3D Coordinate Basics

3D Coordinate Basics

Hello JEE aspirants! Welcome to the world of 3D Coordinate Geometry. This topic is crucial for JEE Main, forming the foundation for Vector Algebra and 3D Geometry problems. Mastering these basics will not only help you score well but also visualize complex problems with ease. So, let's dive in!

1. Three Coordinate Axes: x, y, z

In 2D, we have the x-axis and y-axis. In 3D, we introduce the z-axis, which is perpendicular to both the x and y axes. Imagine the corner of a room where the two walls meet the floor; these represent our three axes. We denote unit vectors along these axes as $\hat{i}$, $\hat{j}$, and $\hat{k}$ respectively.

2. Coordinate Planes: xy-plane, yz-plane, xz-plane

These are the planes formed by any two coordinate axes. The xy-plane is where $z = 0$, the yz-plane is where $x = 0$, and the xz-plane is where $y = 0$. Think of them as the "walls" and "floor" in our room analogy.

• xy-plane: All points in this plane have a z-coordinate of 0, i.e., (x, y, 0).
• yz-plane: All points in this plane have an x-coordinate of 0, i.e., (0, y, z).
• xz-plane: All points in this plane have a y-coordinate of 0, i.e., (x, 0, z).

For example, the point (2, 3, 0) lies on the xy-plane, (0, -1, 4) lies on the yz-plane, and (5, 0, -2) lies on the xz-plane.

3. Octants in 3D Space

Just as the 2D coordinate system is divided into four quadrants, the 3D coordinate system is divided into eight octants. Each octant is defined by the signs of the x, y, and z coordinates.

For example, the octant ( +, +, + ) contains points where all x, y, and z coordinates are positive. The octant ( -, +, - ) contains points where x is negative, y is positive, and z is negative.

4. Distance Formula in 3D

This formula extends the 2D distance formula. Given two points $P(x₁, y₁, z₁)$ and $Q(x₂, y₂, z₂)$, the distance $d$ between them is:

Derivation: Consider a cuboid with P and Q as opposite vertices. The sides of the cuboid are $|x₂-x₁|$, $|y₂-y₁|$, and $|z₂-z₁|$. Applying the Pythagorean theorem twice gives us the distance formula.

Example: Find the distance between (1, 2, 3) and (4, 5, 6). $d = \sqrt{(4-1)² + (5-2)² + (6-3)²} = \sqrt{3² + 3² + 3²} = \sqrt{27} = 3\sqrt{3}$

5. Section Formula in 3D (Internal and External)

The section formula helps find the coordinates of a point that divides the line segment joining two given points in a specific ratio.

Internal Division

If point $R$ divides the line segment joining $P(x₁, y₁, z₁)$ and $Q(x₂, y₂, z₂)$ internally in the ratio $m:n$, then the coordinates of $R$ are:

Derivation: This formula can be derived using similar triangles. Consider projecting the points onto each coordinate plane and applying the 2D section formula.

External Division

If point $R$ divides the line segment joining $P(x₁, y₁, z₁)$ and $Q(x₂, y₂, z₂)$ externally in the ratio $m:n$, then the coordinates of $R$ are:

Derivation: Similar to internal division, this formula can be derived using similar triangles, considering that $R$ lies outside the segment $PQ$.

Example: Find the coordinates of the point that divides the line segment joining A(1, 2, 3) and B(4, 5, 6) internally in the ratio 2:1. Using the internal division formula: $R = \left(\frac{2*4+1*1}{2+1}, \frac{2*5+1*2}{2+1}, \frac{2*6+1*3}{2+1}\right) = \left(\frac{9}{3}, \frac{12}{3}, \frac{15}{3}\right) = (3, 4, 5)$

Midpoint

A special case of the section formula is the midpoint, where $m = n$. The coordinates of the midpoint $M$ of the line segment joining $P(x₁, y₁, z₁)$ and $Q(x₂, y₂, z₂)$ are:

6. Centroid of a Triangle

The centroid of a triangle is the point of intersection of its medians. If the vertices of the triangle are $A(x₁, y₁, z₁)$, $B(x₂, y₂, z₂)$, and $C(x₃, y₃, z₃)$, then the coordinates of the centroid $G$ are:

Derivation: The centroid divides each median in the ratio 2:1. Using the section formula on one of the medians, we can derive the centroid formula.

Example: Find the centroid of a triangle with vertices A(1, 2, 3), B(4, 5, 6), and C(7, 8, 9). $G = \left(\frac{1+4+7}{3}, \frac{2+5+8}{3}, \frac{3+6+9}{3}\right) = \left(\frac{12}{3}, \frac{15}{3}, \frac{18}{3}\right) = (4, 5, 6)$

JEE Specific Trick: Often, JEE problems combine these concepts with vector algebra. Familiarize yourself with converting points to position vectors and vice versa.

That’s it for the basics of the 3D coordinate system! Keep practicing problems to solidify your understanding. Good luck with your JEE preparation!