quote:

Three planes have equations: . $\begin{array}{ccc}x+3y-z \:= \:4\\ 2x+8y \:=\:18 \\x-y-3z \:=\:\text{-}6\end{array}$

a) have no common point of intersection
b) intersect in a point
c) intersect in a line
d) intersect in a plane

We have: . $\begin{pmatrix}1 & 3 & \text{-}1 & | & 4\\ 2 & 8 & 0 & | & 18\\ 1 & \text{-}1 & \text{-}3 & | & \text{-}6\end{pmatrix}$

. . . . $\begin{array}{cccc} \\ \frac{1}{2}R_2\\ \\\end{array}\;\begin{pmatrix}1 & 3 & \text{-}1 & | & 4\\1 & 4 & 0 & | & 9 \\ 1 & \text{-}1 & \text{-}3 & | & \text{-}6\end{pmatrix}$

$\begin{array}{ccc} \\ R_2-R_1\\ R_3 - R_1\end{array}\;\begin{pmatrix}1 & 3 & \text{-}1 & | & 4\\ 0 & 1 & 1 & | & 5\\0 & \text{-}4 & \text{-}2 & | &\text{-}10\end{pmatrix}$

$\begin{array}{ccc}R_1-3R_2\\ \\ R_3+4R_2\end{array}\;\begin{pmatrix}1 & 0 & \text{-}4 & | & \text{-}11 \\ 0 & 1 & 1 & | & 5\\ 0 & 0 & 2 & | & 10\end{pmatrix}$

. . . . $\begin{array}{ccc} \\ \\ \frac{1}{2}R_3\end{array}\;\begin{pmatrix}1 & 0 & \text{-}4 & | & \text{-}11\\0 & 1 & 1 & | & 5\\ 0 & 0 & 1 & | & 5\end{pmatrix}$

. $\begin{array}{cccc} R_1+4R_3 \\ R_2-R_3 \\ \\ \end{array}\;\begin{pmatrix}1 & 0 & 0 & | & 9 \\ 0 & 1 & 0 & | & 0\\0 & 0 & 1 & | & 5\end{pmatrix}$

Therefore: (b) The planes intersect in a point, $\bf{(9,0,5)}$