To find the unit vector in the direction of v = (1, -2), follow these steps:

1. Find the magnitude of vector v: The magnitude $\| \vec{v} \|$ of a vector $\vec{v} = (x, y)$ is given by: $\| \vec{v} \| = \sqrt{x^2 + y^2}$ So, for $\vec{v} = (1, -2)$: $\| \vec{v} \| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$

2. Divide each component of the vector by its magnitude: The unit vector $\hat{v}$ is calculated by dividing each component of $\vec{v}$ by its magnitude: $\hat{v} = \left( \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}} \right)$ You can rationalize the denominator, but the above form is sufficient unless otherwise required.

Thus, the unit vector in the direction of $\vec{v}$ is: $\left( \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}} \right)$

Would you like more details or help with another part?

Here are 5 related questions:

1. What is the significance of a unit vector in physics or geometry?
2. How can you generalize this process for any two-dimensional vector?
3. What happens if the vector components are all negative or zero?
4. How can unit vectors be used in determining direction in 3D space?
5. What does it mean if the magnitude of a vector is 1?

Tip: A unit vector always has a magnitude of 1 and retains the direction of the original vector.