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We are asked to find the second coordinate (i.e., the y-component) of the displacement vector \(\overrightarrow{PQ}\).

To calculate the displacement vector \(\overrightarrow{PQ}\) from point \(P(-3, 1)\) to point \(Q(3, -2)\), we subtract the coordinates of \(P\) from \(Q\). The vector components are given by:

\[
\overrightarrow{PQ} = (x_Q - x_P, y_Q - y_P)
\]

Where:
- \(x_P = -3\), \(y_P = 1\)
- \(x_Q = 3\), \(y_Q = -2\)

So, the second (y-) coordinate of \(\overrightarrow{PQ}\) is:

\[
y_Q - y_P = -2 - 1 = -3
\]

Thus, the second coordinate of \(\overrightarrow{PQ}\) is \(-3\).

Would you like further details or have any questions? Here are some related questions for practice:

1. What is the first (x-) coordinate of the displacement vector \(\overrightarrow{PQ}\)?
2. How do you find the magnitude of the vector \(\overrightarrow{PQ}\)?
3. What is the direction of the vector \(\overrightarrow{PQ}\)?
4. How can you represent the vector \(\overrightarrow{PQ}\) in terms of its unit vector?
5. What are the properties of displacement vectors in general?

**Tip**: A displacement vector describes the change in position from one point to another. It's calculated by subtracting the initial point's coordinates from the final point's coordinates.

What is the second coordinate of the displacement vector \(\overrightarrow{PQ}\) given point P(-3, 1) and point Q(3, -2)?

Learn how to find the second (y-) coordinate of the displacement vector \(\overrightarrow{PQ}\) between points P(-3, 1) and Q(3, -2). The problem involves coordinate geometry and vector subtraction, suitable for students in Grades 8-10.

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