Find a vector $\vec{b}$ that is perpendicular to the vector $\vec{a} = (1, 3)$ and has a magnitude of $\sqrt{5}$.

1. Problem Description

Find a vector $\vec{b}$ that is perpendicular to the vector $\vec{a} = (1, 3)$ and has a magnitude of $\sqrt{5}$.

2. Solution Steps

Let $\vec{b} = (x, y)$. Since $\vec{b}$ is perpendicular to $\vec{a}$, their dot product must be zero.

$$\vec{a} \cdot \vec{b} = 0$$

$$(1, 3) \cdot (x, y) = 0$$

$$1 \cdot x + 3 \cdot y = 0$$

$$x + 3y = 0$$

$$x = -3y$$

Also, the magnitude of $\vec{b}$ is given as $\sqrt{5}$.

$$|\vec{b}| = \sqrt{5}$$

$$\sqrt{x^2 + y^2} = \sqrt{5}$$

$$x^2 + y^2 = 5$$

Substituting $x = -3y$ into the equation $x^2 + y^2 = 5$:

$$(-3y)^2 + y^2 = 5$$

$$9y^2 + y^2 = 5$$

$$10y^2 = 5$$

$$y^2 = \frac{5}{10} = \frac{1}{2}$$

$$y = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$

When $y = \frac{\sqrt{2}}{2}$, $x = -3y = -3 \cdot \frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2}$.

When $y = -\frac{\sqrt{2}}{2}$, $x = -3y = -3 \cdot (-\frac{\sqrt{2}}{2}) = \frac{3\sqrt{2}}{2}$.

Thus, the two possible vectors for $\vec{b}$ are $(-\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ and $(\frac{3\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$.

3. Final Answer

The vectors are $\vec{b} = (-\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ and $\vec{b} = (\frac{3\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$.