part a express the position vector $ \mathbf{r}_{ab} $ in cartesian vector form. (figure 1) express your answer in terms of the unit vectors $ \mathbf{i} $, $ \mathbf{j} $, and $ \mathbf{k} $. to denote vectors in your answers, be sure to select the vec button. part b determine the magnitude of $ \mathbf{r}_{ab} $. express your answer to three significant figures and include the appropriate units.

Question

Accepted Answer

### Part A
# Explanation:
## Step1: Determine coordinates of A and B
From the figure, let's assume the origin is at the intersection point. For point A:
- x - coordinate: Let's see the horizontal (x) direction. The distance along x: From the diagram, the x - component for A: Let's analyze the horizontal (x) and vertical (y, z) directions. Wait, actually, let's define the coordinates properly. Let's find the position vectors of A and B.

Looking at the diagram:
- For point A: Let's find its coordinates. The x - direction: The horizontal (let's say x) has a length. Wait, maybe the coordinates are:

Let's assume the reference point. Let's see the distances:

For point A:
- In the x - y - z system, let's find the coordinates. Let's see the horizontal (x) direction: The distance from the origin (intersection) to A in x: Wait, maybe the coordinates of A: Let's look at the diagram. The x - component: The horizontal (x) has a 3m and 2m? Wait, maybe the coordinates of A are ( - 3, - 2, 0)? No, maybe better to find the displacement from A to B.

Wait, the position vector $\mathbf{r}_{AB}=\mathbf{r}_B-\mathbf{r}_A$

Let's find coordinates of A and B:

From the diagram:

- Point A: Let's see the x, y, z coordinates. Let's assume the origin is at the corner where the 4m, 3m, 3m meet? Wait, no. Let's look at the directions:

The x - direction: The horizontal (x) has a length. Let's see, the distance from A to the y - z plane (x - direction): For A, the x - coordinate: Let's see the diagram, A is at x = - 3m (assuming the direction towards B is positive x? Wait, no. Wait, the diagram has A with a 3m and 2m in x - like direction, and B has 3m, 3m, 4m? Wait, maybe:

Let's define the coordinates:

- Point A: Let's say in x: - 3m (since there's a 3m segment), y: - 2m (2m segment), z: 0? Wait, no. Wait, the vector from A to B: Let's find the differences in x, y, z.

Looking at the diagram:

- In the x - direction (horizontal): The distance from A to B in x: Let's see, the horizontal (x) component: From A to B, the x - difference: 3m (since B is 3m in x from the origin, and A is - 3m? Wait, maybe:

Wait, the diagram shows:

For point A:

- x: Let's see the horizontal (x) has a 3m and 2m? Wait, maybe the coordinates of A are ( - 3, - 2, 0) and B are (3, 3, 4)? Wait, no. Wait, the vertical (z) direction: 4m, y - direction: 3m (from the 3m segment), x - direction: 3m (from the 3m segment). Wait, maybe:

Let's re - examine:

The position vector $\mathbf{r}_{AB}$ is the vector from A to B. So we need to find the coordinates of A and B.

From the diagram:

- Point A: Let's assume the origin is at the intersection of the planes. Then, the coordinates of A:

In the x - direction: The distance from the origin to A in x is - 3m (since there's a 3m segment on the left), in y - direction: - 2m (2m segment), z - direction: 0m.

Point B: In x - direction: 3m, y - direction: 3m, z - direction: 4m.

So, $\mathbf{r}_A=- 3\mathbf{i}-2\mathbf{j}+0\mathbf{k}$

$\mathbf{r}_B = 3\mathbf{i}+3\mathbf{j}+4\mathbf{k}$

Then, $\mathbf{r}_{AB}=\mathbf{r}_B-\mathbf{r}_A=(3 - (-3))\mathbf{i}+(3 - (-2))\mathbf{j}+(4 - 0)\mathbf{k}$

## Step2: Calculate the components
Calculate each component:

- x - component: $3-(-3)=6$? Wait, no, maybe I made a mistake. Wait, maybe the coordinates of A are (0, - 2, 3)? No, this is confusing. Wait, let's look at the diagram again. The diagram has A with a 3m and 2m in the horizontal (x - y) plane? Wait, maybe the correct way is:

Looking at the diagram, the vector from A to B:

- In the x - direction (let's say the horizontal axis): The difference is $3 - (- 3)=6$? No, maybe the x - component is $3 - 0 = 3$? Wait, no. Wait, the diagram shows for A: 3m in one horizontal direction, 2m in another, and for B: 3m, 3m, 4m.

Wait, maybe the coordinates of A are ( - 3, - 2, 0) and B are (0, 3, 4). Then $\mathbf{r}_{AB}=(0 - (-3))\mathbf{i}+(3 - (-2))\mathbf{j}+(4 - 0)\mathbf{k}=3\mathbf{i}+5\mathbf{j}+4\mathbf{k}$? No, that doesn't match. Wait, maybe the correct coordinates:

Wait, the problem is about position vector $\mathbf{r}_{AB}$. Let's look at the diagram:

The horizontal (x - y) plane: A is at a position with x: 3m left, y: 2m back, and B is at x: 3m right, y: 3m front, z: 4m up.

So, displacement in x: $3 - (-3)=6$? No, maybe the x - component is $3 - 0 = 3$? Wait, I think I messed up. Let's start over.

Let's define the coordinate system: Let the origin be at the point where the vertical (z) line meets the horizontal (x - y) plane.

- Point A: In the x - direction: - 3m (3m to the left of origin), y - direction: - 2m (2m behind origin), z - direction: 0m.

- Point B: In the x - direction: 3m (3m to the right of origin), y - direction: 3m (3m in front of origin), z - direction: 4m (4m above origin).

Then, $\mathbf{r}_A=-3\mathbf{i}-2\mathbf{j}+0\mathbf{k}$

$\mathbf{r}_B = 3\mathbf{i}+3\mathbf{j}+4\mathbf{k}$

Then, $\mathbf{r}_{AB}=\mathbf{r}_B-\mathbf{r}_A=(3 - (-3))\mathbf{i}+(3 - (-2))\mathbf{j}+(4 - 0)\mathbf{k}=6\mathbf{i}+5\mathbf{j}+4\mathbf{k}$? No, that can't be. Wait, maybe the x - component is $3 - 0 = 3$, y - component is $3 - (-2)=5$, z - component is $4 - 0 = 4$. Wait, no, maybe the coordinates of A are (0, - 2, 3) and B are (3, 3, 7)? No, this is not working. Wait, maybe the correct approach is to look at the differences:

From the diagram, the horizontal (x - y) plane: The distance from A to B in x: 3m (since B is 3m in x from the origin, and A is 0 in x? No, the diagram shows A with a 3m and 2m in the horizontal. Wait, maybe the vector $\mathbf{r}_{AB}$ has components:

- x: 3m (since from A to B, x increases by 3m)

- y: 3 + 2 = 5m (since A is 2m in negative y, B is 3m in positive y, so 3 - (-2)=5)

- z: 4m (since B is 4m above A)

So, $\mathbf{r}_{AB}=3\mathbf{i}+5\mathbf{j}+4\mathbf{k}$ meters? Wait, no, maybe I made a mistake. Wait, let's check the diagram again. The diagram has A with 3m and 2m in the horizontal (x - y) plane, and B with 3m, 3m, 4m. So the displacement from A to B:

- In x: 3m (from A's x to B's x: 3 - 0 = 3? No, A is at x = - 3, B at x = 0? No, this is too confusing. Wait, maybe the correct components are:

Looking at the figure, the vector $\mathbf{r}_{AB}$ has:

- x - component: 3m (horizontal right)

- y - component: 3 + 2 = 5m (vertical front, since A is 2m back, B is 3m front)

- z - component: 4m (vertical up)

So, $\mathbf{r}_{AB}=3\mathbf{i}+5\mathbf{j}+4\mathbf{k}$ m. Wait, but let's calculate the magnitude for part B to check. The magnitude would be $\sqrt{3^2 + 5^2+4^2}=\sqrt{9 + 25 + 16}=\sqrt{50}\approx7.07$, but that doesn't seem right. Wait, maybe the x - component is 6m? Let's recalculate:

If A is at x = - 3, B at x = 3, then x - component is 3 - (-3)=6.

y - component: B is at y = 3, A at y = - 2, so 3 - (-2)=5.

z - component: B at z = 4, A at z = 0, so 4 - 0 = 4.

Then magnitude is $\sqrt{6^2+5^2 + 4^2}=\sqrt{36 + 25+16}=\sqrt{77}\approx8.77$. But maybe the correct components are:

Wait, the diagram shows for A: 3m in one horizontal direction, 2m in another, and for B: 3m, 3m, 4m. So the displacement from A to B:

- x: 3 - (-3)=6? No, maybe A is at (0, - 2, 3) and B is at (3, 3, 7). No, this is not working. Wait, maybe the problem is in the coordinate system. Let's assume that the position of A is ( - 3, - 2, 0) and B is (3, 3, 4). Then $\mathbf{r}_{AB}=(3 - (-3))\mathbf{i}+(3 - (-2))\mathbf{j}+(4 - 0)\mathbf{k}=6\mathbf{i}+5\mathbf{j}+4\mathbf{k}$ m.

# Answer (Part A):
$\mathbf{r}_{AB}=\boxed{6\mathbf{i} + 5\mathbf{j}+4\mathbf{k}}$ m

### Part B
# Explanation:
## Step1: Recall the formula for magnitude of a vector
The magnitude of a vector $\mathbf{v}=v_x\mathbf{i}+v_y\mathbf{j}+v_z\mathbf{k}$ is given by $|\mathbf{v}|=\sqrt{v_x^2 + v_y^2+v_z^2}$

## Step2: Substitute the components of $\mathbf{r}_{AB}$
From Part A, $\mathbf{r}_{AB}=6\mathbf{i}+5\mathbf{j}+4\mathbf{k}$ m, so $v_x = 6$, $v_y = 5$, $v_z = 4$

Then, $|\mathbf{r}_{AB}|=\sqrt{6^2 + 5^2+4^2}=\sqrt{36 + 25 + 16}=\sqrt{77}\approx8.77$ m (to three significant figures)

# Answer (Part B):
$\boxed{8.77\ \text{m}}$

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QUESTION IMAGE

Question

Explanation:

Part A

Step1: Determine coordinates of A and B

Step2: Calculate the components

Step1: Recall the formula for magnitude of a vector

Step2: Substitute the components of \(\mathbf{r}_{AB}\)

Answer:

Part B