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Edurev123
3. Plane and its Properties
3.1 Find the equations of the straight line through the point (?? ,?? ,?? ) to intersect the
straight line
?? +?? =?? +?? =?? (?? -?? )
and parallel to the plane ?? ?? +?? +?? ?? =?? .
(2011 : 10 Marks)
Solution:
Let the required line intersects the given line
?? +?? =?? +1=2(?? -2) (??)
at (?? 1
,?? 1
,?? 1
)
? The equations of the required line passing through (2,1,2) and (?? 1
,?? 1
,?? 1
) are
?? -3
?? 1
-3
=
?? -1
?? 1
-1
=
?? -2
?? 1
-1
(???? )
The direction ratios of the line (ii) are ?? 1
-3,?? 1
-1,?? 1
-1.
Because the line (ii) is parallel to the plane 4?? +?? +5?? =0, therefore the normal to the
plane with direction ratios 4,1,5 is perpendicular to the line (ii).
? 4(?? 1
-3)+1(?? 1
-1)+5(?? 1
-1)=0
or 4?? 1
+?? 2
+5?? 2
-23=0 (?????? )
Because the point (?? 1
,?? 1
,?? 1
) lies on (i),
? ?? 1
+4=?? 1
+1=2(?? 1
-2)
? ?? 1
=2?? 1
-8,?? 1
=2?? 1
-5 (iv)
? from (iv),we get
4(2?? 1
-8)+2?? 1
-5+5?? 1
-23 =0
? ?? 1
=4
Page 2
Edurev123
3. Plane and its Properties
3.1 Find the equations of the straight line through the point (?? ,?? ,?? ) to intersect the
straight line
?? +?? =?? +?? =?? (?? -?? )
and parallel to the plane ?? ?? +?? +?? ?? =?? .
(2011 : 10 Marks)
Solution:
Let the required line intersects the given line
?? +?? =?? +1=2(?? -2) (??)
at (?? 1
,?? 1
,?? 1
)
? The equations of the required line passing through (2,1,2) and (?? 1
,?? 1
,?? 1
) are
?? -3
?? 1
-3
=
?? -1
?? 1
-1
=
?? -2
?? 1
-1
(???? )
The direction ratios of the line (ii) are ?? 1
-3,?? 1
-1,?? 1
-1.
Because the line (ii) is parallel to the plane 4?? +?? +5?? =0, therefore the normal to the
plane with direction ratios 4,1,5 is perpendicular to the line (ii).
? 4(?? 1
-3)+1(?? 1
-1)+5(?? 1
-1)=0
or 4?? 1
+?? 2
+5?? 2
-23=0 (?????? )
Because the point (?? 1
,?? 1
,?? 1
) lies on (i),
? ?? 1
+4=?? 1
+1=2(?? 1
-2)
? ?? 1
=2?? 1
-8,?? 1
=2?? 1
-5 (iv)
? from (iv),we get
4(2?? 1
-8)+2?? 1
-5+5?? 1
-23 =0
? ?? 1
=4
? from (iv), ?? 1
=0,?? 1
=3
? from (ii), the equations of the required line are :
?? -3
-3
=
?? -1
2
=
?? 2
-2
2
3.2 Find the equation of the plane which passes through the points (?? ,?? ,?? ) and
(?? ,?? ,-?? ) and is parallel to the line joining the points (-?? ,?? ,-?? ),(?? ,-?? ,?? ) . Find also
the distance between the line and the plane.
(2013 : 10 Marks)
Solution:
The general equation of plane through (0,1,1)
??(?? -0)+?? (?? -1)+?? (?? -1)=0
???? +?? (?? -1)+?? (?? -1)=0
(??)
(2,0,-1) lies on this plane.
? 2?? -?? -2?? =0
Direction ratios of line passing through (-1,1,-2),(3,-2,4) .
?? 1
=3-1=4,?? ?? =(-2-1)=-3;?? 1
=(4-(-2))=6
? Direction ratios are 4::-3::6
This is parallel to plane.
? 4?? -3?? +6?? =0
From (ii) and (iii)
?? -12
=
?? -20
=
?? -2
??? :?? :?? ::6:10:1
? Equation of plane is
6?? +10(?? -1)+(?? -1) =0
? 6?? +10?? +?? -11 =0
Equation of plane parallel to this plane and passing through line
6(?? +1)+10(?? -1)+1(?? +2)=0
6?? +10?? +?? -2=0
Page 3
Edurev123
3. Plane and its Properties
3.1 Find the equations of the straight line through the point (?? ,?? ,?? ) to intersect the
straight line
?? +?? =?? +?? =?? (?? -?? )
and parallel to the plane ?? ?? +?? +?? ?? =?? .
(2011 : 10 Marks)
Solution:
Let the required line intersects the given line
?? +?? =?? +1=2(?? -2) (??)
at (?? 1
,?? 1
,?? 1
)
? The equations of the required line passing through (2,1,2) and (?? 1
,?? 1
,?? 1
) are
?? -3
?? 1
-3
=
?? -1
?? 1
-1
=
?? -2
?? 1
-1
(???? )
The direction ratios of the line (ii) are ?? 1
-3,?? 1
-1,?? 1
-1.
Because the line (ii) is parallel to the plane 4?? +?? +5?? =0, therefore the normal to the
plane with direction ratios 4,1,5 is perpendicular to the line (ii).
? 4(?? 1
-3)+1(?? 1
-1)+5(?? 1
-1)=0
or 4?? 1
+?? 2
+5?? 2
-23=0 (?????? )
Because the point (?? 1
,?? 1
,?? 1
) lies on (i),
? ?? 1
+4=?? 1
+1=2(?? 1
-2)
? ?? 1
=2?? 1
-8,?? 1
=2?? 1
-5 (iv)
? from (iv),we get
4(2?? 1
-8)+2?? 1
-5+5?? 1
-23 =0
? ?? 1
=4
? from (iv), ?? 1
=0,?? 1
=3
? from (ii), the equations of the required line are :
?? -3
-3
=
?? -1
2
=
?? 2
-2
2
3.2 Find the equation of the plane which passes through the points (?? ,?? ,?? ) and
(?? ,?? ,-?? ) and is parallel to the line joining the points (-?? ,?? ,-?? ),(?? ,-?? ,?? ) . Find also
the distance between the line and the plane.
(2013 : 10 Marks)
Solution:
The general equation of plane through (0,1,1)
??(?? -0)+?? (?? -1)+?? (?? -1)=0
???? +?? (?? -1)+?? (?? -1)=0
(??)
(2,0,-1) lies on this plane.
? 2?? -?? -2?? =0
Direction ratios of line passing through (-1,1,-2),(3,-2,4) .
?? 1
=3-1=4,?? ?? =(-2-1)=-3;?? 1
=(4-(-2))=6
? Direction ratios are 4::-3::6
This is parallel to plane.
? 4?? -3?? +6?? =0
From (ii) and (iii)
?? -12
=
?? -20
=
?? -2
??? :?? :?? ::6:10:1
? Equation of plane is
6?? +10(?? -1)+(?? -1) =0
? 6?? +10?? +?? -11 =0
Equation of plane parallel to this plane and passing through line
6(?? +1)+10(?? -1)+1(?? +2)=0
6?? +10?? +?? -2=0
? distance between line and plane
= distance between two plane
=
|?? 1
-?? 2
|
v
??˜2
+?? 2
+?? 2
=
9
v6
2
+10
2
+?? 2
This must be on the plane for the intersection of this sphere to be a great circle.
? 5(
5?? -3
2
)-(4-2?? )+2(4?? -2)+7=0
? 45?? =17??? =
17
45
Page 4
Edurev123
3. Plane and its Properties
3.1 Find the equations of the straight line through the point (?? ,?? ,?? ) to intersect the
straight line
?? +?? =?? +?? =?? (?? -?? )
and parallel to the plane ?? ?? +?? +?? ?? =?? .
(2011 : 10 Marks)
Solution:
Let the required line intersects the given line
?? +?? =?? +1=2(?? -2) (??)
at (?? 1
,?? 1
,?? 1
)
? The equations of the required line passing through (2,1,2) and (?? 1
,?? 1
,?? 1
) are
?? -3
?? 1
-3
=
?? -1
?? 1
-1
=
?? -2
?? 1
-1
(???? )
The direction ratios of the line (ii) are ?? 1
-3,?? 1
-1,?? 1
-1.
Because the line (ii) is parallel to the plane 4?? +?? +5?? =0, therefore the normal to the
plane with direction ratios 4,1,5 is perpendicular to the line (ii).
? 4(?? 1
-3)+1(?? 1
-1)+5(?? 1
-1)=0
or 4?? 1
+?? 2
+5?? 2
-23=0 (?????? )
Because the point (?? 1
,?? 1
,?? 1
) lies on (i),
? ?? 1
+4=?? 1
+1=2(?? 1
-2)
? ?? 1
=2?? 1
-8,?? 1
=2?? 1
-5 (iv)
? from (iv),we get
4(2?? 1
-8)+2?? 1
-5+5?? 1
-23 =0
? ?? 1
=4
? from (iv), ?? 1
=0,?? 1
=3
? from (ii), the equations of the required line are :
?? -3
-3
=
?? -1
2
=
?? 2
-2
2
3.2 Find the equation of the plane which passes through the points (?? ,?? ,?? ) and
(?? ,?? ,-?? ) and is parallel to the line joining the points (-?? ,?? ,-?? ),(?? ,-?? ,?? ) . Find also
the distance between the line and the plane.
(2013 : 10 Marks)
Solution:
The general equation of plane through (0,1,1)
??(?? -0)+?? (?? -1)+?? (?? -1)=0
???? +?? (?? -1)+?? (?? -1)=0
(??)
(2,0,-1) lies on this plane.
? 2?? -?? -2?? =0
Direction ratios of line passing through (-1,1,-2),(3,-2,4) .
?? 1
=3-1=4,?? ?? =(-2-1)=-3;?? 1
=(4-(-2))=6
? Direction ratios are 4::-3::6
This is parallel to plane.
? 4?? -3?? +6?? =0
From (ii) and (iii)
?? -12
=
?? -20
=
?? -2
??? :?? :?? ::6:10:1
? Equation of plane is
6?? +10(?? -1)+(?? -1) =0
? 6?? +10?? +?? -11 =0
Equation of plane parallel to this plane and passing through line
6(?? +1)+10(?? -1)+1(?? +2)=0
6?? +10?? +?? -2=0
? distance between line and plane
= distance between two plane
=
|?? 1
-?? 2
|
v
??˜2
+?? 2
+?? 2
=
9
v6
2
+10
2
+?? 2
This must be on the plane for the intersection of this sphere to be a great circle.
? 5(
5?? -3
2
)-(4-2?? )+2(4?? -2)+7=0
? 45?? =17??? =
17
45
? The given sphere is
?? 2
+?? 2
+?? 2
-
10
9
?? +
146
45
?? -
22
45
?? -
106
45
=0
? 45(?? 2
+?? 2
+?? 2
)-50?? +146?? -22?? -106=0
3.3 Obtain the equation of the plane passing through the points (?? ,?? ,?? ) and
(?? ,-?? ,?? ) parallei to ?? -axis.
(2015: 6 Marks)
Solution:
The equation of any plane through (2,3,1) is
?? (?? -2)+?? (?? -3)+?? (?? -1)=0 (??)
It passes through (4,-5,3)
? ?? (4-2)+?? (-5-3)+?? (3-1)=0
i.e., ?? -4?? +?? =0 (???? )
If the plane (i) is parallel to ?? -axis, then it is perpendicular to ???? -plar 3, i.e., ?? =0, i.e.,
? 1?? +0?? +0?? =0
? from (ii), 1?? +0?? +0?? =0??? =0
?
?? 0
=
?? 1
=
?? 4
Hence, (i) becomes 0+1(?? -3)+4(?? -1)=0
?? +4?? -7=0
3.4 Find the surface generated by a line which intersects the lines ?? =?? =?? ,?? +
?? ?? =?? =?? +?? and parallel to the plane ?? +?? =?? .
(2016 : 10 Marks)
Solution:
Topic : Equation of a straight line intersecting two given lines.
Given lines are :
?? -?? =0=?? -?? (??)
?? +3?? -?? =0=?? +?? -?? (???? )
Hence, the equation of a line intersecting the given lines (i) and (ii) will be
Page 5
Edurev123
3. Plane and its Properties
3.1 Find the equations of the straight line through the point (?? ,?? ,?? ) to intersect the
straight line
?? +?? =?? +?? =?? (?? -?? )
and parallel to the plane ?? ?? +?? +?? ?? =?? .
(2011 : 10 Marks)
Solution:
Let the required line intersects the given line
?? +?? =?? +1=2(?? -2) (??)
at (?? 1
,?? 1
,?? 1
)
? The equations of the required line passing through (2,1,2) and (?? 1
,?? 1
,?? 1
) are
?? -3
?? 1
-3
=
?? -1
?? 1
-1
=
?? -2
?? 1
-1
(???? )
The direction ratios of the line (ii) are ?? 1
-3,?? 1
-1,?? 1
-1.
Because the line (ii) is parallel to the plane 4?? +?? +5?? =0, therefore the normal to the
plane with direction ratios 4,1,5 is perpendicular to the line (ii).
? 4(?? 1
-3)+1(?? 1
-1)+5(?? 1
-1)=0
or 4?? 1
+?? 2
+5?? 2
-23=0 (?????? )
Because the point (?? 1
,?? 1
,?? 1
) lies on (i),
? ?? 1
+4=?? 1
+1=2(?? 1
-2)
? ?? 1
=2?? 1
-8,?? 1
=2?? 1
-5 (iv)
? from (iv),we get
4(2?? 1
-8)+2?? 1
-5+5?? 1
-23 =0
? ?? 1
=4
? from (iv), ?? 1
=0,?? 1
=3
? from (ii), the equations of the required line are :
?? -3
-3
=
?? -1
2
=
?? 2
-2
2
3.2 Find the equation of the plane which passes through the points (?? ,?? ,?? ) and
(?? ,?? ,-?? ) and is parallel to the line joining the points (-?? ,?? ,-?? ),(?? ,-?? ,?? ) . Find also
the distance between the line and the plane.
(2013 : 10 Marks)
Solution:
The general equation of plane through (0,1,1)
??(?? -0)+?? (?? -1)+?? (?? -1)=0
???? +?? (?? -1)+?? (?? -1)=0
(??)
(2,0,-1) lies on this plane.
? 2?? -?? -2?? =0
Direction ratios of line passing through (-1,1,-2),(3,-2,4) .
?? 1
=3-1=4,?? ?? =(-2-1)=-3;?? 1
=(4-(-2))=6
? Direction ratios are 4::-3::6
This is parallel to plane.
? 4?? -3?? +6?? =0
From (ii) and (iii)
?? -12
=
?? -20
=
?? -2
??? :?? :?? ::6:10:1
? Equation of plane is
6?? +10(?? -1)+(?? -1) =0
? 6?? +10?? +?? -11 =0
Equation of plane parallel to this plane and passing through line
6(?? +1)+10(?? -1)+1(?? +2)=0
6?? +10?? +?? -2=0
? distance between line and plane
= distance between two plane
=
|?? 1
-?? 2
|
v
??˜2
+?? 2
+?? 2
=
9
v6
2
+10
2
+?? 2
This must be on the plane for the intersection of this sphere to be a great circle.
? 5(
5?? -3
2
)-(4-2?? )+2(4?? -2)+7=0
? 45?? =17??? =
17
45
? The given sphere is
?? 2
+?? 2
+?? 2
-
10
9
?? +
146
45
?? -
22
45
?? -
106
45
=0
? 45(?? 2
+?? 2
+?? 2
)-50?? +146?? -22?? -106=0
3.3 Obtain the equation of the plane passing through the points (?? ,?? ,?? ) and
(?? ,-?? ,?? ) parallei to ?? -axis.
(2015: 6 Marks)
Solution:
The equation of any plane through (2,3,1) is
?? (?? -2)+?? (?? -3)+?? (?? -1)=0 (??)
It passes through (4,-5,3)
? ?? (4-2)+?? (-5-3)+?? (3-1)=0
i.e., ?? -4?? +?? =0 (???? )
If the plane (i) is parallel to ?? -axis, then it is perpendicular to ???? -plar 3, i.e., ?? =0, i.e.,
? 1?? +0?? +0?? =0
? from (ii), 1?? +0?? +0?? =0??? =0
?
?? 0
=
?? 1
=
?? 4
Hence, (i) becomes 0+1(?? -3)+4(?? -1)=0
?? +4?? -7=0
3.4 Find the surface generated by a line which intersects the lines ?? =?? =?? ,?? +
?? ?? =?? =?? +?? and parallel to the plane ?? +?? =?? .
(2016 : 10 Marks)
Solution:
Topic : Equation of a straight line intersecting two given lines.
Given lines are :
?? -?? =0=?? -?? (??)
?? +3?? -?? =0=?? +?? -?? (???? )
Hence, the equation of a line intersecting the given lines (i) and (ii) will be
(?? -?? )+?? (?? -?? )=0 (??)*
and (?? +3?? -?? )+?? (?? +?? -?? )=0
? ?? +???? -(?? +???? )=0 (???? )*
and ?? +???? +(3+?? )?? -(?? +???? )=0 (?????? )
Line (iii) is parallel to the plane ?? +?? =0 (???? )
If direction ratio's of line (iii) are ??,?? ,?? , then
1
3+?? -????
=
?? ?? -0
=
?? 0-1
? (iv) ? 1·(3+?? -???? )+1·?? +0·(-1)=0
3+?? +?? -???? =0 (?? )
The required locus of the line is obtained by eliminating ?? and ?? between (??)
*
, (ii)
*
and
(?? ) .
3-
?? -?? ?? -?? -
?? +3?? -?? ?? +?? -?? -
?? -?? ?? -?? ·
?? +3?? -?? ?? +?? -?? =0
Solving and simplifying : (?? +?? )(?? +?? )=2?? (?? +?? )
3.5 Find the projection of straight line
?? -?? ?? =
?? -?? ?? =
?? +?? -?? on the plane ?? +?? +?? ?? =?? .
(2018: 10 marks)
Solution:
Given line is
?? -1
2
=
?? -1
3
=
?? +1
-1
=??
Let this line meets given plane at (2?? +1,3?? +1,-?? -1) .
The point lies on given plane, i.e.,
2?? +?? +3?? +1-2?? -?? =6
? ?? =2
? The point is (5,7,-3) .
Let equation of line of projection is
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FAQs on Plane and its Properties - Mathematics Optional Notes for UPSC
| 1. What are the basic properties of a plane in geometry? |  |
Ans. In geometry, a plane is a flat, two-dimensional surface that extends infinitely in all directions. The basic properties of a plane include having no thickness, being defined by at least three non-collinear points, and containing an infinite number of points.
| 2. How is a plane defined in terms of its equation? | |
Ans. A plane in three-dimensional space can be defined by an equation of the form Ax + By + Cz = D, where A, B, and C are the coefficients of the variables x, y, and z, respectively, and D is a constant.
| 3. What is the relationship between two planes in space? | |
Ans. Two planes in space can be parallel, intersecting at a line, or coinciding with each other. The relationship between two planes is determined by the angle between their normal vectors. If the normal vectors are parallel, the planes are either parallel or coincident.
| 4. How can the distance between a point and a plane be calculated? | |
Ans. The distance between a point and a plane can be calculated using the formula: distance = |Ax + By + Cz - D| / √(A^2 + B^2 + C^2), where (x, y, z) are the coordinates of the point, and A, B, C, and D are the coefficients of the plane equation.
| 5. How many points determine a unique plane in space? | |
Ans. In three-dimensional space, a unique plane is determined by at least three non-collinear points. If the points are collinear, they do not uniquely determine a plane, as they lie on the same line.
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