Solve the vector problems in the image, including finding norms/magnitudes, vector addition/subtraction, scalar multiplication, and unit vectors.

Steps to Solve

1. Find the norm of vector $u = (1, 1)$

To find the norm (magnitude) of a vector $(x, y)$, we use the formula $\sqrt{x^2 + y^2}$. For $u = (1, 1)$, the norm is $|u| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.

2. Find the norm of vector $v = (-2, 1)$

Using the same formula, for $v = (-2, 1)$, the norm is $|v| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.

3. Find the norm of vector $u = (5, 0)$

For $u = (5, 0)$, the norm is $|u| = \sqrt{5^2 + 0^2} = \sqrt{25 + 0} = \sqrt{25} = 5$.

4. Find the norm of vector $v = (\frac{1}{4}, \frac{3}{2})$

For $v = (\frac{1}{4}, \frac{3}{2})$, the norm is $|v| = \sqrt{(\frac{1}{4})^2 + (\frac{3}{2})^2} = \sqrt{\frac{1}{16} + \frac{9}{4}} = \sqrt{\frac{1}{16} + \frac{36}{16}} = \sqrt{\frac{37}{16}} = \frac{\sqrt{37}}{4}$.

5. Find the norm of vector $u = (\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{5})$

For $u = (\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{5})$, the norm is $|u| = \sqrt{(\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{3}}{5})^2} = \sqrt{\frac{2}{4} + \frac{3}{25}} = \sqrt{\frac{1}{2} + \frac{3}{25}} = \sqrt{\frac{25}{50} + \frac{6}{50}} = \sqrt{\frac{31}{50}} = \frac{\sqrt{31}}{5\sqrt{2}} = \frac{\sqrt{62}}{10}$.

6. Find $|u + v|$ when $u = 3i + 5j$ and $v = 7i - j$

First, find $u + v = (3i + 5j) + (7i - j) = (3+7)i + (5-1)j = 10i + 4j$. Then, find the norm: $|u + v| = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} = 2\sqrt{29}$.

7. Find $|u - v|$ when $u = 3i + 5j$ and $v = 7i - j$

First, find $u - v = (3i + 5j) - (7i - j) = (3-7)i + (5-(-1))j = -4i + 6j$. Then, find the norm: $|u - v| = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

8. Find $r(u + 2v)$ when $u = 3i + 5j$, $v = 7i - j$, and $r \in R$

First, find $u + 2v = (3i + 5j) + 2(7i - j) = (3i + 5j) + (14i - 2j) = (3+14)i + (5-2)j = 17i + 3j$. Then, multiply by $r$: $r(u + 2v) = r(17i + 3j) = 17ri + 3rj$.

9. Find $|\frac{1}{4}u - v|$ when $u = 3i + 5j$ and $v = 7i - j$

First, find $\frac{1}{4}u = \frac{1}{4}(3i + 5j) = \frac{3}{4}i + \frac{5}{4}j$. Then, find $\frac{1}{4}u - v = (\frac{3}{4}i + \frac{5}{4}j) - (7i - j) = (\frac{3}{4} - 7)i + (\frac{5}{4} - (-1))j = (\frac{3}{4} - \frac{28}{4})i + (\frac{5}{4} + \frac{4}{4})j = -\frac{25}{4}i + \frac{9}{4}j$. Finally, find the norm: $|\frac{1}{4}u - v| = \sqrt{(-\frac{25}{4})^2 + (\frac{9}{4})^2} = \sqrt{\frac{625}{16} + \frac{81}{16}} = \sqrt{\frac{706}{16}} = \frac{\sqrt{706}}{4}$.

10. Find a unit vector in the direction of the vector (3, 6)

First, find the magnitude of the vector (3, 6): $\sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$. Then, divide the vector by its magnitude to find the unit vector: $(\frac{3}{3\sqrt{5}}, \frac{6}{3\sqrt{5}}) = (\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}) = (\frac{\sqrt{5}}{5}, \frac{2\sqrt{5}}{5})$.

11. Find a unit vector in the direction opposite to the vector (1, 3)

First, find the magnitude of the vector (1, 3): $\sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$. The unit vector in the direction of (1, 3) is $(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}})$. The unit vector in the opposite direction is $(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}) = (-\frac{\sqrt{10}}{10}, -\frac{3\sqrt{10}}{10})$.

12. Find two unit vectors, one in the same direction and the other opposite to the vector $u = (a, b) \neq 0$

First, find the magnitude of the vector $u = (a, b)$: $|u| = \sqrt{a^2 + b^2}$. The unit vector in the same direction is $(\frac{a}{\sqrt{a^2 + b^2}}, \frac{b}{\sqrt{a^2 + b^2}})$. The unit vector in the opposite direction is $(-\frac{a}{\sqrt{a^2 + b^2}}, -\frac{b}{\sqrt{a^2 + b^2}})$.