Objects in 3-space satisfying a specific condition

Question

1)Which geometric object in 3-space is described by the condition $x^2=y^2=z^2$ ?

I cannot recall a quadric surface described by these equations, so I am obviously missing something... anybody know what?

2)$ x≥0, y≥0, z≥0, 1≥x+y+z$

I have very limited experience in this stuff, so unfortunately I cannot recognize the object described by these conditions.

Answer 1

a) Sometimes a picture is worth a 1000 words:

All your constraints mean is $x = \pm y = \pm z$.

b)

To get intuition, set $z=0$ and find your region. Then $y=0$ and find your region. Then $x=0$ and find your region.

Answer 2

For part a) I will say they are four lines:
$x=y=z$
$x=y=-z$
$x=-y=z$
$x=-y=-z$

For part b) is the tetrahedron with vertices $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$

Answer 3

For first equation, that should be 8 lines that that goes out of (0,0,0). Put a 3D cube with its center at (0,0,0). Each of these line will pass throw the cube's vetercies.

For the second one, imagine putting a triangle on the x,y,z axes, that each of its verticies thouching : (1,0,0), (0,1,0), (0,0,1). The volume that between the triangle and the axes are this second group you mentioned.

Answer 4

(almost) as the figure by David G. Stork shows: (assuming you want both conditions 1) and 2) to be satisfied by one object)

your object is the set of all points in real 3-space where the absolute values of $x,y,z$ are all the same: $|x|=|y|=|z|$. Your additional inequalities define the positive octant where all variables are positive and the object only outside the triangle between $(1,0,0)^T,(0,1,0)^T,(0,0,1)^T$.

As a result, your object is the line segment where $x=y=z$ (diagonal in 3-space) for $0\leq x\leq\frac{1}{3}$.