We are given the following:

• $\vec{AB} = \vec{a}$, $\vec{AC} = \vec{b}$
• $\vec{CD} = 2 \vec{BC}$
• $\vec{NC} \parallel \vec{AB}$
• $\vec{AN} = k \vec{AD}$

From the problem, we also know:

• $\vec{BC} = \vec{b} - \vec{a}$
• $\vec{AD} = 3\vec{b} - 2\vec{a}$

We are tasked with finding the value of $k$.

Step-by-Step Solution:

1. Use the given relation for $\vec{CD}$: Since $\vec{CD} = 2\vec{BC}$, and we know that $\vec{BC} = \vec{b} - \vec{a}$, we substitute: $\vec{CD} = 2(\vec{b} - \vec{a}) = 2\vec{b} - 2\vec{a}$

2. Find the vector $\vec{ND}$: The point $N$ lies on $AD$, and we are given that $\vec{AN} = k \vec{AD}$. This means: $\vec{AN} = k(3\vec{b} - 2\vec{a})$ Hence, the vector $\vec{ND}$ is: $\vec{ND} = \vec{AD} - \vec{AN} = (3\vec{b} - 2\vec{a}) - k(3\vec{b} - 2\vec{a}) = (1 - k)(3\vec{b} - 2\vec{a})$

3. Use the condition that $\vec{NC} \parallel \vec{AB}$: Since $\vec{NC} \parallel \vec{AB}$, we know that $\vec{NC} = \lambda \vec{AB}$ for some scalar $\lambda$, where $\vec{AB} = \vec{a}$. So, $\vec{NC} = \lambda \vec{a}$

4. Find $\vec{NC}$: From the diagram, $\vec{NC} = \vec{AC} - \vec{AN}$, so: $\vec{NC} = \vec{b} - k(3\vec{b} - 2\vec{a}) = \vec{b} - k(3\vec{b}) + k(2\vec{a}) = (1 - 3k)\vec{b} + 2k\vec{a}$

5. Equate $\vec{NC}$ to $\lambda \vec{a}$: From step 3, $\vec{NC} = \lambda \vec{a}$. Therefore, we equate: $(1 - 3k)\vec{b} + 2k\vec{a} = \lambda \vec{a}$ For the vector equation to hold, the coefficients of $\vec{a}$ and $\vec{b}$ must match separately. So, the coefficient of $\vec{b}$ gives: $1 - 3k = 0 \quad \Rightarrow \quad k = \frac{1}{3}$

Thus, the value of $k$ is $\boxed{\frac{1}{3}}$.

Would you like a further breakdown of any steps or have questions?

Additional questions to explore:

1. How do you calculate the position vector of point $N$ given $k$?
2. Why is the condition $\vec{NC} \parallel \vec{AB}$ crucial for determining $k$?
3. What would change if $\vec{CD} \neq 2 \vec{BC}$?
4. How is the vector $\vec{AD}$ calculated from $\vec{AB}$ and $\vec{AC}$?
5. How do we check if two vectors are parallel using their components?

Tip:

Always break down vector equations into their components to handle parallel or perpendicular conditions easily.