Which of the following vectors is a unit vector?Group of answer choices<-1,0,0><-1,-1,-1><-1,0,1><1,1,1>

1. ### Break Down the Problem

To determine which vector is a unit vector, we need to calculate the magnitude of each vector. A vector is considered a unit vector if its magnitude is equal to 1.

2. ### Relevant Concepts

The magnitude $||\mathbf{v}||$ of a vector $\mathbf{v} = \langle x, y, z \rangle$ is calculated using the formula: $||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2}$

3. ### Analysis and Detail

We will calculate the magnitude for each of the given vectors:

1. For the vector $\langle -1, 0, 0 \rangle$: $||\langle -1, 0, 0 \rangle|| = \sqrt{(-1)^2 + 0^2 + 0^2} = \sqrt{1} = 1$

2. For the vector $\langle -1, -1, -1 \rangle$: $||\langle -1, -1, -1 \rangle|| = \sqrt{(-1)^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.732$

3. For the vector $\langle -1, 0, 1 \rangle$: $||\langle -1, 0, 1 \rangle|| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \approx 1.414$

4. For the vector $\langle 1, 1, 1 \rangle$: $||\langle 1, 1, 1 \rangle|| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.732$

4. ### Verify and Summarize

From the calculations, we find:

• $||\langle -1, 0, 0 \rangle|| = 1$ (unit vector)
• $||\langle -1, -1, -1 \rangle|| \approx 1.732$ (not a unit vector)
• $||\langle -1, 0, 1 \rangle|| \approx 1.414$ (not a unit vector)
• $||\langle 1, 1, 1 \rangle|| \approx 1.732$ (not a unit vector)

Final Answer

The vector that is a unit vector is $\langle -1, 0, 0 \rangle$.