Prove that the planes 2x-3y+z=4, x+4y-z=7 and 3x-10y+3z=1 meet in a straight line
Hello, kingkaisai2!
Prove that the planes $\displaystyle \begin{array}{ccc}2x-3y+z\,=\,4 \\ x+4y-z\,=\,7 \\ 3x-10y+3z\,=\,1\end{array}\;\;\hdots\;?$
We have no idea what we're supposed to prove . . .
Its derminant: .$\displaystyle \begin{vmatrix}2 & \text{-}3 & 2 \\ 1 & 4 & \text{-}1 \\ 3 & \text{-}10 & 3\end{vmatrix}\;=\;0$
. . so they do not intersect at a common point.
Hello, kingkaisai2,Originally Posted by kingkaisai2
two planes which are not equal or parallel intersect in a common line. So with your problem you calculate the common line between P1 and P2 and then between P2 and P3. If you get the same line your proof is done:
P1: 2x - 3y = 4 - z
P2: x + 4y = 7 + z
Now calculate for
x= 37/11 - 1/11*z
y= 10/11 + 3/11*z
and set z = 11t
You get the line: (x, y, z) = (37/11, 10/11, 0)+ t*(-1, 3, 11)
P2: x + 4y = 7 + z
P3: 3x - 10y = 1 - 3z
Now calculate for
x = 37/11 - 1/11*z
y = 10/11 + 3/11*z
and set z = 11t
you'll get exactly the same line. So the statement is true.
Greetings
EB
It is necessary and sufficient to show that the system of linear equation has infinitely many solutions.Originally Posted by kingkaisai2
$\displaystyle
\left\{ \begin{array}{c}2x-3y+z=4\\x+4y-z=7\\3x-10y+3z=1 \end{array} \right\}$
The trick is to use Cramer's rule.
As, Soroban said the determinant of this system is $\displaystyle \Delta=0$. Therefore, the system has no solutions (planes do not intersect simultaneously) or has infinitely many solutions (lie on a line). But note that,
$\displaystyle \Delta_x,\Delta_y,\Delta_z=0$. Thus, there are infinitely many solutions. Thus, the planes intersect in a line.