1. Geometry

Can anyone help me with the following question on geometry and vectors:

The line L through the origin is normal to the plane 2x-y-z=4. Find the point in which L meets the plane x+y-2z=2.

Thanks to those who can help.

2. Hello, smiler!

I'm sure you know enough of the basic to solve this.
You just have to put it all together . . .

The line $\displaystyle L$ through the origin is normal to the plane $\displaystyle 2x-y-z\:=\:4$
Find the point in which $\displaystyle L$ meets the plane $\displaystyle x+y-2z\:=\:2$

The plane has the normal vector: .$\displaystyle \vec{n} \:=\:\langle2,-1,-1\rangle$

This is the direction vector of the line through the origin.

Its equations are: .$\displaystyle \begin{array}{ccccc}x & = & 0 + 2t & \Rightarrow & x \:=\:2t \\ y & = & 0 + (\text{-}1)t & \Rightarrow & y \:=\:-t \\ z & = & 0 + (\text{-}1)t & \Rightarrow & z \:=\:-t \end{array}$ . [1]

To find its intersection with the plane: .$\displaystyle x +y -2z \:=\:2$ . [2]
. . substitute [1] into [2]: .$\displaystyle (2t) + (-t) - 2(-t) \:=\:2\quad\Rightarrow\quad 3t \:=\:2\quad\Rightarrow\quad t \:=\:\frac{2}{3}$

Substitute into [1] and we have: . $\displaystyle \boxed{\:x \:=\:\frac{4}{3},\;\;y \:=\:-\frac{2}{3},\;\;z \:=\:-\frac{2}{3}\:}$

3. direction vector

i understand the majority of the answer but how do you work out the direction vectors x=0+2t, y=0+(-t), z=0+(-t)?

thanks