Class XII Mathematics Vector Algebra and 3D

1
VECTOR ALGEBRA & 3D Weightage 17 Marks
SYLLABUS:
1. VECTOR ALGEBRA
Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction
ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors),
position vector of a point, negative of a vector, components of a vector, addition of vectors,
multiplication of a vector by a scalar, position vector of a point dividing a line segment in a
given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross)
product of vectors. Scalar triple product of vectors
SUMMERY OF QUESTIONS (YEAR WISE)
YEAR VSA (1 Mark) SA (4 Marks) LA (6 Marks) Total Marks
2009 3 1 0 7
2010 2 1 0 6
2011 2 1 0 6
2012 2 1 0 6
2013 2 1 0 6
PREVIOUS YEARS QUESTIONS
2009
1. Find projection of a

on b

if . 8 2 6 3a b and b i j k   
   
1
2. Write a unit vector in the direction of 2 6 3a i j k  
 
1
3. Find the value of p for which 3 2 9 3a i j k and b i pj k     
     
are parallel vectors. 1
4. If a b c d  
  
and a c b d  
  
, show that a d

is parallel to b c
 
. 4
2010
1. Write a vector of magnitude 15 units in the direction of vector 2 2i j k 
 
1
2. What is the cosine of the angle which the vector 2 i j k 
 
makes with y-axis. 1
3. Find the position vector of the point R which divides joins of points P and Q whose position
vectors are 2a b

and 3a b

in 1:2 internally. Also show that P is the mid point of RQ. 4
2011
1. For what value of ‘a’ the vectors 2 3 4i j k 
 
and 6 8ai j k 
 
are collinear. 1
2. Find the direction cosines of the vector 2 5i j k  
 
. 1
3. Find the unit vector perpendicular to each of the vectors a b

and a b

where
3 2 2a i j k  
 
and 2 2b i j k  
  
4
2012
1. Write a vector of magnitude 15 units in the direction of vector. 1
2. What is the cosine of the angle which the vector 2 i j k 
 
makes with y-axis? 1
3. Find the position vector of the point R which divides joins of points P and Q whose position
vectors are 2a b

and 3a b

respectively externally in the ratio 1:2 externally.
Also show that P is the mid point of RQ. 4
2013

2
1. If a unit vector a

makes angles
3

with i ,
4

with j and an acute angle  with k

, find
the value of  . 1
2. If 2a xi j z k  
 
and 3b i j k  
  
are two equal vactors , then write value of x + y + z 1
3. If a

and b

are two vectors such that a b a 
 
, then prove that 2a b

is perpendicular
to vector b

4
Topic Wise Distribution of questions
Topic 2009 2010 2011 2012 2013
Introduction 2 6 2 6 1
Scalar Product 1 0 0 0 4
Vector Product 4 0 4 0 0
Triple Scalar Product ----------- ----------- ----------- ----------- -----------
Important results
1. For point A ( x, y, z), OA

= x i y j z k 
 
2. For points A  1 1 1, ,x y z and B 2 2 2, ,x y z ; AB

= 2 1x x i

+  2 1y y j

+  2 1z z k

3. If kcjbiar

 , then r

= 2 2 2
x y z 
4. Two vectors a

and b

are collinear if a

= k b

. Points A, B and C are collinear if AB

= k AC

5.
a
a
a



 ; a

= .a a
 
6.
nm
OAnOBm
OP


 (Internally) ;
nm
OAnOBm
OP


 (externally);
2
OAOB
OP

 (mid point)
7. For OP x i y j z k  
  
vector components are x i

, y j

and z k

& scalar components are x, y & z.
8. For a vector kcjbiar

 a, b, c are direction ratios.
For unit vector knjmilr

 l, m, n are direction cosines.
222
cbar 

and 122
 nml
9. For direction ratios a, b and c, direction cosines can be calculated as
2 2 2
a
l
a b c

 
,
2 2 2
b
m
a b c

 
and
2 2 2
c
n
a b c

 
DOT (SCALAR ) PRODUCT:
1) abba

..  ; Also cabacba

..).( 
2)
22
. aaaa 

; 1...  kkjjii

3) For two perpendicular vectors ;,ba

0. ba

; 0...  ikkjji

4) Projection of .a

on b

=
b
ba


.
; Projection vector of a

on b

= b
b
ba 


.
.
2










5 ) For vectors kcjbiabkcjbiaa

222111 &  ; 212121. ccbbaaba 

6) To find angle  between two vectors
2
2
2
2
2
2
2
1
2
1
2
1
212121.
cos
cbacba
ccbbaa
ba
ba


 

 ;
Condition for two vectors to be perpendicular 0212121  ccbbaa

3
Condition for two vectors to be parallel ba

. OR
2
1
2
1
2
1
c
c
b
b
a
a

7)   bababa

.2
2
22
 ;    22
. bababa


  accbbacbacba

.2.2.2
22
22

VECTOR (CROSS) PRODUCT
ba

 = a. b sin  n

, with  be the angle between the two vectors and n

the unit vector
perpendicular to both the vectors.
1) abbaabba

 ; Also cabacba

 )(
2) 0,0  kkjjiiaa

3) Magnitude of cross-product sin.. baba


4) Unit vector perpendicular to the plane containing vectors a

andb

,
ba
ba
n 





Vectors of magnitude ‘k’ perpendicular to a

andb

is given by
ba
ba
k 




5) For kcjbiabkcjbiaa

222111 &  ;
222
111
cba
cba
kji
ba



6) Geometrically vector product represents the area of the parallelogram whose sides are
represented by the two vectors i.e. Area of parallelogram with consecutive sides represented
by ba

& ; Area = ba


7) Area of a triangle with sides ba

& ; Area =
2
1
ba


8) Area of parallelogram with diagonals represented by 21 ,dd

; Area =
2
1
21 dd


9) Lagrange’s Identity  2222
.. bababa


SCALAR TRIPLE PRODUCT: For vectors cba

&, , scalar triple product is  cba

. OR  cba

. and
is denoted by [ a

b

c

]
Properties:
1. For a parallelepiped having its conterminous edges represented by vectors cba

&,
Volume of parallelepiped = [ a

b

c

]
2.Three vectors cba

&, are coplanar iff [ a

b

c

] = 0
3.Points A, B, C and D are coplanar if [ ADACAB ] = 0
4.For vectors kcjbiabkcjbiaa

222111 &  and kcjbiac

333 
[ a

b

c

] =
333
222
111
cba
cba
cba
5. [ a

b

c

] = - [b

ca

],   0aba

MLM

4
ONE MARKER
1. For the following points A, B find the vector AB

as well as their magnitudes
A (2, -1, 3), B (0, -5, 2)
2. Find the unit vector along the vectors 4 2i j k  
 
3. Find x so that  x i j k 
 
is a unit vector.
4. Find the vector along the direction of the vector 3 2 6i j k 
 
and of magnitude 5
5. Find the scalar and vector components of the vector AB

with A(1, 0, 1), B(2, -2, 4).
6. Find the direction ratios and direction cosines of the line passing through the points
(1, 2, 4) and (-5, 2, 3).
7. Find the scalar product of the vectors 3 5 2i j k and b i j k    
    
8. If , 2 2 4a i j k b i j k and c i j k        
        
find a unit vector parallel to the vector
2 3a b c 
 
.
9. Find the position vector of the point which divides joins of points 2 3a b

and 2a b

in 1:3 internally
and externally using vectors.
10. Find the position vector of the point dividing the join of A ( -1, 2, 4) , B ( 4, -3, 4) in (1) 1:3
internally (2) 1:3 externally using vectors.
11. Find  so that kjia

923  is (1) perpendicular (2) parallel to kjib

3 
12. Find a

and b

, if    . 8 8a b a b and a b   
    
13. For bafindkibkjia

2;3,23  .
14. Show that       0a b c b c a c a b        
       
15. Evaluate      . . .i j k j i k k i j    
       
16. Find the angle between the vectors ba

& if .a b a b 
  
.
17. Find the angle between the vectors kjibandkjia

84623 
FOUR MARKERS
18. Show that the points A (2, -1, 3), B (3, -5, 1) and C (-1, 11, 9) are collinear.
19. Show that points A, B and C with p.v. 3 4 4i j k 
 
. 2i j k 
 
and 3 5i j k 
 
forms a right triangle.
20. If , ,a b c
 
are mutually perpendicular unit vectors and. 0a b c  
 
, show that a b c 
 
3 .
21. If , ,a b c
 
are mutually perpendicular vectors of equal magnitude, show that , ,a b c
 
are equally inclined
to ( )a b c 
 
.
22. If 2 2 3 , 2 3a i j k b i j k and c i j        
       
, find  so that a b

is perpendicular to c

.

5
23. Find projection & projection vector of a

on b

for 5 3 4 , 6 8a i j k b i j k     
     
24. For 3 & 2 3a i j b i j k    
    
, express b

in the form of two vectors 1b

, 2b

such that 1b

is parallel to
a

and 2b

ia perpendicular to a

.
25. If ,7,5,3,0  cbacba

find the angle between ba

& .
26. For vectors , ,a b c
 
if 3, 2, 2,a b c  
 
and. 0a b c  
 
, Find . . .a b b c c a 
    
.
27. Find a vector of magnitude 9 and perpendicular to both the vectors .2&32 kikji


28. Find the area of (i) parallelogram (ii) triangle whose adjacent sides are represented by
2 .i j k and i j k   
    
29. Given bafindbaba

 ;12.&2,10 .
30. Let 4 2 , 3 2 7a i j k b i j k     
    
and 2 4c i j k  
 
. Find a vector d

which is
perpendicular to both &a b

and . 15c d 

.
31. If a b c d  
  
and a c b d  
  
, show that a d

is parallel to b c
 
.
32. Find the volume of the parallelepiped whose sides are given by
kjikjikji

357&375,573 
33. Show that points (6,-7,0) ,(16,-29,-4), ( 0,3,-6) and (2,5,10) are coplanar.
34. Find  so that points kjiandkjikji

5332,2   .are coplanar.
35. Show that    cbaaccbba

2
36. Show that   0 accbba

List of important questions and examples (NCERT)
EXAMPLES: 8,9, 11, 12, 14,15,19,20,26,28,29,30
Ex.10.2: 2,3,5,8,10,11,12,14,15,16,17
Ex. 10.3: 2,4,5,6,10,111,13,15,16
Ex.10.4: 2,3,5,7,9,10,12
MISC: 5,6,7,8,9,11,12,13,14,18,19

6
THREE – DIMENSION GEOMETRY
SYLLABUS
Direction cosines and direction ratios of a line joining two points. Cartesian and vector
equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian
and vector equation of a plane. Angle between (i) two lines, (ii) two planes. (iii) a line and
a plane. Distance of a point from a plane.
2009 ------- 1 1 10
2010 1 1 1 11
2011 1 1 1 11
2012 1 1 1 11
2013 1 1 1 11
2009
1. Find p so that lines are perpendicular
2
3
2
147
3
1 



 z
p
yx
;
5
6
1
5
3
77 zy
p
x 




4
2. Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of 6
the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
2010
1. Write vector equation of the line
5 4 6
3 7 2
x y z  
  1
2. Find the Cartesian equation of the plane passing through the points A(0,0,0) , B (3, -1, 2) and
parallel to line
4 3 1
1 4 7
x y z  
 

4
3. The points A ( 4, 5, 10) , B(2,3,4)and C (1,2,-1) are three vertices of a parallelogram ABCD.
Find the equations of the sides AB and BC and also find the co-ordinates of point D. 6
2011
1. Write the intercept cut off by the plane 2x + y – z = 5 on x-axis. 1
2. Find the angle between the following pair of lines
2 1 3
2 7 3
x y z   
 
 
2 2 8 5
1 4 4
x y z  
 

and check whether the lines are parallel or perpendicular. 4
3. Find the equation of plane which contains the lines of intersection of the planes
.( 2 3 ) 4 0r i j k   
 
; .(2 ) 5 0r i j k   
 
and which is perpendicular to the plane
.(5 3 6 ) 8 0r i j k   
 
6
2012
1. Write the vector equation of the following line:
5 4 6
3 7 2
x y z  
  1
2. Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, –1, 2) and
parallel to the line
4 3 1
1 4 7
x y z  
 

4

7
3. The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, –1) are three vertices of a parallelogram ABCD.
Find the vector equations of the sides AB and BC and also find the coordinates of point D. 6
2013
1. Find the Cartesian equation of the line which passes through the point (-2, 3, -5) and is parallel
to the line
3 4 8
3 5 6
x y z  
  1
2. Find the coordinates of the point, where the line
2 1 2
3 4 2
x y z  
  intersect the plane
x – y + z – 5 = 0. Also find the angle between the line and plane. 4
OR
Find the equation of plane which contains the lines of intersection of the planes
.( 2 3 ) 4 0r i j k   
 
; .(2 ) 5 0r i j k   
 
and which is perpendicular to the plane
.(5 3 6 ) 8 0r i j k   
 
3. Find the vector equation of the plane passing through the points with position vectors 2i j k 
 
;
2i j k 
 
and 2i j k 
 
. Also find the point of intersection of this plane and the line
3 (2 2 )r i j k i j k     
    
6
Topic 2009 2010 2011 2012 2013
St Line 4 7 4 7 5
Plane 6 4 7 4 6
St line & Plane ------------ ------------ ------------ ------------ ------------
Important results
STRAIGHT LINE
1. Line passing through point A (P.V. a

) and parallel to m

is mar


Cartesian form: A ( 111 ,, zyx ) & a, b, c be D. R. of vector parallel to line
c
zz
b
yy
a
xx 111 




2. A general point on the line is
P ( 111 ,, zcybxa   )
3. Line passing through two points A (P.V. a

) & B (P.V.b

) is  abar

 
Cartesian form: For points A ( 111 ,, zyx ) and B ( 222 ,, zyx )
12
1
12
1
12
1
zz
zz
yy
yy
xx
xx








4. For lines having angle  between them
111 mar 

, 222 mar 

OR
1
1
1
1
1
1
c
zz
b
yy
a
xx 




;
2
1
2
1
2
1
c
zz
b
yy
a
xx 




2
2
2
2
2
2
2
1
2
1
2
1
212121
cos
cbacba
ccbbaa


 =
mm
mm


1
21.
For perpendicular lines 0. 21212121  ccbbaamm

For parallel lines
2
1
2
1
2
1
21 ;
c
c
b
b
a
a
mm 


5. Skew lines in space are lines which are neither parallel nor intersecting.
6. Shortest distance between two skew lines.
For two lines 111 mar 

; 222 mar 


8
S. D. =
 2 1 1 2
1 2
( ).( )a a m m
m m
 

   
  OR S.D. =
 
2 1 2 1 2 1
1 1 1
2 2 2
2
1 2 2 1
x x y y z z
a b c
a b c
b c b c
  

Condition for two lines to intersect  2 1 1 2( ).( ) 0a a m m  
   
For parallel lines mar  11

; mar  22

S. D. = 2 1( )a a m
m
 
  

PLANE
1. General equation of plane is a x + b y +c z + d. = 0 Here a, b, c is the d. r. of normal
(perpendicular) to plane.
Vector Form 0.  dnr

, here n

is normal (perpendicular) to plane.
2. Plane through one point A ( 111 ,, zyx ) with p.v. a

and normal to vector kcjbia


(Direction ratios a, b, c)         0.;0111  narzzcyybxxa

OR nanr

.. 
3. Equation of plane through three points A  111, zyx ; B  222 , zyx ; C  333 ,, zyx
0
13121
13121
13121




zzzzzz
yyyyyy
xxxxxx
OR   0 acabar

4. Equation of plane having intercepts a, b, c on the axes 1
c
z
b
y
a
x
5. Plane perpendicular to unit vector n

and at a distance of p from origin
Cartesian Form: l x + m y + n z = p, here l, m, n are d. cosines of normal to plane.
6. Equation of plane through a point A (x 1, y 1, z1) and
(a) perpendicular to planes a 1x + b 1 y + c 1 z + d 1= 0 ; a 2x + b 2 y + c 2 z + d 2= 0 or
(b) parallel to two vectors with direction ratios kcjbia

111  kcjbia

222  or
(c) parallel to two lines having direction ratios a1, b1, c1 & a2, b2, c2
0
222
111
111


cba
cba
zzyyxx
7. Equation if plane through two points A (x 1, y1, z1) , B (x 2, y2, z 2) and
(a) perpendicular to plane a 1x + b 1 y + c 1 z + d 1= 0 OR (b) parallel to one vector kcjbia

 or
(c) parallel to line having D.R.’s a, b, c
0121212
111


cba
zzyyxx
zzyyxx
pnr 

.

9
8. Equation of plane that contains the lines
1
1
1
1
1
1
c
zz
b
yy
a
xx 




&
2
1
2
1
2
1
c
zz
b
yy
a
xx 




is 0
222
111
111


cba
cba
zzyyxx
Two lines
1
1
1
1
1
1
c
zz
b
yy
a
xx 




&
2
1
2
1
2
1
c
zz
b
yy
a
xx 




are coplanar if
0
222
111
121212


cba
cba
zzyyxx
9. Equation of plane that contains parallel lines
c
zz
b
yy
a
xx 111 




&
c
zz
b
yy
a
xx 111 




is 0121212
111


cba
zzyyxx
zzyyxx
10. Equation of plane through the line of intersection of planes
0:&0: 2222211111  dzcybxadzcybxa  is 021  
11. Distance of a point from a plane:
From point A (p.v. a

) From point A ( 111 ,, zyx )
and plane 0.  dnr

and the plane a x + b y + c z + d = 0
p =
n
dna


.
p =
222
111
cba
dzcybxa


12. Angle between the planes: If  be the angle between the planes
0&00.&0. 222211112211  dzcybxadzcybxaORdnrdnr

2
2
2
2
2
2
2
1
2
1
2
1
212121
cos
cbacba
ccbbaa


 =
nn
nn


1
21.
For perpendicular planes 0. 21212121  ccbbaann

;
For parallel planes
2
1
2
1
2
1
21 ;
c
c
b
b
a
a
nn 


LINE AND PLANE
1. Angle between line and plane: For line 0.&  dnrplanemar


OR For line
1
1
1
1
1
1
c
zz
b
yy
a
xx 




& Plane 02222  dzcybxa
2
2
2
2
2
2
2
1
2
1
2
1
212121.
cbacba
ccbbaa
mn
mn
Sin


 


For line to be parallel to plane 0. 212121  ccbbaamn

For line to be perpendicular to plane
2
1
2
1
2
1
;
c
c
b
b
a
a
mn 



10
2. Equation of line perpendicular to plane a x + by + c z + d = 0 and passing through the point
( x1, y1, z1) is
c
zz
b
yy
a
xx 111 




IMPORTANT QUESTIONS (THREE DIMENSIONS)
1. Find the equation of line in cartesian and vector form passing through point kji

432 
and parallel to kji

543  .
2. Find equation of line through (2, -1, 1) and parallel to the line whose equation is
3
2
7
1
2
3





 zyx
.
3. Find vector and Cartesian equation of line passing through the points (2, 3, 4) & ( -1, 2, 3)
4. Find the equation of x –axis.
5. Find the point where line
4
5
3
2
2
1 




 zyx
meets plane 2 x + 4 y – z = 3.
6. Shown that lines intersect  jikjir

 3)(  ;  kikir

32)4(  
7. Find the point of intersection of the lines  kjikjir

432)32(   ;
 kjijir

 25)4( 
8. Find the angle between the lines  kjikjir

2)32(   ;
 kjijir

453)9(  
9. Find p so that lines are perpendicular
2
3
2
147
3
1 



 z
p
yx
;
5
6
1
5
3
77 zy
p
x 




10. Find p so that lines are parallel  kjikjir

2)32(   ;  kpjijir

 33)9( 
11. Find the distance to the line
2
7
2
7
3
6





 zyx
from the point (1, 2, 3)
12. Find the image in the line  kjikjir

11410)8211(   of the point (2, -1, 5)
13. Find the shortest distance between the lines
 kjikjir

 )2( ;  kjikjir

22)2(  
14. Find the equation of plane in Cartesian form 010)23.(  kjir

15. Find the equation of the plane in vector form 2 x + 3 y – 10 = 0
16. Find the equation of plane passing through points (2, 2, -1), (3, 4, 2), (7, 0, 6)
17. Find the equation of plane passing through the point ( 1, 0, -2) and perpendicular to the planes
2x + y –z -2 = 0, x – y – z – 3 = 0
18. Find the plane passing through the point (2, 3, 4) and parallel to the vectors kji

3 and
kji

232  .

11
19. Find the plane passing through the point (2, 1, 4) and parallel to the lines
4
6
3
1
2
1 



 zyx
and
4
7
1
6
3
2 




 zyx
.
20. Find the equation of plane passing through the points ( -1, 2, 3), (2, 3, 7) and perpendicular to the
plane 3x – 2 y + 2 z + 10 = 0
21. Find the equation of plane at a distance of 3 from origin and normal to  kji

 43 .
22. Find the plane passing through the point (1,2,3) and normal to vector kji

3 .
23. Find the plane through (2, -1, 3) and parallel to plane 3x – y + 4 z + 6 = 0.
24. The foot of perpendicular from origin to a plane is (2, 5, 7). Find the equation of plane.
25. Find the direction cosines of normal to the plane 3 x – 6 y + 2 z = 7.
26. Find the equation of plane having intercepts 1, 2 – 3 on the axes.
27. Find ‘p’ so that planes are perpendicular 013)32.(  kjir

; 09)72.(  kjipr

28. Find ‘p’ so that planes are parallel 010)323.(  kpjipr

; 05).(  kjir

29. Find the angle between the line  kjikjir

 22)432(  and plane
010)236.(  kjir

30. Find ‘m’ so that line  kjikjir

22)2(   is parallel plane 012)23.(  kmjir

31. Show that line
1
3
12
1 

 zyx
is lying in the plane x – y – z = 4.
32. Find the distance of the point kji

42  from the plane 9)1243.(  kjir

.
33. Find the distance between the planes 2 x – y + 3 z + 4 = 0; 6 x – 3 y + 4 z – 3 = 0
34. Find p so that distance of point (1,2, p) from plane x + y + z – 10 = 0 is 32 .
35. Find the plane containing the lines  jikjir

 3)(  ;  kikir

32)4(  
36. Show that lines are coplanar  jiikjr

32)32(   ;  kjikjir

432)362(  
37. (a) Find the equation of plane through the line of intersection of planes 2 x – 7 y + 4 z = 3 and
3 x – 5 y + 4 z + 11 = 0 and the point (-2, 1, 3)
(b) Find the equation of plane through the line of intersection of planes x + 2 y + 3 z = 4 and
2 x + y - z + 5 = 0 and perpendicular to the plane 5 x + 3 y – 6 z + 8 = 0.
(c) Find the equation of plane through intersection of planes 04).(  jir

;
01)432.(  kjir

and parallel to the line  kjikjir

 2)3( 
38. Find the length and co-ordinates of foot of perpendicular from the point (7, 14, 5) to the plane
2 x + 4 y – z = 2
EXAMPLES: 4,7,14,16,17,19,21,22,25,26,27,28,30
Ex.11.: 1,3 Ex. 11.2: 2,3,4,6,8,10,11,12,15,17 Ex.11.3:2,4(a),5.(b),6(b),10,11,12,13(c),(e),14(b),(c)
MISC: 3,6,7,9,10,12,13,14,15,16,17,18,19,20,21,22

12
LINEAR PROGRAMMING Weightage 6 Marks
SYLLABUS
Introduction, related terminology such as constraints, objective function, optimization,
different types of linear programming (L.P.) problems, mathematical formulation of L.P.
problems, graphical method of solution for problems in two variables, feasible and infeasible
regions, feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial
constraints)
2009 0 0 1 6
2010 0 0 1 6
2011 0 0 1 6
2012 0 0 1 6
2013 0 0 1 6
2009
1. A diet is to contain at least 80 units of Vitamin A and 100 units of minerals. Two foods F1 and F2 are
available. Food F1 costs Rs. 4 per unit and food F2 costs Rs. 6 per unit. One unit of food F1 constrains 3
units of Vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of Vitamin A and 3 units
of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for
the diet that consists of mixture of these foods and also meets the minimum nutritional requirements.
2010
1. One kind of cake requires 300 g of floor and 15 g of fat, another kind of cake requires 150 g of floor and
30 g of fat. Find the maximum number of cakes which can be made from 7.5 kg of floor and 600 g of fat,
assuming that there is no shortage of ingredients used in making the cakes. Make it LPP and solve ot
graphically.
2011
1. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time
and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine time
and 1 hour of craftsman’s time. In a day, the factory had the availability of not more than 42 hours
of machine time and 24 hours of craftsman’s time. If the profit on a racket and bat is Rs 20 and Rs 10
respectively, find the number of tennis rackets and cricket bats that factory must manufacture to
earn the maximum profit. Make it as an L.P.P. and solve it graphically.
2012
1. A small firm manufactures gold rings and chains. The total number of rings and chains manufactured
per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum
number of hours available per day is 16. If the profit on a ring is Rs. 300 and that on a chain is
Rs. 190, find the number of rings and chains that should be manufactured per day, so as to earn the
maximum profit. Make it as an L.P.P. and solve it graphically.
2013
1. A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from
crops X and Y per hectare are estimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a
liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further,
no more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond

13
which collects drainage from this land. How much land should be allocated to each crop so as to
maximise the total profit of the society?
Topic 2009 2010 2011 2012 2013
Manufacturing Problem ------------ ------------ 6 6 ------------
Diet Problem 6 ------------ ------------ ------------ ------------
Allocation Problems ------------ 6 ------------ ------------ 6
EXAMPLES: 3,6,7,10,11
Ex.12.1, 8
Ex. 12.2: 2,4,5,6,7,8,9,10
Ex.15.3: 2,3,4,6,7,8,9
MISC: 5,7,8,10

14
PROBABILITY Weightage 10 Marks
Conditional probability, multiplication theorem on probability. independent events, total
probability, Baye's theorem, Random variable and its probability distribution, mean and
variance of random variable. Repeated independent (Bernoulli) trials and Binomial
distribution.
2009 0 1 1 10
2010 0 1 1 10
2011 0 1 1 10
2012 0 1 1 10
2013 0 1 1 10
2009
1. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining
the third six in the sixth throw of the die. 4
2. Three bags contains balls as shown in the table below
Bag Number of white balls Number of Black balls Number of Red balls
I 1 2 3
II 2 1 1
III 4 3 2
A bag is chosen at random and two balls are drawn from it. They happen to be white and red.
What is the probability that they are from third bag. 6
2010
1. On a multiple choice examination with three possible answers (out of which only one is correct)
for each of five questions, what is the probability that a candidate would get four or more correct
answers just by guessing? 4
2. One card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are
drawn at random and are found to be both clubs. Find the probability that lost card being of clubs. 6
OR
Out of a lot of 10 bulbs which includes 3 defectives, a sample of 2 bulbs is drawn at random.
Find the probability of the number of defective bulbs. 6
2011
1. Probabilities of solving a specific problem independently by A and B are
1
2
and
1
3
respectively.
If both try solve the problem independently, find the probability that (i) the problem is solved
(ii) exactly one of them solve the problem. 4
2. Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at
random. What is the probability of this person being male? Assume that there are equal number of
males and females. 6
2012
1. A family has 2 children. Find the probability that both are boys, if it is known that
(i) at least one of the children is a boy (ii) the elder child is a boy. 4
2. A bag contains 4 balls. Two balls are drawn at random, and are found to be white.
What is the probability that all balls are white? 6

15
2013
1. A speaks truth in 60% of cases, while B in 90% of the cases. In what percentage of cases are they
likely to contradict each other in stating the same fact? In the cases of contradiction do you think,
the statement of B will carry more weight as he speaks truth in more number of cases than A? 4
2. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a
meditation and yoga course reduce the risk of heart attack by30% and prescription of certain drug
reduces its chances by 25%. At a time a patient can choose any one of the two options with equal
probabilities. It is given that after going through one of the two options the patient selected at
random suffers a heart attack. Find the probability that the patient followed a course of meditation
and yoga? 6
Topic 2009 2010 2011 2012 2013
Conditional Probability -------- -------- -------- 4 --------
Multiplication Th. , Independent Events -------- -------- 4 -------- 4
Bays’ Th. 6 6 6 6 6
Probability Distribution -------- -------- -------- -------- --------
Binomial distribution 4 4 -------- -------- --------
Important Results
Bayes’ theorem:- For the sample space S associated with a random experiment . Let nEEE ,,21, 
are mutually exclusive and exhaustive events associated with a sample space S .If A is a
event of S then
   

 n
i
iI
iI
i
EAPEP
EAPEP
AEP
1
)/().(
/.
)/(
2. Conditional Probability: If A and B are two events of a sample space S associated with a random
experiment, then probability of occurrence of A provided B has already occurred is called conditional
probability and is denoted by P(A/B) and its value is given by
)(
)(
BP
BAP 
.
3. Probability distribution:-If a random variable X takes values nxxx .,,........., 21 with respective
probabilities nppp .....,,........., 21 then following pattern is called probability distribution
X : nxxxx ............................321
P(X) : npppp .........................321
Mean E(X) = 
n
i
ii xp
1
. Variance V(X) =  
22
( ) ( )E X E X S.D. = Variance
4. Binomial distribution:- if an experiment is repeated finite number of times ,events associated are
independent and probability of success or failure is constant for all trials then probability of random
variable X having value r is given by rrn
r
n
pqCrXP .)( 
 .Here n: Total number of trials of
experiments.
.

16
Problems involving Dice (Up to 2 tosses) or Coins (Up to 3 tosses)
Form Sample Space and use the result
( )
( )
( )
n E
P E
n S

Q: A coin is tossed two times. Find the probability of getting at least
one tail.
Sol: S ={HH, HT, TH, TT} E: Getting at least one tail = {HT, TH, TT}
( ) 3
( )
( ) 4
n E
P E
n S
 
Problems involving Cards,
Marbles, Balls, Boys/ Girls etc
Only one object is selected
Use
( ) .
( )
( ) .
n E no of favourable objects
P E
n S Total no of objects
 
Q: If one card is selected from a deck of playing cards,
find the probability that it is a red king.
Sol.
2
( )
52
noof red king cards
P red king
total noof cards
 
1
26

Note: Can use results of algebra of events.
1. ( ) ( ) ( ) ( )P E or F P E P F P E and F  
2. /
( ) 1 ( )P E P E 
3. / / /
( ) ( ) 1 ( )P E F P E F P E F     
More than only one object is selected
Objects are selected one by one
with Replacement
Events are independent. Use individual cases
with rotation or use Binomial Distribution
Q: If two balls are selected with replacement
from a box having 2 red, 5 green and 3 black
balls, find the probability of getting one black
balls and one red balls.
Sol. ( ) ( ) ( )P BRorRB P BR P RB 
( ). ( ) ( ). ( )P B P R P R P B 
3 2 2 3
. .
10 10 10 10
 
12 3
100 25
 
Objects are selected at random
without Replacement
Events are dependent. Use combination(C)
method or conditional probability
Q: If two balls are selected from a box having 2
red, 5 green and 3 black balls, find the probability
of getting one black ball and one red ball.
Sol. ( & )P onered oneblack ball 
3 2
1 1
10
2
.C C
C
 =
3 .2 2
45 15
 
PROBABILITY
PROBLEM SOLVING ALGORITHM
EXAMPLES: 2,3,4,6,7,9,11,18,19,20,21,22,24,26,28,29,32,33,34,35,36
Ex.13.1: 3,5,7,8,9,10,13,14 Ex. 13.2:,3,5,7,8,9,11,13,14,16 Ex.13.3: 1,3,4,5,6,7,9,10,12,13
Ex.13.4: ,4,5,6,7,11,13,15 Ex.13.5: 2,3,5,6,7,8,12,13 MISC: 3,5,6,7,8,10,13,16

Class XII Mathematics Vector Algebra and 3D

Recommended

Recommended

More Related Content

More from FellowBuddy.com

More from FellowBuddy.com (20)

Recently uploaded

Recently uploaded (20)

Class XII Mathematics Vector Algebra and 3D