Detailed Notes: Three Dimensional Geometry

``` 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
.
```

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

## FAQs on Detailed Notes: Three Dimensional Geometry - Mathematics (Maths) for JEE Main & Advanced

 1. How do you find the volume of a three-dimensional shape?
Ans. To find the volume of a three-dimensional shape, you multiply the length, width, and height of the shape. For example, the volume of a rectangular prism can be calculated by V = lwh, where l is the length, w is the width, and h is the height.
 2. What is the difference between two-dimensional and three-dimensional geometry?
Ans. Two-dimensional geometry deals with shapes and figures that have only length and width, like squares and circles. Three-dimensional geometry involves shapes and figures that have length, width, and height, such as cubes and spheres.
 3. How can you determine if two three-dimensional shapes are congruent?
Ans. Two three-dimensional shapes are congruent if they have the same size and shape, and their corresponding angles and sides are equal. You can determine congruence by comparing the dimensions of the shapes.
 4. What is the formula for finding the surface area of a three-dimensional shape?
Ans. The formula for finding the surface area of a three-dimensional shape depends on the shape. For example, the surface area of a cube can be calculated by SA = 6s^2, where s is the length of one side of the cube.
 5. How can you determine the center of mass of a three-dimensional object?
Ans. The center of mass of a three-dimensional object is the point where the object's mass is evenly distributed in all directions. It can be calculated by finding the average position of all the points in the object's volume. This point represents the object's balance point.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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