The problem asks to determine the values of $\alpha$ and $\beta$ for which the vectors
$\vec{a} = 2\vec{i} - \alpha \vec{j} + 4\vec{k} \quad \text{and} \quad \vec{b} = 6\vec{i} + 8\vec{j} + \beta \vec{k}$
are collinear.

Condition for Collinearity

Two vectors $\vec{a}$ and $\vec{b}$ are collinear if there exists a scalar $k$ such that:
$\vec{a} = k \cdot \vec{b}.$
This means that the components of $\vec{a}$ and $\vec{b}$ are proportional: $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3},$ where $a_1, a_2, a_3$ and $b_1, b_2, b_3$ are the components of vectors $\vec{a}$ and $\vec{b}$, respectively.

Given Components

From the problem:

• $\vec{a} = (2, -\alpha, 4)$
• $\vec{b} = (6, 8, \beta)$

Equating Proportions

For collinearity: $\frac{2}{6} = \frac{-\alpha}{8} = \frac{4}{\beta}.$

1. Start with the first ratio: $\frac{2}{6} = \frac{1}{3}.$

Thus: $\frac{-\alpha}{8} = \frac{1}{3} \quad \Rightarrow \quad -\alpha = \frac{8}{3} \quad \Rightarrow \quad \alpha = -\frac{8}{3}.$

2. Now consider the last ratio: $\frac{4}{\beta} = \frac{1}{3}.$

Thus: $4 = \frac{\beta}{3} \quad \Rightarrow \quad \beta = 12.$

Final Answer

The values of $\alpha$ and $\beta$ for which the vectors are collinear are:
$\alpha = -\frac{8}{3}, \quad \beta = 12.$

Do you want me to clarify any step or expand the explanation further?

Related Questions:

1. What is the condition for orthogonality of two vectors?
2. How do you determine the angle between two vectors using their dot product?
3. How do you solve for the scalar triple product of three vectors?
4. What is the geometric meaning of vector collinearity?
5. How do you express a vector in parametric form?

Tip: Always check the proportionality of components when determining collinearity of vectors.