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Page 1
Edurev123
5. Application to Geometry
5.1 A curve in space is defined by the vector equation ?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
. Determine the
angle between the tangents to this curve at the ?? =+1 and ?? =-1.
(2013 : 10 Marks)
Solution:
?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
????
????
=2?? ??ˆ+2??ˆ-3?? 2
??ˆ
????
????
=|
?? ??
????
|=v4?? 2
+4+9?? 4
??ˆ
=
?? ??
????
=
?? ??
????
????
????
=
2+??ˆ+2??ˆ-3?? 2
??ˆ
v4?? 2
+4+9?? 4
where ??ˆ
is the direction of the tangent
??ˆ|
?? =1
=
2??ˆ+2??ˆ-3??ˆ
v17
??ˆ|
?? =-1
=
-2??ˆ+2??ˆ-3??ˆ
v17
Let ?? be the angle
Then
cos ?? =??ˆ
1
·??ˆ
2
=
1
17
(2??ˆ+2??ˆ-3??ˆ
)·(-2??ˆ+2??ˆ-3??ˆ
)=
-4+4+9
17
=
9
17
? ?? =cos
-1
9
17
5.2 Find the angle between the surfaces ?? ?? +?? ?? +?? ?? -?? =?? and ?? =?? ?? +?? ?? -?? at
(?? ,-?? ,?? ) .
(2015 : 10 Marks)
Solution:
Angle between two surfaces at a point is the angle between the normals to the surfaces
at that point.
Let
Page 2
Edurev123
5. Application to Geometry
5.1 A curve in space is defined by the vector equation ?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
. Determine the
angle between the tangents to this curve at the ?? =+1 and ?? =-1.
(2013 : 10 Marks)
Solution:
?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
????
????
=2?? ??ˆ+2??ˆ-3?? 2
??ˆ
????
????
=|
?? ??
????
|=v4?? 2
+4+9?? 4
??ˆ
=
?? ??
????
=
?? ??
????
????
????
=
2+??ˆ+2??ˆ-3?? 2
??ˆ
v4?? 2
+4+9?? 4
where ??ˆ
is the direction of the tangent
??ˆ|
?? =1
=
2??ˆ+2??ˆ-3??ˆ
v17
??ˆ|
?? =-1
=
-2??ˆ+2??ˆ-3??ˆ
v17
Let ?? be the angle
Then
cos ?? =??ˆ
1
·??ˆ
2
=
1
17
(2??ˆ+2??ˆ-3??ˆ
)·(-2??ˆ+2??ˆ-3??ˆ
)=
-4+4+9
17
=
9
17
? ?? =cos
-1
9
17
5.2 Find the angle between the surfaces ?? ?? +?? ?? +?? ?? -?? =?? and ?? =?? ?? +?? ?? -?? at
(?? ,-?? ,?? ) .
(2015 : 10 Marks)
Solution:
Angle between two surfaces at a point is the angle between the normals to the surfaces
at that point.
Let
?? 1
=?? 2
+?? 2
+?? 2
and
?? 2
=?? 2
+?? 2
-??
Then grad (?? 1
)=2???? +2???? +2????
grad (?? 2
)=2???? +2???? -??
?????? ?? 1
=grad ?? 1
at point (2,-1,2)=4?? -2?? +4??
?? 2
=grad ?? 2
at point (2,-1,2)=4?? -2?? -4??
The vectors ?? 1
and ?? 2
are along normals to the two surfaces at the point (2,-1,2) . If ?? is
the angle between these two vectors then,
cos ?? =
?? 1
·?? 2
|?? 1
||?? 2
|
=
16+4-4
v16+4+16v16+4+1
=
16
6v21
?? =cos
-1
8
3v21
5.3 Find the values of ?? and ?? so that the surfaces ?? ?? ?? -?????? =(?? +?? )?? and ?? ?? ?? ?? +
?? ?? =?? may intersect orthogonally at (?? ,-?? ,?? ) .
(2015 : 12 Marks)
Solution:
Let
?? 1
=?? ?? 2
-(?? +2)?? -?????? ?? 2
=4?? 2
?? +?? 3
grad ?? 1
=??? 1
=(2???? -?? -2)?? -?????? -?????? ??? 2
=8?????? +4?? 2
?? +3?? 2
??
At point (1,-1,2)
?? 1
=(?? -2)?? -2???? +????
?? 2
=-8?? +4?? +12?? ?? 1
·?? 2
=0
-8(?? -2)-8?? +12?? =0
(intersect orthogonally)
-8?? +4?? =-16?2?? -?? =4 (??)
Point (1,-1,2) lies on surface ?? ?? 2
-?????? =(?? +2)??
Page 3
Edurev123
5. Application to Geometry
5.1 A curve in space is defined by the vector equation ?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
. Determine the
angle between the tangents to this curve at the ?? =+1 and ?? =-1.
(2013 : 10 Marks)
Solution:
?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
????
????
=2?? ??ˆ+2??ˆ-3?? 2
??ˆ
????
????
=|
?? ??
????
|=v4?? 2
+4+9?? 4
??ˆ
=
?? ??
????
=
?? ??
????
????
????
=
2+??ˆ+2??ˆ-3?? 2
??ˆ
v4?? 2
+4+9?? 4
where ??ˆ
is the direction of the tangent
??ˆ|
?? =1
=
2??ˆ+2??ˆ-3??ˆ
v17
??ˆ|
?? =-1
=
-2??ˆ+2??ˆ-3??ˆ
v17
Let ?? be the angle
Then
cos ?? =??ˆ
1
·??ˆ
2
=
1
17
(2??ˆ+2??ˆ-3??ˆ
)·(-2??ˆ+2??ˆ-3??ˆ
)=
-4+4+9
17
=
9
17
? ?? =cos
-1
9
17
5.2 Find the angle between the surfaces ?? ?? +?? ?? +?? ?? -?? =?? and ?? =?? ?? +?? ?? -?? at
(?? ,-?? ,?? ) .
(2015 : 10 Marks)
Solution:
Angle between two surfaces at a point is the angle between the normals to the surfaces
at that point.
Let
?? 1
=?? 2
+?? 2
+?? 2
and
?? 2
=?? 2
+?? 2
-??
Then grad (?? 1
)=2???? +2???? +2????
grad (?? 2
)=2???? +2???? -??
?????? ?? 1
=grad ?? 1
at point (2,-1,2)=4?? -2?? +4??
?? 2
=grad ?? 2
at point (2,-1,2)=4?? -2?? -4??
The vectors ?? 1
and ?? 2
are along normals to the two surfaces at the point (2,-1,2) . If ?? is
the angle between these two vectors then,
cos ?? =
?? 1
·?? 2
|?? 1
||?? 2
|
=
16+4-4
v16+4+16v16+4+1
=
16
6v21
?? =cos
-1
8
3v21
5.3 Find the values of ?? and ?? so that the surfaces ?? ?? ?? -?????? =(?? +?? )?? and ?? ?? ?? ?? +
?? ?? =?? may intersect orthogonally at (?? ,-?? ,?? ) .
(2015 : 12 Marks)
Solution:
Let
?? 1
=?? ?? 2
-(?? +2)?? -?????? ?? 2
=4?? 2
?? +?? 3
grad ?? 1
=??? 1
=(2???? -?? -2)?? -?????? -?????? ??? 2
=8?????? +4?? 2
?? +3?? 2
??
At point (1,-1,2)
?? 1
=(?? -2)?? -2???? +????
?? 2
=-8?? +4?? +12?? ?? 1
·?? 2
=0
-8(?? -2)-8?? +12?? =0
(intersect orthogonally)
-8?? +4?? =-16?2?? -?? =4 (??)
Point (1,-1,2) lies on surface ?? ?? 2
-?????? =(?? +2)??
? ?? +2?? =(?? +2)??? =1
? 2?? -1=4??? =
5
2
5.4 Find the curvature vector and its magnitude at any point ??? =(?? ) of the curve
??? =(?? ?????? ?? ,?? ?????? ?? , a). Show that the locus of the feet of the perpendicular from the
origin to the tangent is a curve that completely lies on the hyperboloid ?? ?? +?? ?? -
?? ?? =?? ?? .
(2017 : 16 Marks)
Solution:
?? =?? cos ???? +?? sin ???? +??????
?? ??
????
=-?? sin ???? +?? cos ???? +????
????
????
=|
????
????
|=v2??
?? =
?? ??
????
=
?? ??
????
????
=
1
v2
(-sin ???? +cos ???? +?? )
?? ??
????
=
1
v2
(-cos ?? ??ˆ-sin ?? ??ˆ)
?????????????????? ???????????? , ???
=
?? ??
????
=
?? ?? /????
???? /????
=
1
v2?? ·
1
v2
(-cos ???? -sin ???? )
=
-cos ????
2?? ·
sin ????
2??
Magnitude,
|?? |=
1
2??
Locus of feet of perpendicular from origin to tangent
Page 4
Edurev123
5. Application to Geometry
5.1 A curve in space is defined by the vector equation ?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
. Determine the
angle between the tangents to this curve at the ?? =+1 and ?? =-1.
(2013 : 10 Marks)
Solution:
?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
????
????
=2?? ??ˆ+2??ˆ-3?? 2
??ˆ
????
????
=|
?? ??
????
|=v4?? 2
+4+9?? 4
??ˆ
=
?? ??
????
=
?? ??
????
????
????
=
2+??ˆ+2??ˆ-3?? 2
??ˆ
v4?? 2
+4+9?? 4
where ??ˆ
is the direction of the tangent
??ˆ|
?? =1
=
2??ˆ+2??ˆ-3??ˆ
v17
??ˆ|
?? =-1
=
-2??ˆ+2??ˆ-3??ˆ
v17
Let ?? be the angle
Then
cos ?? =??ˆ
1
·??ˆ
2
=
1
17
(2??ˆ+2??ˆ-3??ˆ
)·(-2??ˆ+2??ˆ-3??ˆ
)=
-4+4+9
17
=
9
17
? ?? =cos
-1
9
17
5.2 Find the angle between the surfaces ?? ?? +?? ?? +?? ?? -?? =?? and ?? =?? ?? +?? ?? -?? at
(?? ,-?? ,?? ) .
(2015 : 10 Marks)
Solution:
Angle between two surfaces at a point is the angle between the normals to the surfaces
at that point.
Let
?? 1
=?? 2
+?? 2
+?? 2
and
?? 2
=?? 2
+?? 2
-??
Then grad (?? 1
)=2???? +2???? +2????
grad (?? 2
)=2???? +2???? -??
?????? ?? 1
=grad ?? 1
at point (2,-1,2)=4?? -2?? +4??
?? 2
=grad ?? 2
at point (2,-1,2)=4?? -2?? -4??
The vectors ?? 1
and ?? 2
are along normals to the two surfaces at the point (2,-1,2) . If ?? is
the angle between these two vectors then,
cos ?? =
?? 1
·?? 2
|?? 1
||?? 2
|
=
16+4-4
v16+4+16v16+4+1
=
16
6v21
?? =cos
-1
8
3v21
5.3 Find the values of ?? and ?? so that the surfaces ?? ?? ?? -?????? =(?? +?? )?? and ?? ?? ?? ?? +
?? ?? =?? may intersect orthogonally at (?? ,-?? ,?? ) .
(2015 : 12 Marks)
Solution:
Let
?? 1
=?? ?? 2
-(?? +2)?? -?????? ?? 2
=4?? 2
?? +?? 3
grad ?? 1
=??? 1
=(2???? -?? -2)?? -?????? -?????? ??? 2
=8?????? +4?? 2
?? +3?? 2
??
At point (1,-1,2)
?? 1
=(?? -2)?? -2???? +????
?? 2
=-8?? +4?? +12?? ?? 1
·?? 2
=0
-8(?? -2)-8?? +12?? =0
(intersect orthogonally)
-8?? +4?? =-16?2?? -?? =4 (??)
Point (1,-1,2) lies on surface ?? ?? 2
-?????? =(?? +2)??
? ?? +2?? =(?? +2)??? =1
? 2?? -1=4??? =
5
2
5.4 Find the curvature vector and its magnitude at any point ??? =(?? ) of the curve
??? =(?? ?????? ?? ,?? ?????? ?? , a). Show that the locus of the feet of the perpendicular from the
origin to the tangent is a curve that completely lies on the hyperboloid ?? ?? +?? ?? -
?? ?? =?? ?? .
(2017 : 16 Marks)
Solution:
?? =?? cos ???? +?? sin ???? +??????
?? ??
????
=-?? sin ???? +?? cos ???? +????
????
????
=|
????
????
|=v2??
?? =
?? ??
????
=
?? ??
????
????
=
1
v2
(-sin ???? +cos ???? +?? )
?? ??
????
=
1
v2
(-cos ?? ??ˆ-sin ?? ??ˆ)
?????????????????? ???????????? , ???
=
?? ??
????
=
?? ?? /????
???? /????
=
1
v2?? ·
1
v2
(-cos ???? -sin ???? )
=
-cos ????
2?? ·
sin ????
2??
Magnitude,
|?? |=
1
2??
Locus of feet of perpendicular from origin to tangent
????
?????
=?? -(?? ·?? )??
?? = Foot of perpendicular
?? ·?? =(?? cos ???? +?? sin ???? +?????? )·(-sin ???? +cos ?? +?? )
1
v2
=
????
v2
?? -(?? ·?? )?? =(?? cos ???? +?? sin ???? +?????? )-
????
v2
·
1
v2
(-sin ???? +cos ?? +?? )
=(?? cos ?? +
???? sin ?? 2
)?? +(?? sin ?? -
???? cos ?? 2
)?? +
?? 2
??
Let foot of perpendicular be ?? (?? ,?? ,?? )
?? =?? cos ?? +
???? sin ?? 2
?? =?? sin ?? -
???? cos ?? 2
?? =
????
2
? ?? 2
+?? 2
-?? 2
=?? 2
+
?? 2
?? 2
4
-
?? 2
?? 2
4
=?? 2
?? 2
+?? 2
-?? 2
=?? 2
is the required locus.
5.5 Find the angle between tangent at general point of curve whose equations are
?? =?? ?? ,?? =?? ?? ?? , ?? =?? ?? ?? and the line ?? -?? =?? =?? .
(2018: 10 marks)
Solution:
The radius vector of given curve, ?? can be written as
Page 5
Edurev123
5. Application to Geometry
5.1 A curve in space is defined by the vector equation ?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
. Determine the
angle between the tangents to this curve at the ?? =+1 and ?? =-1.
(2013 : 10 Marks)
Solution:
?? =?? 2
??ˆ+2?? ??ˆ-?? 3
??ˆ
????
????
=2?? ??ˆ+2??ˆ-3?? 2
??ˆ
????
????
=|
?? ??
????
|=v4?? 2
+4+9?? 4
??ˆ
=
?? ??
????
=
?? ??
????
????
????
=
2+??ˆ+2??ˆ-3?? 2
??ˆ
v4?? 2
+4+9?? 4
where ??ˆ
is the direction of the tangent
??ˆ|
?? =1
=
2??ˆ+2??ˆ-3??ˆ
v17
??ˆ|
?? =-1
=
-2??ˆ+2??ˆ-3??ˆ
v17
Let ?? be the angle
Then
cos ?? =??ˆ
1
·??ˆ
2
=
1
17
(2??ˆ+2??ˆ-3??ˆ
)·(-2??ˆ+2??ˆ-3??ˆ
)=
-4+4+9
17
=
9
17
? ?? =cos
-1
9
17
5.2 Find the angle between the surfaces ?? ?? +?? ?? +?? ?? -?? =?? and ?? =?? ?? +?? ?? -?? at
(?? ,-?? ,?? ) .
(2015 : 10 Marks)
Solution:
Angle between two surfaces at a point is the angle between the normals to the surfaces
at that point.
Let
?? 1
=?? 2
+?? 2
+?? 2
and
?? 2
=?? 2
+?? 2
-??
Then grad (?? 1
)=2???? +2???? +2????
grad (?? 2
)=2???? +2???? -??
?????? ?? 1
=grad ?? 1
at point (2,-1,2)=4?? -2?? +4??
?? 2
=grad ?? 2
at point (2,-1,2)=4?? -2?? -4??
The vectors ?? 1
and ?? 2
are along normals to the two surfaces at the point (2,-1,2) . If ?? is
the angle between these two vectors then,
cos ?? =
?? 1
·?? 2
|?? 1
||?? 2
|
=
16+4-4
v16+4+16v16+4+1
=
16
6v21
?? =cos
-1
8
3v21
5.3 Find the values of ?? and ?? so that the surfaces ?? ?? ?? -?????? =(?? +?? )?? and ?? ?? ?? ?? +
?? ?? =?? may intersect orthogonally at (?? ,-?? ,?? ) .
(2015 : 12 Marks)
Solution:
Let
?? 1
=?? ?? 2
-(?? +2)?? -?????? ?? 2
=4?? 2
?? +?? 3
grad ?? 1
=??? 1
=(2???? -?? -2)?? -?????? -?????? ??? 2
=8?????? +4?? 2
?? +3?? 2
??
At point (1,-1,2)
?? 1
=(?? -2)?? -2???? +????
?? 2
=-8?? +4?? +12?? ?? 1
·?? 2
=0
-8(?? -2)-8?? +12?? =0
(intersect orthogonally)
-8?? +4?? =-16?2?? -?? =4 (??)
Point (1,-1,2) lies on surface ?? ?? 2
-?????? =(?? +2)??
? ?? +2?? =(?? +2)??? =1
? 2?? -1=4??? =
5
2
5.4 Find the curvature vector and its magnitude at any point ??? =(?? ) of the curve
??? =(?? ?????? ?? ,?? ?????? ?? , a). Show that the locus of the feet of the perpendicular from the
origin to the tangent is a curve that completely lies on the hyperboloid ?? ?? +?? ?? -
?? ?? =?? ?? .
(2017 : 16 Marks)
Solution:
?? =?? cos ???? +?? sin ???? +??????
?? ??
????
=-?? sin ???? +?? cos ???? +????
????
????
=|
????
????
|=v2??
?? =
?? ??
????
=
?? ??
????
????
=
1
v2
(-sin ???? +cos ???? +?? )
?? ??
????
=
1
v2
(-cos ?? ??ˆ-sin ?? ??ˆ)
?????????????????? ???????????? , ???
=
?? ??
????
=
?? ?? /????
???? /????
=
1
v2?? ·
1
v2
(-cos ???? -sin ???? )
=
-cos ????
2?? ·
sin ????
2??
Magnitude,
|?? |=
1
2??
Locus of feet of perpendicular from origin to tangent
????
?????
=?? -(?? ·?? )??
?? = Foot of perpendicular
?? ·?? =(?? cos ???? +?? sin ???? +?????? )·(-sin ???? +cos ?? +?? )
1
v2
=
????
v2
?? -(?? ·?? )?? =(?? cos ???? +?? sin ???? +?????? )-
????
v2
·
1
v2
(-sin ???? +cos ?? +?? )
=(?? cos ?? +
???? sin ?? 2
)?? +(?? sin ?? -
???? cos ?? 2
)?? +
?? 2
??
Let foot of perpendicular be ?? (?? ,?? ,?? )
?? =?? cos ?? +
???? sin ?? 2
?? =?? sin ?? -
???? cos ?? 2
?? =
????
2
? ?? 2
+?? 2
-?? 2
=?? 2
+
?? 2
?? 2
4
-
?? 2
?? 2
4
=?? 2
?? 2
+?? 2
-?? 2
=?? 2
is the required locus.
5.5 Find the angle between tangent at general point of curve whose equations are
?? =?? ?? ,?? =?? ?? ?? , ?? =?? ?? ?? and the line ?? -?? =?? =?? .
(2018: 10 marks)
Solution:
The radius vector of given curve, ?? can be written as
?? =3??ˆ+3?? 2
??ˆ+3?? 3
??ˆ
?? ??
????
=3??ˆ+6??ˆ+9?? 2
??ˆ
????
????
=|
?? ??
????
|=v9+36?? 2
+81?? 4
=3v1+4?? 2
+9?? 4
Tangent vector,
???
=
?? ??
????
=
?? ?? /????
???? /????
=
??
3v1+4?? 2
+9?? 4
×(3??ˆ+6?? ??ˆ)
? ???
=
??ˆ+2?? ??ˆ+3?? 2
??ˆ
v1+4?? 2
+9?? 4
The given line can be written as
?? 1
=
?? 0
=
?? 1
? angle between line and tangent vector, ??
cos ?? =
1×1+2?? ×0+3?? 2
×1
v2·v1+4?? 2
+9?? 4
=
1+3?? 2
v2·v1+4?? 2
+9?? 4
?? =cos
-1
(
1+3?? 2
v2·v1+4?? 2
+9?? 4
)
5.6 Find the circulation of ????
round the curve ?? , where ????
=(?? ?? +?? ?? )??ˆ+(?? ?? -?? ?? )??ˆ
and ?? is the curve ?? =?? ?? from (?? ,?? ) to (?? ,?? ) and the curve ?? ?? =?? from (?? ,?? ) to
(?? ,?? ) .
(2019: 15 Marks)
Solution:
Here the closed curve ' ?? consists of arcs ?????? and ?????? . Let ?? 1
denote the arc ?????? and
?? 2
denote arc ?????? . Along ?? 1
, we have ?? =?? 2
so that ???? =2?????? and ?? varies from 0?1
along ?? 2
, we have ?? =?? 2
so that ???? = 2?????? and ?? varies from 1 to also.
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Application to Geometry Notes
Application to Geometry Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Application to Geometry.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
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Application to Geometry UPSC Questions
The "Application to Geometry UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Application to Geometry on the App
Students of UPSC can study Application to Geometry alongwith tests & analysis from the EduRev app,
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students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
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The content of Application to Geometry is prepared as per the latest UPSC syllabus.
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