Position Vector in Cartesian coordinate

Introduction

Hello students! Today we are going to learn about Position vector in Cartesian Coordinate. Don't worry if it looks hard, we will make it very simple.

See, the Position Vector is one of the most fundamental concepts in Physics. It is simply a vector that tells us the location, or the position, of any particle or object with respect to a fixed reference point, which we usually call the Origin ($O$). When we use the Cartesian system (that is, $X,Y,$ and $Z$ axes), we are just giving three specific numbers to locate that point in space.

Why do we study this? (Real Life Examples)

Now, why should we study this concept? This is not just theoretical derivation, students. This is essential for understanding motion.

1. Missile Guidance System: For example, when the Army launches a missile, they must know the exact location of the target relative to the launch point. If the target is $500\text{ km}$ East, $20\text{ km}$ North, and $1\text{ km}$ above the sea level, these three distances ($x,y,z$) define the target's position vector from the launch pad. The missile’s computer calculates the flight path using this vector.

2. Using GPS in Mobile Phones: Have you noticed how your mobile phone shows your location on Google Maps? The satellites orbiting the Earth constantly communicate coordinates ($x,y,z$) to your phone. Your location on the map is nothing but the end point of your Position Vector relative to a standard global reference point. Without vectors, GPS cannot work.

3. Robotics in Factories: When a robot arm in a car manufacturing plant needs to pick up a specific component, the robot’s programming relies entirely on knowing the exact position vector of the component and the destination position vector where it needs to place it. These coordinates are fed as inputs to the robot’s control system.

Simplest Definition

Definition:

The Position Vector ($\mathbf{r}$) of a particle is defined as the vector drawn from the origin ($O$) of a coordinate system to the point ($P$) where the particle is located. It describes both the distance of the particle from the origin and the direction in which it lies.

Diagram

Figure: Detailed description of diagram for Position vector in Cartesian Coordinate: A 3D coordinate system (X, Y, Z axes) originating at O. A point P(x, y, z) is shown in the first octant. A vector $\mathbf{r}$ is drawn from O to P. Unit vectors $\hat{i},\hat{j},\hat{k}$ are shown along the positive X, Y, and Z axes respectively. Perpendiculars are shown dropped from P to the axes to illustrate the components x, y, and z.

Key Formula Box

We use the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ to denote the directions along the $X,Y,$ and $Z$ axes, respectively.

Detailed Derivation (Step-by-Step for Exams)

This derivation is very important. It is often asked in 5-mark questions. You must show all the steps properly.

Step-by-Step Derivation

Let us consider a particle situated at point $P$ in a three-dimensional Cartesian coordinate system $OXYZ$.

Let the coordinates of the point $P$ be $(x,y,z)$. The origin is at $O(0,0,0)$.

The vector $\mathbf{OP}$ is the position vector, which we denote as $\mathbf{r}$.

Step 1: Defining Components

The coordinates $(x,y,z)$ tell us that the distance covered along the $X$-axis is $x$, along the $Y$-axis is $y$, and along the $Z$-axis is $z$.

We define the following component vectors:

1. The component vector along the $X$-axis is ${\mathbf{r}}_x$. Since $\hat{i}$ is the unit vector in the $X$ direction, we write: $${\mathbf{r}}_x=x\hat{i}\text{...(1)}$$

2. The component vector along the $Y$-axis is ${\mathbf{r}}_y$. Using the unit vector $\hat{j}$: $${\mathbf{r}}_y=y\hat{j}\text{...(2)}$$

3. The component vector along the $Z$-axis is ${\mathbf{r}}_z$. Using the unit vector $\hat{k}$: $${\mathbf{r}}_z=z\hat{k}\text{...(3)}$$

Step 2: Applying Vector Addition (Polygon Law)

The total position vector $\mathbf{r}$ is the resultant vector of its three orthogonal components (${\mathbf{r}}_x,{\mathbf{r}}_y,{\mathbf{r}}_z$).

According to the Polygon Law of Vector Addition, the resultant vector is the vector sum of its components: $$\mathbf{r}={\mathbf{r}}_x+{\mathbf{r}}_y+{\mathbf{r}}_z$$

Step 3: Final Form of Position Vector

Now, let us substitute the component definitions from equations (1), (2), and (3) into the vector sum equation: $$\mathbf{r}=(x\hat{i})+(y\hat{j})+(z\hat{k})$$

Thus, the Position Vector $\mathbf{r}$ of a particle at $P(x,y,z)$ is: $$\mathbf{r}=x\hat{i}+y\hat{j}+z\hat{k}$$ This is the required expression for the position vector.

Step 4: Derivation of Magnitude of the Position Vector ($|\mathbf{r}|$)

The magnitude of the position vector, denoted by $r$ or $|\mathbf{r}|$, is simply the straight-line distance from the origin $O$ to the point $P$.

We can find this magnitude by applying the Pythagorean Theorem twice in the 3D space.

1. First, consider the components in the $XY$ plane. Let $A$ be the projection of $P$ onto the $XY$ plane. The distance $OA$ is found using Pythagoras theorem in 2D: $$O{A}^2=x^2+y^2$$

2. Now, consider the right-angled triangle formed by $O$, $A$, and $P$. Here, $OA$ is the base, $AP$ is the height (which is $z$), and $OP$ (which is $r$) is the hypotenuse.

   $$O{P}^2=O{A}^2+A{P}^2$$

3. We substitute the value of $O{A}^2$ and note that $AP=z$: $$r^2=(x^2+y^2)+z^2$$

4. Taking the square root on both sides, we get the magnitude: $$|\mathbf{r}|=\sqrt{x^2+y^2+z^2}$$ You must remember this formula for the numerical problems in the exams.

Important Table

Here is a summary of the fundamental components we used in the derivation. This table will help you clarify the role of unit vectors.

Solved Numericals (Exam Style)

Let us solve a few problems based on this concept. You must write the steps clearly, just like shown below.

Solved Numericals (Exam Style)

Q1: A particle is located at point $P(5,-2,1)$. Write the position vector ($\mathbf{r}$) of the particle.

Ans:

Given: Coordinates of the particle $P(x,y,z)=(5,-2,1)$. This means $x=5$, $y=-2$, and $z=1$.

To Find: Position vector $\mathbf{r}$.

Formula: $$\mathbf{r}=x\hat{i}+y\hat{j}+z\hat{k}$$

Calculation: We substitute the given values directly into the formula. $$\mathbf{r}=(5)\hat{i}+(-2)\hat{j}+(1)\hat{k}$$ $$\mathbf{r}=5\hat{i}-2\hat{j}+\hat{k}$$

Final Answer with units: The position vector is $\mathbf{r}=5\hat{i}-2\hat{j}+\hat{k}$. (Units will depend on whether $x,y,z$ are in meters, centimeters, etc.)

Q2: Find the magnitude of the position vector $\mathbf{r}=3\hat{i}+4\hat{j}-12\hat{k}$.

Ans:

Given: Position vector $\mathbf{r}=3\hat{i}+4\hat{j}-12\hat{k}$. Here, $x=3$, $y=4$, and $z=-12$.

To Find: Magnitude of the position vector, $|\mathbf{r}|$.

Formula: $$|\mathbf{r}|=\sqrt{x^2+y^2+z^2}$$

Calculation: We substitute the values: $$|\mathbf{r}|=\sqrt{(3{)}^2+(4{)}^2+(-12{)}^2}$$ $$|\mathbf{r}|=\sqrt{9+16+144}$$ $$|\mathbf{r}|=\sqrt{25+144}$$ $$|\mathbf{r}|=\sqrt{169}$$ $$|\mathbf{r}|=13$$

Final Answer with units: The magnitude of the position vector is $|\mathbf{r}|=13$ units.

Q3: A particle moves from point $P_1(1,2,3)$ to point $P_2(5,5,5)$. Calculate the displacement vector ($\mathrm{\Delta }\mathbf{r}$).

Ans:

Given: Initial position vector $P_1(1,2,3)$, so ${\mathbf{r}}_1=1\hat{i}+2\hat{j}+3\hat{k}$. Final position vector $P_2(5,5,5)$, so ${\mathbf{r}}_2=5\hat{i}+5\hat{j}+5\hat{k}$.

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

Formula: Displacement vector is the change in position: $$\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$$

Calculation: $$\mathrm{\Delta }\mathbf{r}=(5\hat{i}+5\hat{j}+5\hat{k})-(1\hat{i}+2\hat{j}+3\hat{k})$$ We group the components: $$\mathrm{\Delta }\mathbf{r}=(5-1)\hat{i}+(5-2)\hat{j}+(5-3)\hat{k}$$ $$\mathrm{\Delta }\mathbf{r}=4\hat{i}+3\hat{j}+2\hat{k}$$

Final Answer with units: The displacement vector is $\mathrm{\Delta }\mathbf{r}=4\hat{i}+3\hat{j}+2\hat{k}$ units.

Exam Corner: Questions & Answers

Short Answer Questions (2 Marks Each)

Q: What is the physical significance of the origin ($O$) when defining a position vector?

Ans: The origin ($O$) serves as the fixed reference point or the coordinate axis zero point. The position vector always starts from this origin. Its physical significance is that all measurements of distance and direction (position) are defined relative to this single, constant reference point. This allows different observers, using the same origin, to consistently measure the location of a particle.

Q: Write the difference between a position vector and a displacement vector.

Ans: The Position Vector ($\mathbf{r}$) defines the location of a single point $P$ relative to the fixed origin $O$. $\mathbf{r}$ shows where the object is. The Displacement Vector ($\mathrm{\Delta }\mathbf{r}$) defines the change in position. It is the vector drawn from the initial position $P_1$ to the final position $P_2$. $\mathrm{\Delta }\mathbf{r}$ shows how much the object moved and in which direction. Mathematically, $\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$.

Q: If the position vector of a particle is $\mathbf{r}=8\hat{i}+6\hat{j}$. What are the coordinates of the particle and what is its distance from the origin?

Ans:

1. Coordinates: Comparing the vector with the general form $\mathbf{r}=x\hat{i}+y\hat{j}$, the coordinates are $x=8$ and $y=6$. The particle coordinates are $(8,6,0)$.
2. Distance from Origin (Magnitude): $$|\mathbf{r}|=\sqrt{x^2+y^2}$$ $$|\mathbf{r}|=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{ units.}$$

Long Answer Questions (5 Marks Each)

Q: Explain the position vector in Cartesian coordinates. Derive the mathematical expression for the position vector and its magnitude in three dimensions.

Ans:

Position Vector in Cartesian Coordinates

The position vector $\mathbf{r}$ is a vector quantity that specifies the exact location of a particle $P$ relative to the origin $O$ of a chosen coordinate system. In the Cartesian system, we use three mutually perpendicular axes ($X,Y,Z$) and their corresponding unit vectors ($\hat{i},\hat{j},\hat{k}$) to represent space.

Derivation of Position Vector ($\mathbf{r}$)

1. Setup: Let $P$ be a point with coordinates $(x,y,z)$. The position vector is the vector $\mathbf{r}$ running from $O(0,0,0)$ to $P(x,y,z)$.
2. Components: The vector $\mathbf{r}$ can be resolved into three mutually perpendicular component vectors along the axes: ${\mathbf{r}}_x,{\mathbf{r}}_y,$ and ${\mathbf{r}}_z$.
3. Expression for Components: Since unit vectors give direction, the component vectors are written as: $${\mathbf{r}}_x=x\hat{i}$$ $${\mathbf{r}}_y=y\hat{j}$$ $${\mathbf{r}}_z=z\hat{k}$$
4. Vector Addition: Using the law of vector addition (specifically, the Polygon Law), the position vector is the sum of its components: $$\mathbf{r}={\mathbf{r}}_x+{\mathbf{r}}_y+{\mathbf{r}}_z$$
5. Final Position Vector: Substituting the components: $$\mathbf{r}=x\hat{i}+y\hat{j}+z\hat{k}$$

Derivation of Magnitude ($|\mathbf{r}|$)

The magnitude $|\mathbf{r}|$ is the distance $OP$. We apply the Pythagorean theorem.

1. XY Plane Distance ($r_{xy}$): The distance of the projection of $P$ onto the $XY$ plane from the origin ($r_{xy}$) is: $$r_{xy}^2=x^2+y^2\text{...(1)}$$
2. 3D Distance: Now, considering the right triangle formed by $r_{xy}$, the $z$-component, and the final vector $\mathbf{r}$, we have: $$|\mathbf{r}{|}^2=r_{xy}^2+z^2$$
3. Substitution: Substituting Eq. (1) into the expression: $$|\mathbf{r}{|}^2=x^2+y^2+z^2$$
4. Final Magnitude: Taking the square root gives the magnitude: $$|\mathbf{r}|=\sqrt{x^2+y^2+z^2}$$

Multiple Choice Questions (MCQs)

Q1: The position vector of a particle is given by $\mathbf{r}=2\hat{i}-3\hat{j}+6\hat{k}$. What is its magnitude? A) 7 B) 11 C) 36 D) 49 Ans: A) - $|\mathbf{r}|=\sqrt{2^2+(-3{)}^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$.

Q2: The direction of the unit vector $\hat{k}$ is along which axis? A) Negative Y-axis B) Positive Z-axis C) Positive X-axis D) Negative X-axis Ans: B) - $\hat{k}$ is conventionally the unit vector along the Positive Z-axis.

Q3: If a particle is located exactly on the Y-axis at $y=4$, its position vector in Cartesian coordinates is: A) $4\hat{i}$ B) $4\hat{j}$ C) $4\hat{k}$ D) $4\hat{i}+4\hat{j}$ Ans: B) - Since $x=0$ and $z=0$, the vector is $\mathbf{r}=0\hat{i}+4\hat{j}+0\hat{k}=4\hat{j}$.

Q4: A displacement from point $A(1,1,1)$ to point $B(2,3,4)$ is given by: A) $\hat{i}+2\hat{j}+3\hat{k}$ B) $3\hat{i}+4\hat{j}+5\hat{k}$ C) $-\hat{i}-2\hat{j}-3\hat{k}$ D) $2\hat{i}+3\hat{j}+4\hat{k}$ Ans: A) - Displacement $\mathrm{\Delta }\mathbf{r}={\mathbf{r}}_B-{\mathbf{r}}_A=(2-1)\hat{i}+(3-1)\hat{j}+(4-1)\hat{k}$.

Q5: Which of the following defines the unit vector in the direction of position vector $\mathbf{r}$? A) $|\mathbf{r}|/\mathbf{r}$ B) $\mathbf{r}\cdot |\mathbf{r}|$ C) $\mathbf{r}\times |\mathbf{r}|$ D) $\mathbf{r}/|\mathbf{r}|$ Ans: D) - The unit vector ($\hat{r}$) is always defined as the vector divided by its own magnitude.

Summary

I hope this concept is clear now. See, the position vector is just a tool to specify location in space using a fixed reference point (the origin). We found that in Cartesian coordinates, the vector is simply $\mathbf{r}=x\hat{i}+y\hat{j}+z\hat{k}$, and its length (magnitude) is calculated by the 3D Pythagoras theorem: $|\mathbf{r}|=\sqrt{x^2+y^2+z^2}$.

This understanding is essential for the next topics, especially velocity and acceleration vectors. Practice the derivation twice and work out the solved numericals on your own. All the best for your exams!