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A position vector describes the location of a point in space relative to an origin. It is typically represented using coordinates in a 2D or 3D space. For a point with coordinates \(x\) and \(y\) in 2D, the position vector \vec{r}\ is given by:
\[ \vec{r} = x\hat{i} + y\hat{j} \] Here, \(x\) and \(y\) are the coordinates, while \(\hat{i}\) and \(\hat{j}\) are the unit vectors along the \(\hat{x}\) and \(\hat{y}\) axes respectively. For example, if a watermelon seed is at coordinates \(-5.0\mathrm{~m}, 8.0\mathrm{~m}\), its position vector becomes:
\[ \vec{r} = -5.0 \hat{i} + 8.0 \hat{j} \] This notation provides a clear, easy-to-understand representation of the point's location in space. For visual understanding, one can sketch the vector on a coordinate system by plotting the point and drawing an arrow from the origin to that point.

The magnitude of a vector, often referred to as the vector's length, measures how far the point is from the origin. To determine the magnitude of a position vector \vec{r}\, we use the Pythagorean theorem. The formula to compute the magnitude \| \vec{r} \| is: \[ \| \vec{r} \| = \sqrt{x^2 + y^2} \] Plugging in the coordinates of the watermelon seed (\(-5.0\ \mathrm{~m}, 8.0\ \mathrm{~m}\)), we get:
\[ \| \vec{r} \| = \sqrt{(-5.0)^2 + (8.0)^2} = \sqrt{25 + 64} = \sqrt{89} \, or approximately 9.43 \mathrm{~m} \] This magnitude tells us the straight-line distance between the point and the origin.

To find the angle a vector makes relative to the positive \(x\)-axis, we use the arctangent function. Knowing the \(x\) and \(y\) coordinates, the formula for the angle \(\theta\) is:
\[ \theta\ = \tan^{-1} \(\frac{y}{x}\)\ \] For the watermelon seed at \(-5.0\ \mathrm{~m}, 8.0\ \mathrm{~m} \), we find:
\[ \theta = \tan^{-1} \left( \frac{8.0}{-5.0} \right) \approx -58.0^\circ \] However, because our vector is in the second quadrant, we adjust the angle:
\[ \theta = 180^\circ - 58.0^\circ = 122.0^\circ \] This adjusted angle accurately reflects the vector's direction in standard position, measured counterclockwise from the positive \(x\)-axis.

The displacement vector indicates the change in position from an initial point to a final point. It is calculated by subtracting the initial position vector from the final position vector. For an initial position \( (x_i, y_i) \) and a final position \( (x_f, y_f) \), the displacement vector \( \vec{d} \) is found as:
\[ \vec{d} = (x_f - x_i) \hat{i} + (y_f - y_i) \hat{j} \] In our example, if the watermelon seed moves from \(-5.0 \mathrm{~m}, 8.0 \mathrm{~m}\) to \(3.0 \mathrm{~m}, 0 \mathrm{~m} \), we calculate:
\[ \vec{d} = (3.0 - (-5.0)) \hat{i} + (0 - 8.0) \hat{j} = 8.0 \hat{i} - 8.0 \hat{j} \] This tells us the vector pointing from the initial to the final position.

A unit vector is a vector with a magnitude of one. It is used to specify direction. To express any vector in unit-vector notation, we divide the vector by its magnitude. For any vector \vec{A} = x\hat{i} + y\hat{j}, the unit vector \(\hat{A}\) is:
\[ \hat{A} = \frac{\vec{A}}{\|\vec{A}\|} \] Taking our displacement vector \vec{d} = 8.0\hat{i} - 8.0\hat{j}, with magnitude \|\vec{d}\| \, we find:

First, calculate the magnitude:

\[ \|\vec{d}\| = \sqrt{(8.0)^2 + (-8.0)^2} = \sqrt{64 + 64} = 11.31 \]
Now, express the unit vector:
\[ \hat{d} = \frac{8.0\hat{i} - 8.0\hat{j}}{11.31} \approx 0.71\hat{i} - 0.71\hat{j} \]
This unit vector shows direction without considering the magnitude, making it particularly useful for many physics and engineering applications.