Given the vector [−5, 12], find the following vectors:
(a) same direction, twice as long
(b) same direction, length 1

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.

Given the vector [−5, 12], find the following vectors: (a) same direction, twice as long (b) same direction, length 1

Learn how to calculate a vector twice as long and a unit vector in the same direction as [-5, 12]. This solution demonstrates vector scalar multiplication, magnitude calculation, and the creation of unit vectors. Ideal for students in Grades 9-12.

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