To find two unit vectors in 3-space that are perpendicular to the xy-plane, we need to understand that vectors perpendicular to the xy-plane must be parallel to the z-axis.

Explanation:

• A vector perpendicular to the xy-plane will have no components in the x or y directions, meaning its components in the $x$- and $y$-directions will be 0.
• This type of vector can only have a component in the $z$-direction.

So, the vectors we are looking for will be of the form $(0, 0, z)$.

Step-by-Step:

1. The magnitude of a unit vector must be 1. Therefore, we must ensure that the magnitude of the vector satisfies: $\sqrt{x^2 + y^2 + z^2} = 1$ For vectors perpendicular to the xy-plane, both $x = 0$ and $y = 0$, so: $\sqrt{0^2 + 0^2 + z^2} = 1 \quad \Rightarrow \quad |z| = 1$

2. Thus, the possible vectors are:

   • $(0, 0, 1)$ (a unit vector pointing in the positive z-direction)
   • $(0, 0, -1)$ (a unit vector pointing in the negative z-direction)

Conclusion:

The two unit vectors that are perpendicular to the xy-plane are: $\mathbf{v}_1 = (0, 0, 1) \quad \text{and} \quad \mathbf{v}_2 = (0, 0, -1)$

Would you like further details on this explanation or have any other questions?

Here are five related questions you might consider:

1. How do you find a vector perpendicular to the yz-plane?
2. How do you calculate the dot product of two vectors in 3-space?
3. What is the significance of unit vectors in physics and engineering?
4. How do you verify that two vectors are perpendicular?
5. What is the geometric interpretation of the magnitude of a vector?

Tip: In 3-space, a vector is said to be perpendicular to a plane if its dot product with any vector on the plane is zero.