# Thread: intersection of planes

1. ## intersection of planes

1. The line of intersection of the planes 2x+y-3z=3 and x-2y+z=-1 is L. If L meets the xy plane at Point A and the z-axis at Point B, determine the length of line segment AB.

2. Determine the Cartesian equation of the plane that is parallel to the line with equation x = -2y = 3z and that contains the line of intersection of the planes with equation x-y+z=1 and 2y-z=0

Thanks, any help is greatly appreciated!

2. Hello, checkmarks!

1. The line of intersection of the planes: $\displaystyle 2x+y-3z\:=\:3\:\text{ and }\:x-2y+z\:=\:-1\:\text{ is }L.$
If $\displaystyle L$ meets the $\displaystyle xy$-plane at point $\displaystyle A$ and the $\displaystyle z$-axis at point $\displaystyle B,$
determine the length of line segment $\displaystyle AB.$
First, find the equation of the plane.

We have: .$\displaystyle \begin{array}{cccc}2x + y - 3x &=& 3 & [1] \\ x - 2y + z &=& \text{-}1 & [2] \end{array}$

$\displaystyle \begin{array}{ccccc}\text{Multiply [1] by 2:} & 4x + 2y - 6x &=& 6 \\ \text{Add [2]:} & x - 2y + z &=& \text{-}1 \end{array}$

And we have: .$\displaystyle 5x - 5z \:=\:5 \quad\Rightarrow\quad x \:=\:z+1$

Substitute into [2]: .$\displaystyle (z+1) - 2y + z \:=\:\text{-}1 \quad\Rightarrow\quad y \:=\:z+1$

We have these equations: .$\displaystyle \begin{Bmatrix}x &=& z+1 \\ y &=& z+1 \\ z &=&z \end{Bmatrix}$

On the right, replace $\displaystyle z$ with a parameter $\displaystyle t\!:\quad \begin{Bmatrix}x &=& t+1 \\ y &=& t+1 \\ z &=& t\end{Bmatrix}$

These are the parametric equations of the line of intersection.

Line $\displaystyle L$ intersects the $\displaystyle xy$-plane at $\displaystyle A$ . . . Then: .$\displaystyle z = 0.$

Then $\displaystyle t = 0 \quad\Rightarrow\quad x = 1,\;y = 1$

. . We have point $\displaystyle A(1,1,0)$

Line $\displaystyle L$ intersected the $\displaystyle z$-axis at $\displaystyle B$ . . . Then: .$\displaystyle x = 0,\:y = 0$

Then: .$\displaystyle t = \text{-1} \quad\Rightarrow\quad z \:=\:\text{-1}$

. . We have point $\displaystyle B(0,0,\text{-}1)$

Therefore, length of $\displaystyle AB \;=\;\sqrt{(0-1)^2 + (0-1)^2 + (-1-0)^2} \;=\;\sqrt{3}$