Introduction to Three Dimensional Geometry For Class 11 Notes are available for students at BYJU’S. The notes for Class 11 Maths Chapter 12 Introduction to three-dimensional geometry are prepared as per the latest exam pattern. These notes will help students to have a quick look at the Chapter 12 Introduction to Three Dimensional Geometry for exams. These notes are provided with reference to NCERT textbooks and the CBSE syllabus (2022-2023).
Class 11 Introduction to Three Dimensional Geometry
In Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Notes, we have given all the important concepts related to 3D geometry explained in the chapter. The topics here include:
- Coordinate axes
- Coordinate planes
- Coordinate of point in space
- Distance between two points
- Section formula
Also check: Equation Of A Line In Three Dimensions
Coordinate Axes
In three-dimensional geometry, the x-axis, y-axis and z-axis are the three coordinate axes of a rectangular Cartesian coordinate system. These lines are three mutually perpendicular lines. The values of the coordinate axes determine the location of the point in the coordinate plane.
Coordinate Planes
The three planes (XY, YZ and ZX) determined by the pair of axes are the coordinate planes. All three planes divide the space into eight parts, called octants.
Coordinate of a Point in Space
In three-dimensional geometry, the coordinates of a point P is always written in the form of P(x, y, z), where x, y and z are the distances of the point, from the YZ, ZX and XY-planes.
- The coordinates of any point at the origin is (0,0,0)
- The coordinates of any point on the x-axis is in the form of (x,0,0)
- The coordinates of any point on the y-axis is in the form of (0,y,0)
- The coordinates of any point on the z-axis is in the form of (0,0,z)
- The coordinates of any point on the XY-plane is in the form (x, y, 0)
- The coordinates of any point on the YZ-plane is in the form (0, y, z)
- The coordinates of any point on the ZX-plane is in the form (x, 0, z)
Sign of Coordinates in Different Octants:
The sign (+ or -) of the coordinates of a point determines the octant in which the point lies.
| Octants→ | I | II | III | IV | V | VI | VII | VIII |
|---|---|---|---|---|---|---|---|---|
| x | + | – | – | + | + | – | – | + |
| y | + | + | – | – | + | + | – | – |
| z | + | + | + | + | – | – | – | – |
Distance Between Two Points
If P (x1, y1, z1) and Q (x2, y2, z2) are the two points, then the distance between P and Q is given by:
\(\begin{array}{l}P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\end{array} \) |
|---|
The distance of the point from the origin will be:
\(\begin{array}{l}OP=\sqrt{x^{2}+y^{2}+z^{2}}\end{array} \) |
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Section Formula
The coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n are represented by:
\(\begin{array}{l}\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)\end{array} \) |
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The coordinates of the point R which divides the line segment joining two points P (x1, y1, z1) and Q (x2, y2 ,z2) externally in the ratio m : n are represented by:
\(\begin{array}{l}\left(\frac{m x_{2}-n x_{1}}{m-n}, \frac{m y_{2}-n y_{1}}{m-n}, \frac{m z_{2}-n z_{1}}{m-n}\right)\end{array} \) |
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The coordinates of the mid-point of the line segment joining two points P (x1, y1, z1) and Q (x2, y2 ,z2) are given by:
\(\begin{array}{l}\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\end{array} \) |
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The coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1) (x2, y2, z2) and (x3, y3, z3), are given by:
List of Class 11 Maths Notes
Solved Examples
Q.1 If PA2 + PB2 = 2k2 and A and B are the points (3, 4, 5) and (–1, 3, –7), respectively. Find the equation of the set of points P.
Solution: Suppose, the point P(x,yz,)
So by distance formula,
PA2 = (x – 3)2 + (y – 4)2 + ( z – 5)2
PB2 = (x + 1)2 + (y – 3)2 + (z + 7)2
As per the given condition,
PA2 + PB2 = 2k2
By putting the value we get;
(x – 3)2 + (y – 4)2 + (z – 5)2 + (x + 1)2 + (y – 3)2 + (z + 7)2 = 2k2
2x2 + 2y2 + 2z2 – 4x – 14y + 4z = 2k2 – 109
This is the required equation.
Q.2: Find the distance between the points (2, –1, 3) and (–2, 1, 3).
Solution: Suppose the two points are P(2, –1, 3) and Q(–2, 1, 3)
By using the distance formula,
PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
Hence, the distance between P and Q is 2√5 units.
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