To find an expression for the position vector along the straight line connecting the origin $\mathbf{0} = 0\hat{i} + 0\hat{j} + 0\hat{k}$ and the point $\mathbf{P} = 3\hat{i} + \hat{j} + 2\hat{k}$, we can parametrize the line using a single parameter $t$, which ranges from 0 to 1.

The general approach for parametrizing a line segment is:

$\mathbf{r}(t) = (1 - t) \mathbf{0} + t \mathbf{P}$ Here:

• $\mathbf{0}$ is the position vector of the origin.
• $\mathbf{P}$ is the position vector of the point $(3, 1, 2)$.
• $t$ is a parameter, where $t = 0$ corresponds to the origin, and $t = 1$ corresponds to the point $\mathbf{P}$.

Simplifying the equation:

$\mathbf{r}(t) = (1 - t)(0\hat{i} + 0\hat{j} + 0\hat{k}) + t(3\hat{i} + \hat{j} + 2\hat{k})$ $\mathbf{r}(t) = t(3\hat{i} + \hat{j} + 2\hat{k})$

Thus, the position vector $\mathbf{r}(t)$ of any point on the line is:

$\mathbf{r}(t) = 3t\hat{i} + t\hat{j} + 2t\hat{k}$

This gives the expression for the position vector at any point on the line, where $t$ varies between 0 and 1.

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Would you like further clarification on any of the steps or additional details?

Here are 5 related questions:

1. How can you extend this parametrization to a line in 3D that doesn't pass through the origin?
2. What happens to the parametrization if the point is in a different quadrant (e.g., negative coordinates)?
3. How would you modify the expression to represent the position vector at any point on the line segment from the point $\mathbf{P}$ to the origin, using a reverse parameter?
4. How do you calculate the distance between two points using a parameterized line?
5. Can you find a parametrization for a line that connects two arbitrary points in 3D space?

Tip: When working with line parametrizations, always think of $t$ as a scale factor that linearly interpolates between two endpoints, with $t = 0$ representing the first point and $t = 1$ representing the second point.