To find two vectors parallel to $\mathbf{V} = (8, -6, 0)$ with a length of 15, we first need to normalize $\mathbf{V}$ and then scale it to the desired length.

Step 1: Find the magnitude of $\mathbf{V}$

The magnitude (or length) of $\mathbf{V} = (8, -6, 0)$ is calculated using the formula:

$\|\mathbf{V}\| = \sqrt{8^2 + (-6)^2 + 0^2} = \sqrt{64 + 36 + 0} = \sqrt{100} = 10$

Step 2: Normalize $\mathbf{V}$

The unit vector $\mathbf{u}$ in the direction of $\mathbf{V}$ is obtained by dividing each component of $\mathbf{V}$ by its magnitude:

$\mathbf{u} = \left(\frac{8}{10}, \frac{-6}{10}, \frac{0}{10}\right) = \left(\frac{4}{5}, -\frac{3}{5}, 0\right)$

Step 3: Scale to the desired length

To find a vector with a length of 15, we scale the unit vector by 15:

$\mathbf{V_1} = 15 \times \mathbf{u} = 15 \times \left(\frac{4}{5}, -\frac{3}{5}, 0\right) = \left(12, -9, 0\right)$

This is one of the parallel vectors.

Step 4: Find the opposite vector

To find the second vector, simply take the negative of $\mathbf{V_1}$:

$\mathbf{V_2} = -\mathbf{V_1} = \left(-12, 9, 0\right)$

Final Answer:

The two vectors parallel to $\mathbf{V} = (8, -6, 0)$ with a length of 15 are:

$\mathbf{V_1} = (12, -9, 0) \quad \text{and} \quad \mathbf{V_2} = (-12, 9, 0)$

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How do you find the angle between two vectors?
2. What is the significance of a unit vector?
3. How would you compute the cross product of two vectors?
4. What are the properties of parallel vectors in space?
5. How do you find the projection of one vector onto another?

Tip: When scaling vectors, always make sure you first normalize the vector to ensure you maintain its direction while adjusting its length.