To find the norm (or magnitude) of each vector $\mathbf{v}$, we use the formula:

$||\mathbf{v}|| = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2}$

where $x_1, x_2, \dots, x_n$ are the components of the vector $\mathbf{v}$.

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

$||\mathbf{v}|| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$

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

Here, the vector components are $-9$ and $3$. So,

$||\mathbf{v}|| = \sqrt{(-9)^2 + 3^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}$

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

$||\mathbf{v}|| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$

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

Here, the vector components are $-1$, $2$, and $3$. So,

$||\mathbf{v}|| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

Thus, the norms are:

• (a) $||\mathbf{v}|| = \sqrt{2}$
• (b) $||\mathbf{v}|| = 3\sqrt{10}$
• (c) $||\mathbf{v}|| = \sqrt{14}$
• (d) $||\mathbf{v}|| = \sqrt{14}$

Would you like more details on any of these calculations?

Here are five related questions to deepen your understanding:

1. How would the norm formula change for 4-dimensional vectors?
2. What is the geometrical meaning of the norm of a vector?
3. How does the norm relate to the distance between two points in space?
4. Can you derive the norm for complex vectors?
5. How would you normalize each of these vectors?

Tip: The norm of a vector gives its length in the corresponding space, and it's always a non-negative value.