# Sketching a simple plane in R3 (should be easy, but not for me :P)

Well, I'm learning plane sketching in R3.

One of the questions is 5y- z - 10 = 0

It is my understanding that the y intercept will be 2, and the z intercept will be -10.

With the absence of an x-term it should be parallel (?) to the x-axis.

I sketch the three axis, note the two points (0,2,0) and (0,0,-10) as the intercepts, but I have no idea how to actually draw the plane itself from this point on.

Unfortunately I have no scanner available to me to even show you what I have so far.

If you can even (verbally) suggest how I draw it, that'd be great.

To clarify, I have the two points selected, but I just don't understand how I could draw a rectangular section that would be parallel to the x-axis.

It looks....skewed

x can be anything. if you draw the line 5y-z=10 in the yz-plane & then imagine lifting it straight upwards you'll get your plane.

rachmaninoff
An easy way I keep track of planes: if the equation is
$$a x + b y + c z = d$$

then a normal vector is
$$\vec{n} = a \hat{x} + b \hat{y} + c \hat{z}$$

If you think about it long enough it'll become intuitive. A plane z=constant, is parallel to the xy-plane, so the normal vector goes straight up in the z-direction.

fourier jr said:
x can be anything. if you draw the line 5y-z=10 in the yz-plane & then imagine lifting it straight upwards you'll get your plane.
aha!

I think I understand it now. I believe that I was going about it the wrong way. It just looked a little funny with the way I drew the x axis. Hard to comprehend three dimensions on a two dimensional paper :D

Later today I'll try and draw it in MSPaint :D