The position vector of the point (1, 2, 0) is....Ai + j +kBi + 2j + kCi + 2jD2j + k

To determine the position vector of the point (1, 2, 0), we need to express it in terms of the unit vectors $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$. The coordinates of the point (1, 2, 0) correspond to the coefficients of these unit vectors.

Here's the step-by-step process:

1. Identify the coordinates: The point given is (1, 2, 0).

   • The x-coordinate is 1.
   • The y-coordinate is 2.
   • The z-coordinate is 0.

2. Express the position vector: The position vector $\mathbf{r}$ can be written as: $\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$ Substituting the coordinates (1, 2, 0) into this formula, we get: $\mathbf{r} = 1 \mathbf{i} + 2 \mathbf{j} + 0 \mathbf{k}$

3. Simplify the expression: Since $0 \mathbf{k}$ is zero, it can be omitted: $\mathbf{r} = \mathbf{i} + 2 \mathbf{j}$

Now, let's match this with the given options:

• A: $\mathbf{i} + \mathbf{j} + \mathbf{k}$
• B: $\mathbf{i} + 2 \mathbf{j} + \mathbf{k}$
• C: $\mathbf{i} + 2 \mathbf{j}$
• D: $2 \mathbf{j} + \mathbf{k}$

The correct answer is:

C: $\mathbf{i} + 2 \mathbf{j}$