This image presents several vector magnitude calculations in different dimensions. Let's go through each one step by step to verify the results.

(a) Vector $\mathbf{v} = \langle 1, -1 \rangle$

• The magnitude of a 2D vector $\mathbf{v} = \langle v_1, v_2 \rangle$ is calculated as: $\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}$ For $\mathbf{v} = \langle 1, -1 \rangle$: $\|\mathbf{v}\| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$ Answer is correct: $\|\mathbf{v}\| = \sqrt{2}$.

(b) Vector $\mathbf{v} = -9\hat{i} + 3\hat{j}$

• This is also a 2D vector $\mathbf{v} = \langle -9, 3 \rangle$. The magnitude is: $\|\mathbf{v}\| = \sqrt{(-9)^2 + 3^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}$ Answer is correct: $\|\mathbf{v}\| = 3\sqrt{10}$.

(c) Vector $\mathbf{v} = \langle -1, 3, 2 \rangle$

• For a 3D vector $\mathbf{v} = \langle v_1, v_2, v_3 \rangle$, the magnitude is: $\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ For $\mathbf{v} = \langle -1, 3, 2 \rangle$: $\|\mathbf{v}\| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$ Answer is correct: $\|\mathbf{v}\| = \sqrt{14}$.

(d) Vector $\mathbf{v} = -\hat{i} + 2\hat{j} + 3\hat{k}$

• This is another 3D vector. We calculate its magnitude similarly: $\|\mathbf{v}\| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$ Answer is correct: $\|\mathbf{v}\| = \sqrt{14}$.

All answers provided in the image are correct!

Do you need further explanations or have any specific questions?

Here are 5 related questions to extend this topic:

1. How would you calculate the unit vector for each of these vectors?
2. What happens to the magnitude of a vector if you multiply it by a scalar?
3. Can you compute the magnitude of a vector in higher dimensions, say 4D or 5D?
4. How does the magnitude formula change when dealing with complex numbers?
5. What is the physical interpretation of the magnitude of a vector in physics?

Tip: Always double-check signs and square each component carefully when calculating magnitudes!