Let’s solve the questions step by step based on the given quadrilateral $OABC$:

(a) Find $\vec{AB}$ in terms of $\vec{a}$ and $\vec{b}$

Using the vector addition and subtraction properties: $\vec{AB} = \vec{OB} - \vec{OA}$ Given $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$, we get: $\vec{AB} = \vec{b} - \vec{a}$

(b) Show that $\vec{PQ} = \mu\vec{b}$, where $\mu$ is a scalar, and find the value of $\mu$.

The midpoints $P$ and $Q$ are defined as follows:

• $P$ is the midpoint of $\vec{OA}$, so: $\vec{OP} = \frac{\vec{OA}}{2} = \frac{\vec{a}}{2}$
• $Q$ is the midpoint of $\vec{AB}$, so: $\vec{OQ} = \vec{OP} + \frac{\vec{AB}}{2}$ Substituting $\vec{AB} = \vec{b} - \vec{a}$ and $\vec{OP} = \frac{\vec{a}}{2}$, we have: $\vec{OQ} = \frac{\vec{a}}{2} + \frac{\vec{b} - \vec{a}}{2} = \frac{\vec{a} + \vec{b} - \vec{a}}{2} = \frac{\vec{b}}{2}$ Thus, $\vec{PQ} = \vec{OQ} - \vec{OP} = \frac{\vec{b}}{2} - \frac{\vec{a}}{2}$. Rearranging: $\vec{PQ} = \frac{\vec{b}}{2}$ This implies that $\mu = \frac{1}{2}$.

(c) Show that $\vec{PQ}$ is parallel to $\vec{SR}$.

From the question:

• $\vec{OS} = \lambda \vec{OC} = \lambda \vec{c}$,
• $\vec{BR} = \lambda \vec{BC}$, and $\vec{BC} = \vec{c} - \vec{b}$.

We can calculate $\vec{SR}$ as: $\vec{SR} = \vec{OR} - \vec{OS}.$ Express $\vec{OR}$ using $\vec{BR} = \lambda \vec{BC}$: $\vec{OR} = \vec{OB} + \lambda (\vec{c} - \vec{b}).$ So: $\vec{SR} = \left(\vec{OB} + \lambda (\vec{c} - \vec{b})\right) - \lambda \vec{c}.$ Simplify: $\vec{SR} = \vec{b} + \lambda \vec{c} - \lambda \vec{b} - \lambda \vec{c}.$ This reduces to: $\vec{SR} = (1 - \lambda) \vec{b}.$ Since $\vec{PQ} = \frac{1}{2} \vec{b}$, both vectors are scalar multiples of $\vec{b}$, so $\vec{PQ} \parallel \vec{SR}$.

(d) Find the value of $\lambda$.

It is given that: $\vec{PQ} = \frac{3}{2} \vec{SR}.$ Substitute: $\frac{\vec{b}}{2} = \frac{3}{2} (1 - \lambda) \vec{b}.$ Cancel $\vec{b}$ (non-zero vector) and simplify: $\frac{1}{2} = \frac{3}{2} (1 - \lambda).$ Multiply through by 2: $1 = 3(1 - \lambda).$ Distribute: $1 = 3 - 3\lambda.$ Solve for $\lambda$: $3\lambda = 3 - 1 \quad \Rightarrow \quad \lambda = \frac{2}{3}.$

Final Answers:

(a) $\vec{AB} = \vec{b} - \vec{a}$
(b) $\mu = \frac{1}{2}$
(c) $\vec{PQ} \parallel \vec{SR}$
(d) $\lambda = \frac{2}{3}$

Would you like detailed explanations for any part?

Here are five related questions to explore:

1. What are the properties of parallel vectors in vector algebra?
2. How do you find midpoints of line segments using vectors?
3. What are scalar multiples, and how do they determine parallelism?
4. How can vector projections help verify relationships like $\vec{PQ} \parallel \vec{SR}$?
5. How does the choice of $\lambda$ affect other geometric properties of the quadrilateral?

Tip: Always verify vector relationships using geometric intuition or diagrams to check your results!