Given the vector $\mathbf{v} = [-5, 12]$, we can find the requested vectors by applying basic vector operations.

(a) Same direction, twice as long:

To find a vector in the same direction but twice as long, multiply each component of the vector by 2:

[ \mathbf{v}_{\text{twice}} = 2 \times \mathbf{v} = 2 \times [-5, 12] = [-10, 24] ]

(b) Same direction, length 1 (unit vector):

To find a unit vector in the same direction, we first calculate the magnitude (length) of the vector $\mathbf{v}$:

$|\mathbf{v}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Now, divide each component of $\mathbf{v}$ by its magnitude to get the unit vector:

[ \mathbf{v}_{\text{unit}} = \frac{1}{13} \times [-5, 12] = \left[ \frac{-5}{13}, \frac{12}{13} \right] ]

Thus, the unit vector is:

$\mathbf{v}_{\text{unit}} = \left[ \frac{-5}{13}, \frac{12}{13} \right]$

Summary:

• (a) Same direction, twice as long: $[-10, 24]$
• (b) Same direction, length 1: $\left[ \frac{-5}{13}, \frac{12}{13} \right]$

Would you like further explanation or details on any steps?

Related Questions:

1. How can you calculate the magnitude of a 3D vector?
2. What does it mean for a vector to be normalized?
3. How do you find a vector perpendicular to a given vector in 2D?
4. What is the difference between scalar multiplication and dot product?
5. How do you compute the cross product of two vectors in 3D?

Tip:

Always double-check the magnitude of your unit vector to ensure it equals 1, as a small mistake in calculation can affect the result.