Detailed Notes: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


Three Dimensional Geometry 
 
A PLANE 
If line joining any two points on a surface lies completely on it then the surface is a plane. 
OR 
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this 
surface is called a plane. This fixed line is called the normal to the plane. 
Equation of a plane : 
(i) Vector form : The equation (??? - ???
0
) · ??? ? = 0 represents a plane containing the point with position 
vector is a vector normal to the plane. 
The above equation can also be written as ??? · ??? ? = ?? , where ?? = ???
0
· ??? ? 
(ii) Cartesian form : The equation of a plane passing through the point (?? 1
, ?? 1
, ?? 1
) is given by 
?? (?? - ?? 1
) + ?? (?? - ?? 1
) + ?? (?? - ?? 1
) = 0 where ?? , ?? , ?? are the direction ratios of the normal to the plane. 
(iii) Normal form : Vector equation of a plane normal to unit vector and at a distance ?? from the 
origin is ??? · ??? ? = ?? . Normal form of the equation of a plane is ???? + ???? + ???? = ?? , where, ?? , ?? , ?? are the 
direction cosines of the normal to the plane and ?? is the distance of the plane from the origin. 
(iv) General form : ???? + ???? + ???? + ?? = 0 is the equation of a plane, where ?? , ?? , ?? are the direction 
ratios of the normal to the plane. 
(v) Plane through three points : The equation of the plane through three non-collinear points 
(?? 1
, ?? 1
, ?? 1
), (?? 2
, ?? 2
, ?? 2
), (?? 3
, ?? 3
, ?? 3
) ???? |?? - ?? 3
 ?? - ?? 3
 ?? - ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 | = 0 
(vi) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is 
?? ?? +
?? ?? +
?? ?? = 1 
Note : 
Equation of ???? -plane, ???? -plane and ???? -plane is ?? = 0, ?? = 0 and ?? = 0 
Transformation of the equation of a plane to the normal form: To reduce any equation ???? + ???? +
???? - ?? = 0 to the normal form, first write the constant term on the right hand side and make it 
positive, then divide each term by v?? 2
+ ?? 2
+ ?? 2
, where ?? , ?? , ?? are coefficients of ?? , ?? and ?? 
respectively e.g. 
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
=
?? ±v?? 2
+ ?? 2
+ ?? 2
 
Where (+) sign is to be taken if ?? > 0 and (-) sign is to be taken if ?? < 0. 
Page 2


Three Dimensional Geometry 
 
A PLANE 
If line joining any two points on a surface lies completely on it then the surface is a plane. 
OR 
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this 
surface is called a plane. This fixed line is called the normal to the plane. 
Equation of a plane : 
(i) Vector form : The equation (??? - ???
0
) · ??? ? = 0 represents a plane containing the point with position 
vector is a vector normal to the plane. 
The above equation can also be written as ??? · ??? ? = ?? , where ?? = ???
0
· ??? ? 
(ii) Cartesian form : The equation of a plane passing through the point (?? 1
, ?? 1
, ?? 1
) is given by 
?? (?? - ?? 1
) + ?? (?? - ?? 1
) + ?? (?? - ?? 1
) = 0 where ?? , ?? , ?? are the direction ratios of the normal to the plane. 
(iii) Normal form : Vector equation of a plane normal to unit vector and at a distance ?? from the 
origin is ??? · ??? ? = ?? . Normal form of the equation of a plane is ???? + ???? + ???? = ?? , where, ?? , ?? , ?? are the 
direction cosines of the normal to the plane and ?? is the distance of the plane from the origin. 
(iv) General form : ???? + ???? + ???? + ?? = 0 is the equation of a plane, where ?? , ?? , ?? are the direction 
ratios of the normal to the plane. 
(v) Plane through three points : The equation of the plane through three non-collinear points 
(?? 1
, ?? 1
, ?? 1
), (?? 2
, ?? 2
, ?? 2
), (?? 3
, ?? 3
, ?? 3
) ???? |?? - ?? 3
 ?? - ?? 3
 ?? - ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 | = 0 
(vi) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is 
?? ?? +
?? ?? +
?? ?? = 1 
Note : 
Equation of ???? -plane, ???? -plane and ???? -plane is ?? = 0, ?? = 0 and ?? = 0 
Transformation of the equation of a plane to the normal form: To reduce any equation ???? + ???? +
???? - ?? = 0 to the normal form, first write the constant term on the right hand side and make it 
positive, then divide each term by v?? 2
+ ?? 2
+ ?? 2
, where ?? , ?? , ?? are coefficients of ?? , ?? and ?? 
respectively e.g. 
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
=
?? ±v?? 2
+ ?? 2
+ ?? 2
 
Where (+) sign is to be taken if ?? > 0 and (-) sign is to be taken if ?? < 0. 
A plane ???? + ???? + ???? + ?? = 0 divides the line segment joining (?? 1
, ?? 1
, ?? 1
) and (?? 2
, ?? 2
, ?? 2
). in the ratio 
(-
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? ?? ?? 2
+?? ?? 2
+?? ?? 2
+?? ) 
Coplanarity of four points 
The points ?? (?? 1
?? 1
?? 1
), ?? (?? 2
?? 2
?? 2
)?? (?? 3
?? 3
?? 3
) and ?? (?? 4
?? 4
?? 4
) are coplanar then 
|?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 | = 0 
Problem 41 : Find the equation of the plane upon which the length of normal from origin is 10 and 
direction ratios of this normal are 3,2,6. 
Solution: If ?? be the length of perpendicular from origin to the plane and ?? , ?? , ?? be the direction 
cosines of this normal, then its equation is 
???? + ???? + ???? = 10 
Direction ratios of normal to the plane are 3, 2, 6 
?  Direction cosines of normal to the required plane are ?? =
3
7
, ?? =
2
7
, ?? =
6
7
 
Equation of required plane is 
3
7
?? +
2
7
?? +
6
7
?? = 10 or,  3?? + 2?? + 6?? = 70 
Problem 42 :Find the plane through the points (2, -3,3), (-5,2,0), (1, -7,1) 
Solution :  |?? - 2 ?? + 3 ?? - 3 - 5 - 2 2 + 3 0 - 3 1 - 2 - 7 + 3 1 - 3 | = 0  or  |?? - 2 ?? + 3 ?? - 3 -
7 5 - 3 - 1 - 4 - 2 | = 0 ? 2?? + ?? - 3?? + 8 = 0 
Problem 43 : If ?? be any point on the plane ???? + ???? + ???? = ?? and ?? be a point on the line ???? such 
that 
???? . ???? = ?? 2
, show that the locus of the point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Solution :   Let ?? = (?? , ?? , ?? ), ?? = (?? 1
, ?? 1
, ?? 1
) 
Direction ratios of OP are ?? , ?? , ?? and direction ratios of ???? are ?? 1
, ?? 1
, ?? 1
. 
Since ?? , ?? , ?? are collinear, we have 
?? ?? 1
=
?? ?? 1
=
?? ?? 1
= ?? (say) 
As ?? (?? , ?? , ?? ) lies on the plane ???? + ???? + ???? = ?? , 
???? + ???? + ???? = ??  or  ?? (?? ?? 1
+ ????
1
+ ???? ?? 1
) = ?? 
Given ???? . ???? = ?? 2
 ? v?? 2
+ ?? 2
+ ?? 2
v?? 1
2
+ ?? 1
2
+ ?? 1
2
 = ?? 2
 
or, v?? 2
(?? 1
2
+ ?? 1
2
+ ?? 1
2
)v?? 1
2
+ ?? 1
2
+ ?? 1
2
= ?? 2
  or,  ?? (?? 1
2
+ ?? 1
2
+ ?? 1
2
) = ?? 2
 
 
Page 3


Three Dimensional Geometry 
 
A PLANE 
If line joining any two points on a surface lies completely on it then the surface is a plane. 
OR 
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this 
surface is called a plane. This fixed line is called the normal to the plane. 
Equation of a plane : 
(i) Vector form : The equation (??? - ???
0
) · ??? ? = 0 represents a plane containing the point with position 
vector is a vector normal to the plane. 
The above equation can also be written as ??? · ??? ? = ?? , where ?? = ???
0
· ??? ? 
(ii) Cartesian form : The equation of a plane passing through the point (?? 1
, ?? 1
, ?? 1
) is given by 
?? (?? - ?? 1
) + ?? (?? - ?? 1
) + ?? (?? - ?? 1
) = 0 where ?? , ?? , ?? are the direction ratios of the normal to the plane. 
(iii) Normal form : Vector equation of a plane normal to unit vector and at a distance ?? from the 
origin is ??? · ??? ? = ?? . Normal form of the equation of a plane is ???? + ???? + ???? = ?? , where, ?? , ?? , ?? are the 
direction cosines of the normal to the plane and ?? is the distance of the plane from the origin. 
(iv) General form : ???? + ???? + ???? + ?? = 0 is the equation of a plane, where ?? , ?? , ?? are the direction 
ratios of the normal to the plane. 
(v) Plane through three points : The equation of the plane through three non-collinear points 
(?? 1
, ?? 1
, ?? 1
), (?? 2
, ?? 2
, ?? 2
), (?? 3
, ?? 3
, ?? 3
) ???? |?? - ?? 3
 ?? - ?? 3
 ?? - ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 | = 0 
(vi) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is 
?? ?? +
?? ?? +
?? ?? = 1 
Note : 
Equation of ???? -plane, ???? -plane and ???? -plane is ?? = 0, ?? = 0 and ?? = 0 
Transformation of the equation of a plane to the normal form: To reduce any equation ???? + ???? +
???? - ?? = 0 to the normal form, first write the constant term on the right hand side and make it 
positive, then divide each term by v?? 2
+ ?? 2
+ ?? 2
, where ?? , ?? , ?? are coefficients of ?? , ?? and ?? 
respectively e.g. 
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
=
?? ±v?? 2
+ ?? 2
+ ?? 2
 
Where (+) sign is to be taken if ?? > 0 and (-) sign is to be taken if ?? < 0. 
A plane ???? + ???? + ???? + ?? = 0 divides the line segment joining (?? 1
, ?? 1
, ?? 1
) and (?? 2
, ?? 2
, ?? 2
). in the ratio 
(-
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? ?? ?? 2
+?? ?? 2
+?? ?? 2
+?? ) 
Coplanarity of four points 
The points ?? (?? 1
?? 1
?? 1
), ?? (?? 2
?? 2
?? 2
)?? (?? 3
?? 3
?? 3
) and ?? (?? 4
?? 4
?? 4
) are coplanar then 
|?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 | = 0 
Problem 41 : Find the equation of the plane upon which the length of normal from origin is 10 and 
direction ratios of this normal are 3,2,6. 
Solution: If ?? be the length of perpendicular from origin to the plane and ?? , ?? , ?? be the direction 
cosines of this normal, then its equation is 
???? + ???? + ???? = 10 
Direction ratios of normal to the plane are 3, 2, 6 
?  Direction cosines of normal to the required plane are ?? =
3
7
, ?? =
2
7
, ?? =
6
7
 
Equation of required plane is 
3
7
?? +
2
7
?? +
6
7
?? = 10 or,  3?? + 2?? + 6?? = 70 
Problem 42 :Find the plane through the points (2, -3,3), (-5,2,0), (1, -7,1) 
Solution :  |?? - 2 ?? + 3 ?? - 3 - 5 - 2 2 + 3 0 - 3 1 - 2 - 7 + 3 1 - 3 | = 0  or  |?? - 2 ?? + 3 ?? - 3 -
7 5 - 3 - 1 - 4 - 2 | = 0 ? 2?? + ?? - 3?? + 8 = 0 
Problem 43 : If ?? be any point on the plane ???? + ???? + ???? = ?? and ?? be a point on the line ???? such 
that 
???? . ???? = ?? 2
, show that the locus of the point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Solution :   Let ?? = (?? , ?? , ?? ), ?? = (?? 1
, ?? 1
, ?? 1
) 
Direction ratios of OP are ?? , ?? , ?? and direction ratios of ???? are ?? 1
, ?? 1
, ?? 1
. 
Since ?? , ?? , ?? are collinear, we have 
?? ?? 1
=
?? ?? 1
=
?? ?? 1
= ?? (say) 
As ?? (?? , ?? , ?? ) lies on the plane ???? + ???? + ???? = ?? , 
???? + ???? + ???? = ??  or  ?? (?? ?? 1
+ ????
1
+ ???? ?? 1
) = ?? 
Given ???? . ???? = ?? 2
 ? v?? 2
+ ?? 2
+ ?? 2
v?? 1
2
+ ?? 1
2
+ ?? 1
2
 = ?? 2
 
or, v?? 2
(?? 1
2
+ ?? 1
2
+ ?? 1
2
)v?? 1
2
+ ?? 1
2
+ ?? 1
2
= ?? 2
  or,  ?? (?? 1
2
+ ?? 1
2
+ ?? 1
2
) = ?? 2
 
 
Hence the locus of point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Problem 44 : A moving plane passes through a fixed point (?? , ?? , ?? ) and cuts the coordinate axes A, B, 
C . Find the locus of the centroid of the tetrahedron OABC. 
Solution : Let the plane be 
?? ?? +
?? ?? +
?? ?? = 1,0(0,0,0), ?? (?? , 0,0), ?? (0, ?? , 0) 
C (0,0, ?? ). Centroid of ???????? is (
?? 4
,
?? 4
,
?? 4
) 
The plane passes through (?? , ?? , ?? ) ? 
?? ?? +
?? ?? +
?? ?? = 1 
Centroid, ?? =
?? 4
, ?? =
?? 4
, ?? =
?? 4
  or  ?? = 4?? , ?? = 4?? , ?? = 4?? 
Now (1) gives the locus of ?? as 
?? ?? +
?? ?? +
?? ?? = 4 
Position of point with respect to plane : 
A plane divides the three dimensional space in two equal parts. Two points ?? (?? 1
?? 1
?? 1
) and ?? (?? 2
?? 2
?? 2
) 
are on the same side of the plane ???? + ???? + ???? + ?? = 0 if ?? ?? 1
+ ?? ?? 1
+ ?? ?? 1
+ ?? and ????
2
+ ????
2
+ ????
2
+ ?? 
are both positive or both negative and are opposite side of plane if both of these values are in 
opposite sign 
Problem 45 : Show that the points (1,2,3) and (2, -1,4) lie on opposite sides of the plane ?? + 4?? + ?? -
3 = 0 
Solution : Since the numbers 1 + 4 × 2 + 3 - 3 = 9 and 2 - 4 + 4 - 3 = -1 are of opposite sign, then 
points are on opposite sides of the plane. 
A plane & a point 
 
Let ?? = ???? + ???? + ???? + ?? = 0 is a given plane and ?? (?? 1
, ?? 1
, ?? 1
) is given point as shown in figure. 
Let ?? (?? '
, ?? '
, ?? '
) be the foot of the point ?? (?? 1
, ?? 1
, ?? 1
) with respect to the plane ?? . 
And ?? (?? ''
, ?? ''
, ?? ''
) be the reflection of point ?? (?? 1
, ?? 1
, ?? 1
) with respect to the plane ?? . 
(i) Distance of the point (?? 1
, ?? 1
, ?? 1
) from the plane ???? + ???? + ???? + ?? = 0 is given by |
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? v?? 2
+?? 2
+?? 2
|. 
Page 4


Three Dimensional Geometry 
 
A PLANE 
If line joining any two points on a surface lies completely on it then the surface is a plane. 
OR 
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this 
surface is called a plane. This fixed line is called the normal to the plane. 
Equation of a plane : 
(i) Vector form : The equation (??? - ???
0
) · ??? ? = 0 represents a plane containing the point with position 
vector is a vector normal to the plane. 
The above equation can also be written as ??? · ??? ? = ?? , where ?? = ???
0
· ??? ? 
(ii) Cartesian form : The equation of a plane passing through the point (?? 1
, ?? 1
, ?? 1
) is given by 
?? (?? - ?? 1
) + ?? (?? - ?? 1
) + ?? (?? - ?? 1
) = 0 where ?? , ?? , ?? are the direction ratios of the normal to the plane. 
(iii) Normal form : Vector equation of a plane normal to unit vector and at a distance ?? from the 
origin is ??? · ??? ? = ?? . Normal form of the equation of a plane is ???? + ???? + ???? = ?? , where, ?? , ?? , ?? are the 
direction cosines of the normal to the plane and ?? is the distance of the plane from the origin. 
(iv) General form : ???? + ???? + ???? + ?? = 0 is the equation of a plane, where ?? , ?? , ?? are the direction 
ratios of the normal to the plane. 
(v) Plane through three points : The equation of the plane through three non-collinear points 
(?? 1
, ?? 1
, ?? 1
), (?? 2
, ?? 2
, ?? 2
), (?? 3
, ?? 3
, ?? 3
) ???? |?? - ?? 3
 ?? - ?? 3
 ?? - ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 | = 0 
(vi) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is 
?? ?? +
?? ?? +
?? ?? = 1 
Note : 
Equation of ???? -plane, ???? -plane and ???? -plane is ?? = 0, ?? = 0 and ?? = 0 
Transformation of the equation of a plane to the normal form: To reduce any equation ???? + ???? +
???? - ?? = 0 to the normal form, first write the constant term on the right hand side and make it 
positive, then divide each term by v?? 2
+ ?? 2
+ ?? 2
, where ?? , ?? , ?? are coefficients of ?? , ?? and ?? 
respectively e.g. 
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
=
?? ±v?? 2
+ ?? 2
+ ?? 2
 
Where (+) sign is to be taken if ?? > 0 and (-) sign is to be taken if ?? < 0. 
A plane ???? + ???? + ???? + ?? = 0 divides the line segment joining (?? 1
, ?? 1
, ?? 1
) and (?? 2
, ?? 2
, ?? 2
). in the ratio 
(-
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? ?? ?? 2
+?? ?? 2
+?? ?? 2
+?? ) 
Coplanarity of four points 
The points ?? (?? 1
?? 1
?? 1
), ?? (?? 2
?? 2
?? 2
)?? (?? 3
?? 3
?? 3
) and ?? (?? 4
?? 4
?? 4
) are coplanar then 
|?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 | = 0 
Problem 41 : Find the equation of the plane upon which the length of normal from origin is 10 and 
direction ratios of this normal are 3,2,6. 
Solution: If ?? be the length of perpendicular from origin to the plane and ?? , ?? , ?? be the direction 
cosines of this normal, then its equation is 
???? + ???? + ???? = 10 
Direction ratios of normal to the plane are 3, 2, 6 
?  Direction cosines of normal to the required plane are ?? =
3
7
, ?? =
2
7
, ?? =
6
7
 
Equation of required plane is 
3
7
?? +
2
7
?? +
6
7
?? = 10 or,  3?? + 2?? + 6?? = 70 
Problem 42 :Find the plane through the points (2, -3,3), (-5,2,0), (1, -7,1) 
Solution :  |?? - 2 ?? + 3 ?? - 3 - 5 - 2 2 + 3 0 - 3 1 - 2 - 7 + 3 1 - 3 | = 0  or  |?? - 2 ?? + 3 ?? - 3 -
7 5 - 3 - 1 - 4 - 2 | = 0 ? 2?? + ?? - 3?? + 8 = 0 
Problem 43 : If ?? be any point on the plane ???? + ???? + ???? = ?? and ?? be a point on the line ???? such 
that 
???? . ???? = ?? 2
, show that the locus of the point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Solution :   Let ?? = (?? , ?? , ?? ), ?? = (?? 1
, ?? 1
, ?? 1
) 
Direction ratios of OP are ?? , ?? , ?? and direction ratios of ???? are ?? 1
, ?? 1
, ?? 1
. 
Since ?? , ?? , ?? are collinear, we have 
?? ?? 1
=
?? ?? 1
=
?? ?? 1
= ?? (say) 
As ?? (?? , ?? , ?? ) lies on the plane ???? + ???? + ???? = ?? , 
???? + ???? + ???? = ??  or  ?? (?? ?? 1
+ ????
1
+ ???? ?? 1
) = ?? 
Given ???? . ???? = ?? 2
 ? v?? 2
+ ?? 2
+ ?? 2
v?? 1
2
+ ?? 1
2
+ ?? 1
2
 = ?? 2
 
or, v?? 2
(?? 1
2
+ ?? 1
2
+ ?? 1
2
)v?? 1
2
+ ?? 1
2
+ ?? 1
2
= ?? 2
  or,  ?? (?? 1
2
+ ?? 1
2
+ ?? 1
2
) = ?? 2
 
 
Hence the locus of point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Problem 44 : A moving plane passes through a fixed point (?? , ?? , ?? ) and cuts the coordinate axes A, B, 
C . Find the locus of the centroid of the tetrahedron OABC. 
Solution : Let the plane be 
?? ?? +
?? ?? +
?? ?? = 1,0(0,0,0), ?? (?? , 0,0), ?? (0, ?? , 0) 
C (0,0, ?? ). Centroid of ???????? is (
?? 4
,
?? 4
,
?? 4
) 
The plane passes through (?? , ?? , ?? ) ? 
?? ?? +
?? ?? +
?? ?? = 1 
Centroid, ?? =
?? 4
, ?? =
?? 4
, ?? =
?? 4
  or  ?? = 4?? , ?? = 4?? , ?? = 4?? 
Now (1) gives the locus of ?? as 
?? ?? +
?? ?? +
?? ?? = 4 
Position of point with respect to plane : 
A plane divides the three dimensional space in two equal parts. Two points ?? (?? 1
?? 1
?? 1
) and ?? (?? 2
?? 2
?? 2
) 
are on the same side of the plane ???? + ???? + ???? + ?? = 0 if ?? ?? 1
+ ?? ?? 1
+ ?? ?? 1
+ ?? and ????
2
+ ????
2
+ ????
2
+ ?? 
are both positive or both negative and are opposite side of plane if both of these values are in 
opposite sign 
Problem 45 : Show that the points (1,2,3) and (2, -1,4) lie on opposite sides of the plane ?? + 4?? + ?? -
3 = 0 
Solution : Since the numbers 1 + 4 × 2 + 3 - 3 = 9 and 2 - 4 + 4 - 3 = -1 are of opposite sign, then 
points are on opposite sides of the plane. 
A plane & a point 
 
Let ?? = ???? + ???? + ???? + ?? = 0 is a given plane and ?? (?? 1
, ?? 1
, ?? 1
) is given point as shown in figure. 
Let ?? (?? '
, ?? '
, ?? '
) be the foot of the point ?? (?? 1
, ?? 1
, ?? 1
) with respect to the plane ?? . 
And ?? (?? ''
, ?? ''
, ?? ''
) be the reflection of point ?? (?? 1
, ?? 1
, ?? 1
) with respect to the plane ?? . 
(i) Distance of the point (?? 1
, ?? 1
, ?? 1
) from the plane ???? + ???? + ???? + ?? = 0 is given by |
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? v?? 2
+?? 2
+?? 2
|. 
(ii) The length of the perpendicular from a point having position vector ?? ? to plane ??? · ??? ? = ?? is ?? =
|?? ? ?·?? ? ?-?? |
|?? ? ?|
. 
(iii) The coordinates of the foot (F) of perpendicular from the point (?? 1
, ?? 1
, ?? 1
) to the plane 
???? + ???? + ???? + ?? = 0 are 
?? -?? 1
?? =
?? -?? 1
?? =
?? -?? 1
?? = -
(?? ?? 1
+?? ?? 1
+?? ?? 1
+?? )
?? 2
+?? 2
+?? 2
 
(iv) The coordinates of the Image (R) of point (?? 1
, ?? 1
, ?? 1
) to the plane 
???? + ???? + ???? + ?? = 0 are 
?? -?? 1
?? =
?? -?? 1
?? =
?? -?? 1
?? = -
2(?? ?? 1
+?? ?? 1
+?? ?? 1
+?? )
?? 2
+?? 2
+?? 2
 
Problem 46 : Find the image of the point ?? (3,5,7) in the plane 2?? + ?? + ?? = 0. 
Solution:   Given plane is 2?? + ?? + ?? = 0 
Direction ratios of normal to plane (1) are 2,1,1 
Let ?? be the image of point ?? in plane (1). Let ???? meet plane (1) in ?? then ???? ? plane (1) 
Let  ?? = (2?? + 3, ?? + 5, ?? + 7) 
Since ?? lies on plane (1) 
? 2(2?? + 3) + ?? + 5 + ?? + 7 = 0  or,  6?? + 18 = 0 ? ?? = -3 
? ?? = (-3,2,4) 
Let  ?? = (?? , ?? , ?? ) 
Since ?? is the middle point of ???? 
? -3 =
?? + 3
2
? ?? = -9 ?????? 2 =
?? + 5
2
? ?? = -1 ?????? 4 =
?? + 7
2
 ? ?? = 1 ? ?? = (-9, -1,1).  
Problem 47 : A plane passes through a fixed point (?? , ?? , ?? ). Show that the locus of the foot of 
perpendicular to it from the origin is the sphere ?? 2
+ ?? 2
+ ?? 2
- ???? - ???? - ???? = 0 
Solution : Let the equation of the variable plane be ???? + ???? + ???? + ?? = 0 
Plane passes through the fixed point (?? , ?? , ?? ) ? ???? + ???? + ???? + ?? = 0 
Let ?? (?? , ?? , ?? ) be the foot of perpendicular from origin to plane (1). 
Direction ratios of OP are 
Page 5


Three Dimensional Geometry 
 
A PLANE 
If line joining any two points on a surface lies completely on it then the surface is a plane. 
OR 
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this 
surface is called a plane. This fixed line is called the normal to the plane. 
Equation of a plane : 
(i) Vector form : The equation (??? - ???
0
) · ??? ? = 0 represents a plane containing the point with position 
vector is a vector normal to the plane. 
The above equation can also be written as ??? · ??? ? = ?? , where ?? = ???
0
· ??? ? 
(ii) Cartesian form : The equation of a plane passing through the point (?? 1
, ?? 1
, ?? 1
) is given by 
?? (?? - ?? 1
) + ?? (?? - ?? 1
) + ?? (?? - ?? 1
) = 0 where ?? , ?? , ?? are the direction ratios of the normal to the plane. 
(iii) Normal form : Vector equation of a plane normal to unit vector and at a distance ?? from the 
origin is ??? · ??? ? = ?? . Normal form of the equation of a plane is ???? + ???? + ???? = ?? , where, ?? , ?? , ?? are the 
direction cosines of the normal to the plane and ?? is the distance of the plane from the origin. 
(iv) General form : ???? + ???? + ???? + ?? = 0 is the equation of a plane, where ?? , ?? , ?? are the direction 
ratios of the normal to the plane. 
(v) Plane through three points : The equation of the plane through three non-collinear points 
(?? 1
, ?? 1
, ?? 1
), (?? 2
, ?? 2
, ?? 2
), (?? 3
, ?? 3
, ?? 3
) ???? |?? - ?? 3
 ?? - ?? 3
 ?? - ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 1
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 ?? 2
- ?? 3
 | = 0 
(vi) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is 
?? ?? +
?? ?? +
?? ?? = 1 
Note : 
Equation of ???? -plane, ???? -plane and ???? -plane is ?? = 0, ?? = 0 and ?? = 0 
Transformation of the equation of a plane to the normal form: To reduce any equation ???? + ???? +
???? - ?? = 0 to the normal form, first write the constant term on the right hand side and make it 
positive, then divide each term by v?? 2
+ ?? 2
+ ?? 2
, where ?? , ?? , ?? are coefficients of ?? , ?? and ?? 
respectively e.g. 
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
+
????
±v?? 2
+ ?? 2
+ ?? 2
=
?? ±v?? 2
+ ?? 2
+ ?? 2
 
Where (+) sign is to be taken if ?? > 0 and (-) sign is to be taken if ?? < 0. 
A plane ???? + ???? + ???? + ?? = 0 divides the line segment joining (?? 1
, ?? 1
, ?? 1
) and (?? 2
, ?? 2
, ?? 2
). in the ratio 
(-
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? ?? ?? 2
+?? ?? 2
+?? ?? 2
+?? ) 
Coplanarity of four points 
The points ?? (?? 1
?? 1
?? 1
), ?? (?? 2
?? 2
?? 2
)?? (?? 3
?? 3
?? 3
) and ?? (?? 4
?? 4
?? 4
) are coplanar then 
|?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 2
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 3
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 ?? 4
- ?? 1
 | = 0 
Problem 41 : Find the equation of the plane upon which the length of normal from origin is 10 and 
direction ratios of this normal are 3,2,6. 
Solution: If ?? be the length of perpendicular from origin to the plane and ?? , ?? , ?? be the direction 
cosines of this normal, then its equation is 
???? + ???? + ???? = 10 
Direction ratios of normal to the plane are 3, 2, 6 
?  Direction cosines of normal to the required plane are ?? =
3
7
, ?? =
2
7
, ?? =
6
7
 
Equation of required plane is 
3
7
?? +
2
7
?? +
6
7
?? = 10 or,  3?? + 2?? + 6?? = 70 
Problem 42 :Find the plane through the points (2, -3,3), (-5,2,0), (1, -7,1) 
Solution :  |?? - 2 ?? + 3 ?? - 3 - 5 - 2 2 + 3 0 - 3 1 - 2 - 7 + 3 1 - 3 | = 0  or  |?? - 2 ?? + 3 ?? - 3 -
7 5 - 3 - 1 - 4 - 2 | = 0 ? 2?? + ?? - 3?? + 8 = 0 
Problem 43 : If ?? be any point on the plane ???? + ???? + ???? = ?? and ?? be a point on the line ???? such 
that 
???? . ???? = ?? 2
, show that the locus of the point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Solution :   Let ?? = (?? , ?? , ?? ), ?? = (?? 1
, ?? 1
, ?? 1
) 
Direction ratios of OP are ?? , ?? , ?? and direction ratios of ???? are ?? 1
, ?? 1
, ?? 1
. 
Since ?? , ?? , ?? are collinear, we have 
?? ?? 1
=
?? ?? 1
=
?? ?? 1
= ?? (say) 
As ?? (?? , ?? , ?? ) lies on the plane ???? + ???? + ???? = ?? , 
???? + ???? + ???? = ??  or  ?? (?? ?? 1
+ ????
1
+ ???? ?? 1
) = ?? 
Given ???? . ???? = ?? 2
 ? v?? 2
+ ?? 2
+ ?? 2
v?? 1
2
+ ?? 1
2
+ ?? 1
2
 = ?? 2
 
or, v?? 2
(?? 1
2
+ ?? 1
2
+ ?? 1
2
)v?? 1
2
+ ?? 1
2
+ ?? 1
2
= ?? 2
  or,  ?? (?? 1
2
+ ?? 1
2
+ ?? 1
2
) = ?? 2
 
 
Hence the locus of point ?? is ?? (???? + ???? + ???? ) = ?? 2
+ ?? 2
+ ?? 2
. 
Problem 44 : A moving plane passes through a fixed point (?? , ?? , ?? ) and cuts the coordinate axes A, B, 
C . Find the locus of the centroid of the tetrahedron OABC. 
Solution : Let the plane be 
?? ?? +
?? ?? +
?? ?? = 1,0(0,0,0), ?? (?? , 0,0), ?? (0, ?? , 0) 
C (0,0, ?? ). Centroid of ???????? is (
?? 4
,
?? 4
,
?? 4
) 
The plane passes through (?? , ?? , ?? ) ? 
?? ?? +
?? ?? +
?? ?? = 1 
Centroid, ?? =
?? 4
, ?? =
?? 4
, ?? =
?? 4
  or  ?? = 4?? , ?? = 4?? , ?? = 4?? 
Now (1) gives the locus of ?? as 
?? ?? +
?? ?? +
?? ?? = 4 
Position of point with respect to plane : 
A plane divides the three dimensional space in two equal parts. Two points ?? (?? 1
?? 1
?? 1
) and ?? (?? 2
?? 2
?? 2
) 
are on the same side of the plane ???? + ???? + ???? + ?? = 0 if ?? ?? 1
+ ?? ?? 1
+ ?? ?? 1
+ ?? and ????
2
+ ????
2
+ ????
2
+ ?? 
are both positive or both negative and are opposite side of plane if both of these values are in 
opposite sign 
Problem 45 : Show that the points (1,2,3) and (2, -1,4) lie on opposite sides of the plane ?? + 4?? + ?? -
3 = 0 
Solution : Since the numbers 1 + 4 × 2 + 3 - 3 = 9 and 2 - 4 + 4 - 3 = -1 are of opposite sign, then 
points are on opposite sides of the plane. 
A plane & a point 
 
Let ?? = ???? + ???? + ???? + ?? = 0 is a given plane and ?? (?? 1
, ?? 1
, ?? 1
) is given point as shown in figure. 
Let ?? (?? '
, ?? '
, ?? '
) be the foot of the point ?? (?? 1
, ?? 1
, ?? 1
) with respect to the plane ?? . 
And ?? (?? ''
, ?? ''
, ?? ''
) be the reflection of point ?? (?? 1
, ?? 1
, ?? 1
) with respect to the plane ?? . 
(i) Distance of the point (?? 1
, ?? 1
, ?? 1
) from the plane ???? + ???? + ???? + ?? = 0 is given by |
?? ?? 1
+?? ?? 1
+?? ?? 1
+?? v?? 2
+?? 2
+?? 2
|. 
(ii) The length of the perpendicular from a point having position vector ?? ? to plane ??? · ??? ? = ?? is ?? =
|?? ? ?·?? ? ?-?? |
|?? ? ?|
. 
(iii) The coordinates of the foot (F) of perpendicular from the point (?? 1
, ?? 1
, ?? 1
) to the plane 
???? + ???? + ???? + ?? = 0 are 
?? -?? 1
?? =
?? -?? 1
?? =
?? -?? 1
?? = -
(?? ?? 1
+?? ?? 1
+?? ?? 1
+?? )
?? 2
+?? 2
+?? 2
 
(iv) The coordinates of the Image (R) of point (?? 1
, ?? 1
, ?? 1
) to the plane 
???? + ???? + ???? + ?? = 0 are 
?? -?? 1
?? =
?? -?? 1
?? =
?? -?? 1
?? = -
2(?? ?? 1
+?? ?? 1
+?? ?? 1
+?? )
?? 2
+?? 2
+?? 2
 
Problem 46 : Find the image of the point ?? (3,5,7) in the plane 2?? + ?? + ?? = 0. 
Solution:   Given plane is 2?? + ?? + ?? = 0 
Direction ratios of normal to plane (1) are 2,1,1 
Let ?? be the image of point ?? in plane (1). Let ???? meet plane (1) in ?? then ???? ? plane (1) 
Let  ?? = (2?? + 3, ?? + 5, ?? + 7) 
Since ?? lies on plane (1) 
? 2(2?? + 3) + ?? + 5 + ?? + 7 = 0  or,  6?? + 18 = 0 ? ?? = -3 
? ?? = (-3,2,4) 
Let  ?? = (?? , ?? , ?? ) 
Since ?? is the middle point of ???? 
? -3 =
?? + 3
2
? ?? = -9 ?????? 2 =
?? + 5
2
? ?? = -1 ?????? 4 =
?? + 7
2
 ? ?? = 1 ? ?? = (-9, -1,1).  
Problem 47 : A plane passes through a fixed point (?? , ?? , ?? ). Show that the locus of the foot of 
perpendicular to it from the origin is the sphere ?? 2
+ ?? 2
+ ?? 2
- ???? - ???? - ???? = 0 
Solution : Let the equation of the variable plane be ???? + ???? + ???? + ?? = 0 
Plane passes through the fixed point (?? , ?? , ?? ) ? ???? + ???? + ???? + ?? = 0 
Let ?? (?? , ?? , ?? ) be the foot of perpendicular from origin to plane (1). 
Direction ratios of OP are 
 
i.e. ?? , ?? , ?? 
From equation (1), it is clear that the direction ratios of normal to the plane i.e. OP are ?? , ?? , ?? ; ?? , ?? , ?? 
and ?? , ?? , ?? are the direction ratios of the same line OP 
? 
?? ?? =
?? ?? =
?? ?? =
1
?? (?????? )  ? ?? = ???? , ?? = ???? , ?? = ???? #(3)  
Putting the values of ?? , ?? , ?? in equation (2), we get ka ?? + ?????? + ?????? + ?? = 0 
Since ?? , ?? , ?? lies in plane (1)  ? ???? + ???? + ???? + ?? = 0 
Putting the values of ?? , ?? , ?? from (3) in (5), we get ?? ?? 2
+ ?? ?? 2
+ ?? ?? 2
+ ?? = 0 
or 
?? ?? 2
+ ?? ?? 2
+ ?? ?? 2
- ?????? - ?????? - ?????? = 0  [putting the value of ?? from (4) in (6)] 
or ?? 2
+ ?? 2
+ ?? 2
- ???? - ???? - ???? = 0 
Therefore, locus of foot of perpendicular ?? (?? , ?? , ?? ) is ?? 2
+ ?? 2
+ ?? 2
- ???? - ???? - ???? = 0 
Angle between two planes : 
(i) Consider two planes ???? + ???? + ???? + ?? = 0 and ?? '
?? + ?? '
?? + ?? '
?? + ?? '
= 0. Angle between these 
planes is the angle between their normals. Since direction ratios of their normals are (?? , ?? , ?? ) and 
(?? '
, ?? '
, ?? '
 ) respectively, hence ?? , the angle between them, is given by 
?????? ?? =
?? ?? '
+ ?? ?? '
+ ?? ?? '
v?? 2
+ ?? 2
+ ?? 2
v?? '2
+ ?? '2
+ ?? '2
 
Planes are perpendicular if ????
'
+ ????
'
+ ????
'
= 0 and planes are parallel if 
?? ?? '
=
?? ?? '
=
?? ?? '
 
(ii) The angle ?? between the planes ??? · ??? ?
1
= ?? 1
 and ??? · ??? ?
2
= ?? 2
 is given by, ?????? ?? =
?? ? ?
1
·?? ? ?
2
|?? ? ?
1
||?? ? ?
2
|
 
Planes are perpendicular if ??? ?
1
· ??? ?
2
= 0 & planes are parallel if ??? ?
1
= ?? ??? ?
2
.