Let’s solve this step by step.

We are tasked to find $\| c\mathbf{u} \|$, where:

$\mathbf{u} = \langle 5, -12 \rangle \quad \text{and} \quad c = -3.$

Step 1: Definition of $\| c\mathbf{u} \|$

The magnitude (norm) of a scaled vector $c\mathbf{u}$ is given by: $\| c\mathbf{u} \| = |c| \cdot \|\mathbf{u}\|.$

Step 2: Find $\|\mathbf{u}\|$

The magnitude of $\mathbf{u}$ is: $\|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2},$ where $u_1 = 5$ and $u_2 = -12$. Substituting: $\|\mathbf{u}\| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.$

Step 3: Scale the magnitude by $|c|$

The absolute value of $c = -3$ is $|c| = 3$. Thus: $\| c\mathbf{u} \| = |c| \cdot \|\mathbf{u}\| = 3 \cdot 13 = 39.$

Final Answer:

$\| c\mathbf{u} \| = 39.$

Would you like a detailed breakdown or further examples?

5 Related Questions:

1. How is the magnitude of a vector calculated in general?
2. What happens to a vector's magnitude if scaled by a constant $c$?
3. Can the magnitude of a vector ever be negative?
4. How do you compute the direction of a scaled vector?
5. How can this concept be applied to 3D vectors?

Tip:

Remember, scaling a vector by $c$ affects its magnitude by $|c|$, but its direction changes depending on the sign of $c$.