Displacement in Cartesian coordinate System

Displacement in Cartesian Coordinate

Hello students! Today we are going to learn about displacement in Cartesian coordinate. Don’t worry if it looks hard, we will make it very simple.

See, the logic is very simple. When an object moves, it changes its position. We need a proper system to track this change. That system is the Cartesian coordinate system, which uses the $X$, $Y$, and $Z$ axes. Displacement is simply the measurement of how far and in what direction the object has moved, measured in a straight line, using these $X,Y,Z$ axes. This topic is fundamental for Kinematics, so pay close attention.

Why do we study this? (Real Life Examples)

Physics is not just formulas, beta. It is about understanding the world around us. Why do we need displacement? Because we need to know the net change in position, not just the path taken.

For example, have you noticed how GPS in your mobile phone works? When you start driving from point A to point B, the road distance might be 10 km. But the displacement, which is the straight line from A to B, might only be 7 km. Engineers use the Cartesian system to calculate this shortest distance (the displacement vector) instantly, even if you are moving in three dimensions (like climbing a hill).

Another example is launching a weather rocket or a drone. Suppose a drone needs to lift off from the coordinates $(x_1,y_1,z_1)$ and deliver a package to the coordinates $(x_2,y_2,z_2)$. The flight control system must calculate the precise vector displacement $\mathrm{\Delta }\mathbf{r}$ to know exactly how much thrust and in which direction (vector) is required. If they only calculated distance, the drone might drift off course.

Even in something simple like a billiards table (which is a 2D plane), when a ball is hit, its initial position $(x_1,y_1)$ and final position $(x_2,y_2)$ define the displacement vector. This helps us predict its trajectory after collision.

Simplest Definition

Definition: Displacement ($\mathrm{\Delta }\mathbf{r}$) is defined as the vector quantity representing the shortest straight-line change in position of a body from its initial position (${\mathbf{r}}_1$) to its final position (${\mathbf{r}}_2$). It depends only on the start and end points, not on the path followed.

Diagram

Figure: Detailed description of diagram for displacement in Cartesian coordinate

*A diagram showing a standard 3D Cartesian coordinate system (X, Y, Z axes intersecting at Origin O). *Two points $P_1$ (initial position) and $P_2$ (final position) are marked. *The position vector ${\mathbf{r}}_1$ is drawn from O to $P_1$. *The position vector ${\mathbf{r}}_2$ is drawn from O to $P_2$. *The displacement vector $\mathrm{\Delta }\mathbf{r}$ is drawn as a straight line from $P_1$ to $P_2$. *This diagram clearly illustrates the vector addition: ${\mathbf{r}}_1+\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2$.

Key Formula Box

This is the main formula, you must remember this point for exams.

Detailed Derivation (Step-by-Step for Exams)

Step-by-Step Derivation

This derivation is very important. It is often asked in 5-mark questions. We will use the concept of position vectors and the Triangle Law of Vector Addition.

Step 1: Defining Initial and Final Position Vectors

Let us consider a body moving from an initial point $P_1$ to a final point $P_2$ in the 3D Cartesian coordinate system ($X,Y,Z$). The origin is denoted by $O$.

The position vector for the initial point $P_1(x_1,y_1,z_1)$ is ${\mathbf{r}}_1$.

$${\mathbf{r}}_1=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}\text{(Eq. 1)}$$

The position vector for the final point $P_2(x_2,y_2,z_2)$ is ${\mathbf{r}}_2$.

$${\mathbf{r}}_2=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}\text{(Eq. 2)}$$

Here, $\hat{i}$, $\hat{j}$, and $\hat{k}$ are the unit vectors along the $X$, $Y$, and $Z$ axes respectively.

Step 2: Applying the Vector Law

Now, see the diagram. According to the Triangle Law of Vector Addition, the initial position vector plus the displacement vector must equal the final position vector.

$${\mathbf{r}}_1+\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2$$

Therefore, the displacement vector $\mathrm{\Delta }\mathbf{r}$ is given by:

$$\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1\text{(Eq. 3)}$$

Step 3: Substitution and Calculation of the Displacement Vector

Now, let us put the values of Eq. 1 and Eq. 2 into Eq. 3. Remember, we substitute the entire vector expression.

$$\mathrm{\Delta }\mathbf{r}=(x_2\hat{i}+y_2\hat{j}+z_2\hat{k})-(x_1\hat{i}+y_1\hat{j}+z_1\hat{k})$$

We must now group the terms based on the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$.

$$\mathrm{\Delta }\mathbf{r}=(x_2\hat{i}-x_1\hat{i})+(y_2\hat{j}-y_1\hat{j})+(z_2\hat{k}-z_1\hat{k})$$

Taking the unit vectors common:

$$\mathrm{\Delta }\mathbf{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$$

This is the required formula for the displacement vector in Cartesian coordinates.

Step 4: Finding the Magnitude of Displacement

Since displacement is a vector, we often need to find its length, or magnitude, $|\mathrm{\Delta }\mathbf{r}|$. The magnitude is calculated using the generalized Pythagorean theorem in three dimensions.

If a vector $\mathbf{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$, then its magnitude is $|\mathbf{A}|=\sqrt{A_x^2+A_y^2+A_z^2}$.

Here, the components of $\mathrm{\Delta }\mathbf{r}$ are: $\mathrm{\Delta }x=(x_2-x_1)$ $\mathrm{\Delta }y=(y_2-y_1)$ $\mathrm{\Delta }z=(z_2-z_1)$

Therefore, the magnitude of the displacement is:

$$|\mathrm{\Delta }\mathbf{r}|=\sqrt{(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2+(\mathrm{\Delta }z{)}^2}$$

$$|\mathrm{\Delta }\mathbf{r}|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$$

This is the magnitude of the displacement, which is the shortest distance between $P_1$ and $P_2$.

Important Table

Solved Numericals (Exam Style)

Solved Numericals (Exam Style)

You must know how to apply these formulas. Practice these three problems well.

Q1: A particle moves in a 2D plane from point A (2 m, 3 m) to point B (8 m, 11 m). Calculate the displacement vector ($\mathrm{\Delta }\mathbf{r}$).

Ans: Given: Initial Position, $P_1(x_1,y_1)=(2,3)$ Final Position, $P_2(x_2,y_2)=(8,11)$

To Find: Displacement Vector $\mathrm{\Delta }\mathbf{r}$.

Formula: $$\mathrm{\Delta }\mathbf{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}$$

Calculation:

1. Calculate the change in $X$ component ($\mathrm{\Delta }x$): $$\mathrm{\Delta }x=8-2=6\text{ m}$$
2. Calculate the change in $Y$ component ($\mathrm{\Delta }y$): $$\mathrm{\Delta }y=11-3=8\text{ m}$$
3. Substitute these values into the formula: $$\mathrm{\Delta }\mathbf{r}=(6)\hat{i}+(8)\hat{j}$$

Final Answer with units: The displacement vector is $\mathrm{\Delta }\mathbf{r}=6\hat{i}+8\hat{j}\text{ m}$.

Q2: An aeroplane moves from an initial location $P_1$ defined by position vector ${\mathbf{r}}_1=(3\hat{i}-2\hat{j}+5\hat{k})$ m to a final location $P_2$ defined by position vector ${\mathbf{r}}_2=(-1\hat{i}+4\hat{j}+2\hat{k})$ m. Calculate the magnitude of the displacement.

Ans: Given: Initial Position Vector: ${\mathbf{r}}_1=3\hat{i}-2\hat{j}+5\hat{k}$ Final Position Vector: ${\mathbf{r}}_2=-1\hat{i}+4\hat{j}+2\hat{k}$

To Find: Magnitude of Displacement $|\mathrm{\Delta }\mathbf{r}|$.

Formula: $$\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$$ $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2+(\mathrm{\Delta }z{)}^2}$$

Calculation:

1. First, find the displacement vector $\mathrm{\Delta }\mathbf{r}$: $$\mathrm{\Delta }\mathbf{r}=(-1\hat{i}+4\hat{j}+2\hat{k})-(3\hat{i}-2\hat{j}+5\hat{k})$$ $$\mathrm{\Delta }\mathbf{r}=(-1-3)\hat{i}+(4-(-2))\hat{j}+(2-5)\hat{k}$$ $$\mathrm{\Delta }\mathbf{r}=-4\hat{i}+6\hat{j}-3\hat{k}\text{ m}$$

2. Now, calculate the magnitude using the components: $\mathrm{\Delta }x=-4$, $\mathrm{\Delta }y=6$, $\mathrm{\Delta }z=-3$. $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{(-4{)}^2+(6{)}^2+(-3{)}^2}$$ $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{16+36+9}$$ $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{61}$$ $$|\mathrm{\Delta }\mathbf{r}\approx 7.81\text{ m}$$

Final Answer with units: The magnitude of the displacement is $|\mathrm{\Delta }\mathbf{r}|=\sqrt{61}\text{ m}$ (or $7.81\text{ m}$).

Q3: A bug crawls from $P_1(1,1,1)$ to $P_2(3,4,-1)$. Find the displacement vector and the unit vector in the direction of displacement ($\hat{r}$).

Ans: Given: $P_1(x_1,y_1,z_1)=(1,1,1)$ $P_2(x_2,y_2,z_2)=(3,4,-1)$

To Find: $\mathrm{\Delta }\mathbf{r}$ and the unit vector $\hat{r}$.

Formula: $$\mathrm{\Delta }\mathbf{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$$ $$\hat{r}=\frac{\mathrm{\Delta }\mathbf{r}}{|\mathrm{\Delta }\mathbf{r}|}$$

Calculation:

1. Calculate the displacement vector $\mathrm{\Delta }\mathbf{r}$: $$\mathrm{\Delta }\mathbf{r}=(3-1)\hat{i}+(4-1)\hat{j}+(-1-1)\hat{k}$$ $$\mathrm{\Delta }\mathbf{r}=2\hat{i}+3\hat{j}-2\hat{k}$$

2. Calculate the magnitude $|\mathrm{\Delta }\mathbf{r}|$: $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{(2{)}^2+(3{)}^2+(-2{)}^2}$$ $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{4+9+4}=\sqrt{17}$$

3. Calculate the unit vector $\hat{r}$: $$\hat{r}=\frac{2\hat{i}+3\hat{j}-2\hat{k}}{\sqrt{17}}$$

Final Answer with units: Displacement vector: $2\hat{i}+3\hat{j}-2\hat{k}$ units. Unit Vector: $\hat{r}=\frac{2}{\sqrt{17}}\hat{i}+\frac{3}{\sqrt{17}}\hat{j}-\frac{2}{\sqrt{17}}\hat{k}$.

Exam Corner: Questions & Answers

Short Answer Questions (2 Marks)

Q1: Distinguish clearly between Distance and Displacement.

Ans:

Q2: Can displacement be zero even if the distance covered is large? Give an example.

Ans: Yes, displacement can be zero even if the distance covered is large. This happens when the initial position (${\mathbf{r}}_1$) and the final position (${\mathbf{r}}_2$) are the same. For example: If a runner completes one full lap around a circular track of 400 meters, the total distance covered is 400 meters. However, since the runner returns to the starting point, the initial and final positions coincide, making the net displacement zero.

Q3: What is the physical significance of the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ in the context of displacement?

Ans: The unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ define the specific direction of the displacement components in the 3D space.

1. $\hat{i}$ indicates displacement along the positive $X$-axis.
2. $\hat{j}$ indicates displacement along the positive $Y$-axis.
3. $\hat{k}$ indicates displacement along the positive $Z$-axis. They provide the necessary directional information, ensuring displacement is correctly treated as a vector.

Long Answer Questions (5 Marks)

Q1: Derive the complete formula for the displacement vector $\mathrm{\Delta }\mathbf{r}$ of a particle moving from $P_1(x_1,y_1,z_1)$ to $P_2(x_2,y_2,z_2)$.

Ans: (This answer requires the student to reproduce the derivation from Section 6.)

Concept: We define the movement using position vectors relative to the origin O, and then use the Triangle Law of Vector Addition to find the change in position.

1. Position Vector Definitions: The position vector for $P_1$ is: $${\mathbf{r}}_1=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$$ The position vector for $P_2$ is: $${\mathbf{r}}_2=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}$$

2. Displacement Definition: The displacement vector $\mathrm{\Delta }\mathbf{r}$ is the difference between the final and initial position vectors: $$\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$$

3. Substitution and Grouping: Substituting the component forms: $$\mathrm{\Delta }\mathbf{r}=(x_2\hat{i}+y_2\hat{j}+z_2\hat{k})-(x_1\hat{i}+y_1\hat{j}+z_1\hat{k})$$

We collect the coefficients of the respective unit vectors ($\hat{i},\hat{j},\hat{k}$): $$\mathrm{\Delta }\mathbf{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$$ This gives the required expression for the displacement vector in Cartesian coordinates.

Q2: Starting with the displacement vector formula, derive the expression for the magnitude of displacement.

Ans: 1. Starting Formula: We know the displacement vector $\mathrm{\Delta }\mathbf{r}$ is: $$\mathrm{\Delta }\mathbf{r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$$

Let the components of displacement be: $A_x=(x_2-x_1)$ $A_y=(y_2-y_1)$ $A_z=(z_2-z_1)$ So, $\mathrm{\Delta }\mathbf{r}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}$.

2. Application of Pythagorean Theorem: The magnitude of any vector $\mathbf{A}$ in 3D space is found by taking the square root of the sum of the squares of its components.

$$|\mathrm{\Delta }\mathbf{r}|=\sqrt{A_x^2+A_y^2+A_z^2}$$

3. Final Substitution: Substituting the component definitions back: $$|\mathrm{\Delta }\mathbf{r}|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$$

This derived expression gives the magnitude (length) of the displacement vector, which is always positive and measured in meters.

Multiple Choice Questions (MCQs)

These types of questions are common in IIT JAM and other entrance tests.

Q1: The initial position vector of a particle is ${\mathbf{r}}_1=2\hat{i}-5\hat{j}$ m and the final position vector is ${\mathbf{r}}_2=5\hat{i}-\hat{j}$ m. The displacement vector is: A) $7\hat{i}-6\hat{j}$ m B) $3\hat{i}+4\hat{j}$ m C) $-3\hat{i}-4\hat{j}$ m D) $3\hat{i}-6\hat{j}$ m

Ans: B) – Reasoning: $\mathrm{\Delta }\mathbf{r}=(5-2)\hat{i}+(-1-(-5))\hat{j}=3\hat{i}+4\hat{j}$.

Q2: If a particle moves such that its displacement vector is $\mathrm{\Delta }\mathbf{r}=4\hat{i}-3\hat{j}+12\hat{k}$ meters, what is the magnitude of the displacement? A) 19 m B) 13 m C) $\sqrt{169}$ m D) Both B and C

Ans: D) – Reasoning: Magnitude $=\sqrt{4^2+(-3{)}^2+{12}^2}=\sqrt{16+9+144}=\sqrt{169}=13\text{ m}$.

Q3: Which statement is always true regarding distance ($D$) and magnitude of displacement ($|\mathrm{\Delta }\mathbf{r}|$)? A) $D<|\mathrm{\Delta }\mathbf{r}|$ B) $D=|\mathrm{\Delta }\mathbf{r}|$ C) $D\ge |\mathrm{\Delta }\mathbf{r}|$ D) $D\le |\mathrm{\Delta }\mathbf{r}|$

Ans: C) – Reasoning: Distance must always be greater than or equal to the straight-line displacement (equality only occurs for straight-line motion without reversal).

Q4: A stone is thrown straight up and returns to the hand of the thrower. If the maximum height reached was 5 m, the net displacement of the stone is: A) 10 m B) 5 m C) -5 m D) Zero

Ans: D) – Reasoning: Initial and final positions are the same (the hand), so displacement is zero.

Q5: The position vector of a body changes from ${\mathbf{r}}_1=\hat{i}$ to ${\mathbf{r}}_2=\hat{j}$. The displacement vector is: A) $\hat{i}-\hat{j}$ B) $\hat{j}-\hat{i}$ C) $\sqrt{2}$ D) $\hat{i}+\hat{j}$

Ans: B) – Reasoning: $\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1=\hat{j}-\hat{i}$.

Summary

So, students, we have covered the concept of displacement vector thoroughly. Remember that displacement is the change in the position vector, and it is crucial because it accounts for direction. In 3D Cartesian coordinates, we simply subtract the initial components from the final components: $\mathrm{\Delta }x=x_2-x_1$, $\mathrm{\Delta }y=y_2-y_1$, $\mathrm{\Delta }z=z_2-z_1$.

I hope this concept is clear now. Practice the derivation twice, and make sure you understand how to calculate the magnitude using the Pythagorean theorem. All the best for your upcoming exams!