# Thread: vector calculus 3D question

1. ## vector calculus 3D question

We just started vector calculus and in trying to get us thinking in 3 dimensions, we are asked to describe the set of points whose distance from the x axis is the same as the distance from the zy plane.

I am thinking that the points would represent a plane parallel with the zy plane with the same dimensions. Can someone tell me if I am even close to right here? I know it is a simple concept but spacial relationships are not simple to me!!!! Thank you FrostKing

2. Unfortunately, no, you are not even close to right. A plane parallel to the yz plane would have constant x but the distance of each point on the plane to the x axis would NOT be constant.

Take the general point (x, y, z). Its distance to the yz-plane (that is, the distance from (x,y,z) to (0,y,z)) is just x. Its distance from the x-axis is the distance from (x,y,z) to (x, 0, 0).
Find that distance and set it equal to x.

3. ## Vector 3d question

So, I need to use (x,y,z) and (x,0,0) and set value equal to x.

(x-x)^2 + (y - 0)^2 + (z - 0)^2 = x^2 Therefore, x^2 = y^2 + z^2 and x = y + z?????????

4. Originally Posted by Frostking
So, I need to use (x,y,z) and (x,0,0) and set value equal to x.

(x-x)^2 + (y - 0)^2 + (z - 0)^2 = x^2 Therefore, x^2 = y^2 + z^2 and x = y + z?????????
The first part of your considerations is OK.

Take those points which have the same perpendicular distance to the x-axis. All those points are located on a circle-line which lies in a plane parallel to the yz-plane. The radius of this circle must be as long as the distance of the circle from the yz-plane.

If the distance from the yz-plane increases the radius of the circle increases too.

I've attached a sketch of the situation and I've drawn 2 examples to demonstrate the relationship between the distance from the x-axis (=the radius) and the distance from the yz-plane.