Vector components worksheet with written exercises and a coordinate grid for graphing vectors.

Let's solve the problem step by step based on the image you provided. The worksheet is titled "Components of Vectors" and has two parts:
- Part A: Write the component form of each vector, and find its initial point and terminal point.
- Part B: Analyze a graph and write the vectors shown in component form.

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Part A: Component Form of Vectors

We are given vectors with initial and terminal points, and we need to write them in component form:

> The component form of a vector $\vec{v}$ from point $A(x_1, y_1)$ to point $B(x_2, y_2)$ is:
> $$
> \vec{v} = \langle x_2 - x_1, y_2 - y_1 \rangle
> $$

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#### a) $\overrightarrow{AB}$, $A(2, 4)$, $B(5, 8)$

$$
\overrightarrow{AB} = \langle 5 - 2, 8 - 4 \rangle = \langle 3, 4 \rangle
$$

✔ Answer: $\boxed{\langle 3, 4 \rangle}$

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#### b) $\overrightarrow{CD}$, $C(3, -2)$, $D(6, 5)$

$$
\overrightarrow{CD} = \langle 6 - 3, 5 - (-2) \rangle = \langle 3, 7 \rangle
$$

✔ Answer: $\boxed{\langle 3, 7 \rangle}$

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#### c) $\overrightarrow{MN}$, $M(4, 5)$, $N(5, -8)$

$$
\overrightarrow{MN} = \langle 5 - 4, -8 - 5 \rangle = \langle 1, -13 \rangle
$$

✔ Answer: $\boxed{\langle 1, -13 \rangle}$

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#### d) $\overrightarrow{PQ}$, $P(-1, -6)$, $Q(8, 6)$

$$
\overrightarrow{PQ} = \langle 8 - (-1), 6 - (-6) \rangle = \langle 9, 12 \rangle
$$

✔ Answer: $\boxed{\langle 9, 12 \rangle}$

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#### e) $\overrightarrow{RS}$

This one is missing initial and terminal points. But since it's not specified, perhaps it was meant to be read from the graph in Part B? Let's move to Part B and see if we can identify it.

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Part B: Graph Analysis

We are given a coordinate grid with several vectors drawn. We need to determine their component forms.

Let’s analyze each labeled vector:

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#### i) $\vec{r}$

From the graph:
- Starts at origin $(0, 0)$
- Ends at point $(3, 2)$

So,
$$
\vec{r} = \langle 3 - 0, 2 - 0 \rangle = \langle 3, 2 \rangle
$$

✔ Answer: $\boxed{\langle 3, 2 \rangle}$

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#### ii) $\vec{q}$

From the graph:
- Starts at $(0, 0)$
- Ends at $(-2, 3)$

$$
\vec{q} = \langle -2 - 0, 3 - 0 \rangle = \langle -2, 3 \rangle
$$

✔ Answer: $\boxed{\langle -2, 3 \rangle}$

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#### iii) $\vec{s}$

From the graph:
- Starts at $(0, 0)$
- Ends at $(-3, -2)$

$$
\vec{s} = \langle -3 - 0, -2 - 0 \rangle = \langle -3, -2 \rangle
$$

✔ Answer: $\boxed{\langle -3, -2 \rangle}$

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#### iv) $\vec{p}$

From the graph:
- Starts at $(0, 0)$
- Ends at $(2, -3)$

$$
\vec{p} = \langle 2 - 0, -3 - 0 \rangle = \langle 2, -3 \rangle
$$

✔ Answer: $\boxed{\langle 2, -3 \rangle}$

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Now, going back to Part A, question e) $\overrightarrow{RS}$ — this might have been intended to be read from the graph, but since no points were given for $R$ and $S$, and there's no label "RS" on the graph, it may be an error or omission.

Alternatively, maybe the vectors in Part B are meant to be used as examples, but since they’re labeled $\vec{r}, \vec{q}, \vec{s}, \vec{p}$, not $\overrightarrow{RS}$, it's likely that $\overrightarrow{RS}$ was not included in the image or is missing.

But if we assume $\overrightarrow{RS}$ refers to one of the vectors in the graph, and if we suppose $R$ and $S$ are endpoints of a vector, then we’d need to identify them.

However, none of the vectors in the graph are labeled as $RS$. So unless there's a typo, we cannot compute $\overrightarrow{RS}$ without more information.

But based on the graph, all four vectors are already accounted for: $\vec{r}, \vec{q}, \vec{s}, \vec{p}$.

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✔ Final Answers:

#### Part A:
a) $\overrightarrow{AB} = \langle 3, 4 \rangle$
b) $\overrightarrow{CD} = \langle 3, 7 \rangle$
c) $\overrightarrow{MN} = \langle 1, -13 \rangle$
d) $\overrightarrow{PQ} = \langle 9, 12 \rangle$
e) $\overrightarrow{RS}$: Not enough information (no coordinates given)

#### Part B:
i) $\vec{r} = \langle 3, 2 \rangle$
ii) $\vec{q} = \langle -2, 3 \rangle$
iii) $\vec{s} = \langle -3, -2 \rangle$
iv) $\vec{p} = \langle 2, -3 \rangle$

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Let me know if you'd like help identifying $\overrightarrow{RS}$ if it's supposed to be one of the vectors in the graph!