To solve this problem, we need to calculate the magnitude (or length) of the vector $\vec{a} = 2\vec{i} + \vec{j} - 2\vec{k}$ in 3-dimensional space.

The formula for the magnitude of a vector $\vec{a} = a_x \vec{i} + a_y \vec{j} + a_z \vec{k}$ is:

$|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$

In this case:

• $a_x = 2$
• $a_y = 1$
• $a_z = -2$

Substitute these values into the formula:

$|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2}$

$|\vec{a}| = \sqrt{4 + 1 + 4}$

$|\vec{a}| = \sqrt{9}$

$|\vec{a}| = 3$

So, the correct answer is 3.

Would you like more details or have any further questions?

Related Questions

1. What is the magnitude of a vector in 2D space, and how does it differ from 3D?
2. How would the magnitude change if one component of $\vec{a}$ were negative?
3. How is the direction of a vector related to its components?
4. Can you determine a vector's direction angle in 3D space?
5. How does vector addition affect the magnitudes of vectors?

Tip

Remember that the magnitude of a vector is always non-negative, regardless of the signs of its components.