Given the vector $\mathbf{v} = \left(-2\sqrt{4}, -2\sqrt{4}\right)$, let's calculate its magnitude $|\mathbf{v}|$ and direction $\theta$.

Step 1: Express the Vector Components

The vector components are given as: $\mathbf{v} = (-2 \times 2, -2 \times 2) = (-4, -4)$

Step 2: Calculate the Magnitude

The magnitude $|\mathbf{v}|$ of a vector $(x, y)$ is given by: $|\mathbf{v}| = \sqrt{x^2 + y^2}$

Substituting the values of $x$ and $y$: $|\mathbf{v}| = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66$

Step 3: Determine the Direction (Angle)

The direction $\theta$ is calculated using: $\theta = \tan^{-1}\left(\frac{y}{x}\right)$

Since both $x$ and $y$ are negative, the vector is located in the third quadrant. Therefore, the angle $\theta$ should be adjusted to be in this quadrant.

$\theta = \tan^{-1}\left(\frac{-4}{-4}\right) = \tan^{-1}(1) = 45^\circ$

However, because the vector lies in the third quadrant, the direction must be: $\theta = 180^\circ + 45^\circ = 225^\circ$

Final Answer:

• Magnitude $|\mathbf{v}|$ ≈ 5.66
• Direction $\theta$ = 225°

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you find the magnitude of a 3D vector?
2. What happens to the direction calculation when the vector is in different quadrants?
3. What is the relationship between the angle and the components of the vector?
4. How would you determine the unit vector in the same direction as $\mathbf{v}$?
5. How can you find the scalar projection of one vector onto another?

Tip: Always check the quadrant of your vector when calculating the direction to ensure the angle is accurate.