The problem asks us to find the value of $x$ such that two given vectors $\vec{a}$ and $\vec{b}$ are perpendicular. There are two sub-problems: (1) $\vec{a} = (2, -3)$ and $\vec{b} = (1, x)$ (2) $\vec{a} = (x, -5)$ and $\vec{b} = (x-4, 1)$

1. Problem Description

The problem asks us to find the value of $x$ such that two given vectors $\vec{a}$ and $\vec{b}$ are perpendicular. There are two sub-problems:

(1) $\vec{a} = (2, -3)$ and $\vec{b} = (1, x)$

(2) $\vec{a} = (x, -5)$ and $\vec{b} = (x-4, 1)$

2. Solution Steps

Two vectors are perpendicular if their dot product is zero.

(1) Given $\vec{a} = (2, -3)$ and $\vec{b} = (1, x)$, their dot product is

$$\vec{a} \cdot \vec{b} = (2)(1) + (-3)(x) = 2 - 3x$$

For $\vec{a}$ and $\vec{b}$ to be perpendicular, we must have

$$2 - 3x = 0$$

Solving for $x$:

$$3x = 2$$

$$x = \frac{2}{3}$$

(2) Given $\vec{a} = (x, -5)$ and $\vec{b} = (x-4, 1)$, their dot product is

$$\vec{a} \cdot \vec{b} = (x)(x-4) + (-5)(1) = x^2 - 4x - 5$$

For $\vec{a}$ and $\vec{b}$ to be perpendicular, we must have

$$x^2 - 4x - 5 = 0$$

Factoring the quadratic equation:

$$(x - 5)(x + 1) = 0$$

So, $x = 5$ or $x = -1$.

3. Final Answer

(1) $x = \frac{2}{3}$

(2) $x = 5$ or $x = -1$