To find a vector a that has the same direction as $\langle -10, 3, 10 \rangle$ but with a length of 5, we can follow these steps:

Step 1: Find the magnitude of the original vector

The magnitude of $\langle -10, 3, 10 \rangle$ is calculated by: $\| \langle -10, 3, 10 \rangle \| = \sqrt{(-10)^2 + 3^2 + 10^2} = \sqrt{100 + 9 + 100} = \sqrt{209}$

Step 2: Create a unit vector in the same direction

To get a unit vector (a vector of length 1) in the same direction, we divide each component by the magnitude: $\text{Unit vector} = \left\langle \frac{-10}{\sqrt{209}}, \frac{3}{\sqrt{209}}, \frac{10}{\sqrt{209}} \right\rangle$

Step 3: Scale the unit vector to have length 5

To get a vector with length 5, we multiply the unit vector by 5: $\text{Vector } \mathbf{a} = 5 \times \left\langle \frac{-10}{\sqrt{209}}, \frac{3}{\sqrt{209}}, \frac{10}{\sqrt{209}} \right\rangle = \left\langle \frac{-50}{\sqrt{209}}, \frac{15}{\sqrt{209}}, \frac{50}{\sqrt{209}} \right\rangle$

So, the vector a that has the same direction as $\langle -10, 3, 10 \rangle$ and a length of 5 is: $\mathbf{a} = \left\langle \frac{-50}{\sqrt{209}}, \frac{15}{\sqrt{209}}, \frac{50}{\sqrt{209}} \right\rangle$

Let me know if you'd like further details or have any questions!

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Here are five related questions you might consider:

1. How do you find the unit vector in the opposite direction of a given vector?
2. What is the significance of a unit vector in vector operations?
3. How do you determine if two vectors are parallel?
4. How can you find the angle between two vectors using their dot product?
5. What is the difference between scaling a vector and changing its direction?

Tip: Always check the magnitude of your final answer to ensure it matches the desired length!