# Thread: planes, lines and points type of question.

1. ## 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

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

3. when i used gaussian elimination i found that on the 3rd equation near the bottom that 0= 8, which shows its inconsistant.

4. 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

5. 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

6. See attached

7. Hello, sterps!

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

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

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

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

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

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

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

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

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

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