MATHEMATICS
II PUC
VECTOR ALGEBRA
QUESTIONS & ANSWER
I
One Mark Question
1) Find the unit vector in the direction of 2i + 3j + k.
r
Let a = 2i + 3j + k
r
a = 2 2 + 32 + 12 =
r
rˆ a 2i + 3 j + k
\a = r =
a
14
4 + 9 + 1 = 14
r
2
3
1
\aˆ =
i+
j+
k
14
14
14
r
r
r r
r r
2) Let a = i + 2j & b = 2i + j. If a = b . Are the vectors a & b equal?
r
a = a 2 + 22 = 5,
r r
\a=b
r
b = 22 + a 2 = 5
But vectors are not equal since the corresponding components are distinct i.e. directions
are different.
3) Find the values of x & y so that vectors 2i+3j and xi+4j are equal.
r
r
a = 2i + 3 j
b = xi + yj
r r
Given a = b \2i + 3 j = xi + yi
\ x = 2, y = 3
4) Find the scalar or dot product of vectors i + 2j - 3k & 2i - j + k.
(i + 2j - 3k ) . (2i - j + k ) = 1(2 ) + 2(- 1) - 3(1) = 2 - 2 - 3 = -3
1
i- j
is a unit vector.
2
r i- j
1
1
a=
=
ij
2
2
2
5) Show that
2
2
r
1 1
æ 1 ö æ 1 ö
a= ç
+ = 1
÷ +ç÷ =
2 2
2ø
è 2ø è
r
r
\ a =1
\ a is a unit vector.
II
Two Marks Questions:
1) Find the vector parallel to the vector i - 2j and has magnitude 10 units.
r
Let a = i - 2 j
r
2
a = 12 + (- 2)
r
\a = 5
r
rˆ a i - 2 j
\a = r =
a
5
r
r
Let b be the vectors parallel to a having magnitude 10 units
r
\ b = 10
Q If vectors are parallel then unit
r r r
vector along & parallel vectors are
Now b = b aˆ
same
i-2j
=10 .
5
r 10i 20 j
\Reqd vector b =
5
5
r
2) Find the direction ratios and direction cosines of the vector a = i + j - 2k .
r
a = i + j - 2k
r
2
a = 12 + 12 + (- 2 )
r
\a = 6
r
rˆ a i + j - 2k
a= r =
a
6
r 1
1
2
\aˆ =
i+
jk
6
6
6
2
r
Here direction ratios are components of a i.e.(1, 1, - 2 )
r
æ 1 1 -2
direction cosines are components of â i.e. çç
,
,
è 6 6 6
ö
÷÷
ø
3) Show the vectors 2i-3j+4k and -4i+6j-8k are collinear.
r
a = 2i - 3 j + 4k
r
r
b = -4i + 6 j - 8k = - 2(2i - 3 j + 4k ) = -2a
\One vector can be expressed interms of another
r r
\a & b are collinear.
r r r r r
r
4) Find x , if for a unit vector a, (x - a ) . (x + a ) = 12
r
Given a = 1
(xr - ar ) . (xr + ar ) = 12
r2 r2
x - a = 12
r2
x - 12 = 12
r2
x = 13
r
\ x = 13
r
r
r r
r
r
r r
5) Find a - b , if two vectors a and b are such that a = 2, b = 3 and a .b = 4
r
r
r r
Given a = 2, b = 3 and a .b = 4
r r2 r2
r r r2
w.k .t. a - b = a - 2 a . b + b
( )
= 2 2 - 2(4 ) + 32 = 4 - 8 + 9
r r2
a -b = 5
r r
\ a -b = 5
3
r
r
r
r r r
6) For any two vectors a and b prove that a · b £ a · b
r r
a .b
w.k .t. cos q = r r
a b
r r
a. b
for all values of q , - 1 £ cosq £ 1
r r = cosq
a b
\ cosq £ 1
r r
a. b
\ r r £1
a b
r r
r r
\ a. b £ a b
r
r
r
r r r
7) For any two vectors a and b prove that a + b £ a + b (Triangle in equality )
r r2 r
r r r2
a + b = a 2 + 2 a. b + b
r
r r r2
rr r r
£ a 2 + 2 a. b + b
Qa.b £ a . b
r
r r r2
r r r r
£ a 2+ 2 a b + b
From previous properties a . b £ a b
r r 2
£ a+b
r r r r
\ a +b £ a + b
( )
{
}
( )(
)(
)
)
r r r r
8) Evaluate 3a - 5b . 2a + 7b
r r
r r
r r r r
r r
r r
3a - 5b . 2a + 7b = 6 a . a + 21 a . b - 10 b . a - 35 b . b
r
r
r r
r r
= 6a 2 + 21 a . b - 10 a . b - 35b 2
r
r
r r
= 6a 2 + 11a . b - 35b 2
r
r2
r r
= 6 a + 11a . b - 35b 2
(
( ) ( ) ( )
( ) ( )
r
r
r r
r r
9) If a = i - 7j + 7k & b = 3i - 2j + 2k find a ´ b and a ´ b
i
j k
r r
a´b = 1 - 7 7
3 -2 2
= i{- 7(2) - (- 2)(7 )}- j{2 - 21} + k {- 2 + 21}
= i (- 14 + 14) - j{- 19}+ k (19)
r r
\a ´ b = 19 j + 19k
4
r
10) Find l & m if (2i + 6j + 27k )´ (i + lj + mk ) = o
r
Given (2i + 6j + 27k )´ (i + lj + mk ) = o
i j k
r
\ 2 6 27 = o
1 l m
r
i(6 m - 27l ) - j (2 m - 27 ) + k (2l - 6) = o
Equating cofficient s
6m - 27l = 0,
2 m - 27 = 0
2l - 6 = 0
27
\m =
and l = 3
2
r r
r r
r r
11) Show that ( a - b ) ´ ( a + b ) = 2( a ´ b )
r r
r r
Consider LHS ( a - b ) ´ ( a + b )
r r r r r r r r
= a ´ a + a ´b - b ´ a - b ´b
r r r
r r r
= o + (a ´ b ) + ( a ´ b ) - o
r r
r r
r r
\ ( a - b ) ´ ( a + b ) = 2( a ´ b )
12) Find the scalar triple product of vectors i + 2j + 3k, - i - j + k and i + j + k
1 2 3
Scalar triple product = - 1 - 1 1
1 1 1
= 1 (- 1 - 1) - 2 (- 1 - 1) + 3(- 1 + 1)
=-2+4+0 = 2
13) Find l if the vectors i + j + 2k, l i - j + k & 3i - 2j - k are coplanar.
Given that vectors are coplanar
1
\ l
3
1
2
-1 1 = 0
- 2 -1
= 1(1 + 2) - 1(- l - 3) + 2(- 2l + 3) = 0
= 3 + l + 3 - 4l + 6 = 0
= -3l + 12 = 0
\l = 4
5
III
Three Marks Questions:
r
r r
r
1) Consider t he points P and Q with position v ectors OP = 3a - 2b and OQ = a + b .
Find the position vector of a point R which divides line joining
the points P and Q in the ratio 2 : 1 internally and externally respectively.
Solution
r
r
r r
given OP = 3a - 2b , OQ = a + b
m : n = 2 :1
Internally,
mOQ + nOP
m+n
r
r r
r
2 a + b + 1 3a - 2b
=
2 +1
r
r
2a + 3a + 2b - 2b
=
2 +1
r
5a
OR =
3
OR =
(
) (
)
mOQ - nOP
m-n
r
r r
r
2 a + b - 1 3a - 2b
=
2 -1
r
r
2a - 3a + 2b + 2b
=
1
r r
OR = 4b - a
externally OR =
(
) (
)
3) Find the vector joining the points P(2, 3, 0) and Q(- 1, - 2, 4) and also direction cosines of PQ
6
Given P º (2, 3, 0)
Q º (- 1, - 2, 4 )
Let i, j, k be unit vectors along axes
then OP = 2i + 3 j
OQ = -i - 2 j + 4k
PQ = OQ - OP
= -i - 2 j + 4k - (2i + 3 j )
= - i - 2i + 4 j - 2i - 3 j
\PQ = -3i - 5 j + 4k
(- 3)2 + (- 5)2 + (4 )2
PQ =
= 9 + 25 + 16 = 50
\ PQ = 5 2
^
PQ =
PQ
PQ
=
- 3i - 5 j + 4k
5 2
-3
5/
4
ij+
k
5 2 5/ 2
5 2
1
4 ö
æ -3
\ direction cosines are ç
,
,
÷
2 5 2 ø
è5 2
^
\PQ =
4) Show that points A(1, 2, 7) , B(2, 6,3) & C(3, 10, -1) are collinear
OR
Show that the points with position vectors i+2j+7k, 2i+6j+3k and 3i+10j-k are
collinear.
OA = (1, 2, 7 )
OB = (2, 6, 3) OC = (3,10, - 1)
AB = OB - OA = (2 - 1, 6 - 2, 3 - 7 ) = (1, 4, - 4 )
BC = OC - OB = (3,10, - 1) - (2, 6, 3) = (1, 4, - 4 )
AC = OC - OA = (3,10, - 1) - (1, 2, 7 ) = (2, 8, - 8)
AB = 33
BC = 33
AC = 132
\ AC = 2 33
Here AB + BC = AC
\ Collinear condition is satisfied.
\A, B, C are collinear.
7
5) Find the angle between the vectors i - 2j + 3k & 3i - 2j + k.
r
r
Let a = i - 2j + 3k
b = 3i - 2j + k
r r
a . b = (i - 2j + 3k ) . (3i - 2j + k ) = 3 + 4 + 3 = 10
r
2
a = 12 + (- 2 ) + 32 = 14
r
b = 32 + 22 + 12
= 14
r r
r r
a.b
10
10
Let q be angle between a & b then cosq = r r =
=
14 14 14
a b
cosq =
5
7
q = cos-1
7
5
1
(2i + 3 j + 6k ), 1 (3i - 6 j + 2k ) and 1 (6i + 2 j + 3k )
7
7
7
are mutually perpendicu lar.
r 1
Let a = (2i + 3 j + 6k )
7
r 1
b = (3i - 6 j + 2k )
7
r 1
c = (6i + 2 j + 3k )
7
r r 1
1
consider a . b = (2i + 3 j + 6k ) . (3i - 6 j + 2k )
7
7
1
= {(2i + 3 j + 6k ) . (3i - 6 j + 2k )}
49
1
1
= {6 - 18 + 12} =
{0}
49
49
r
r r
r
\a . b = 0
\ a is perpendicu lar to b
r r
r r
/// ly We can show that b . c = 0 & c . a = 0
rr r
\ a, b, c are mutually perpendicular vectors.
6) Show that the vectors
8
r
r
r r r r
7) If a = 5i - j - 3k , b = i + 3 j - 5k then show that the vectors a + b & a - b are perpendicu lar.
r r
a + b = 5i - j - 3k + i + 3 j - 5k = 6i + 2 j - 8k
r r
a - b = 5i - j - 3k - (i + 3 j - 5k ) = 4i - 4 j + 2k
r r
r r
consider a + b . a - b = (6i + 2 j - 8k ). (4i - 4 j + 2k ) = 24 - 8 - 16 = 0
r r
r r
r r
r r
\ a+b . a-b =0
\ a + b is perpendicu lar to a - b
(
) (
) ( )
(
)
r
r
r
r
r
8) If a = 2i + 2 j + 3k , b = -i + 2 j + k and c = 3i + j and such that a + lb is
r
perpendicu lar to c then find l .
r
r
a + lb = 2i + 2 j + 3k + l (- i + 2 j + k )
r
r
\ a + lb = (2 - l )i + (2 + 2l ) j + (3 + l )k
r
r
r
Given a + lb is ^ r c.
r r
r
a + lb . c = 0
{(2 - l )i + (2 + 2l ) j + (3 + l )k } . {3i + j} = 0
6 - 3l + 2 + 2l = 0 Þ -l + 8 = 0
\l = 8
(
)
r r r
r r r r
r r rr r r
9) If a , b , c are unit vectors such that a + b + c = o then find a . b + b . c + c . a
r
r
r
a = 1, b = 1, c = 1
r r r r rr
a . a + b + c = a. o
r r r r r
a 2 + a .b + a . c = 0
r r rr
r
a . b + a .c = - a 2
r r r r
r
/// ly b . c + c . a = -b 2
r r r r
r
c . a + a . b = -c 2
rr rr rr
r r r
Adding 2 a.b + b .c + c .a = - a 2 - b 2 - c 2 = -12 - 12 - 12
r r r r r r
2 a . b + b . c + c . a = -3
rr rr rr
Q a.b + b .c + c .a = - 3 2
(
(
)
(
)
)
9
r r
r r
10) Find a vector and unit vector perpendicu lar to each of the vector a + b and a - b
r
r
where a = 3i + 2j + 2k & b = i + 2j - 2k
r r
a + b = 3i + 2j + 2k + i + 2j - 2k
r r
\ a + b = 4i + 4 j
r r
a - b = 3i + 2j + 2k - i - 2j + 2k
r r
\ a - b = 2i + 4k
(
) (
)
(
) (
r
r r
r r
r r r r r
Let c be the vector perpendicu lar to a + b and a - b then c = a + b ´ a - b
+ - +
i j k
r
c = 4 4 0 = i (16 - 0 ) - j (16 - 0 ) + k (0 - 8)
2 0 4
r
\ c = 16i - 16 j - 8k
r
2
c = 16 2 + 16 2 + (- 8) = 256 + 256 + 64 = 576 = 24
r
r r
r r
Let ĉ be the unit vector perpendicu lar to a + b and a - b
r
r c
then ĉ = r
c
(
)
r 16i -16j - 8k 2i 2 j k
ĉ =
= 3 3 3
24
10
)
r
11) If a unit vector a makes angles p 3 with i, p 4 with j and and acute angle
r
q with k then find q and hence components of a.
r
Let a , b , g be the angles made by a with i, j, k then
a = p 3 , b = p 4, g = q
r
Let a = a1i + a2 j + a3k
r
Given a = 1
then cosa =
a1
a1
cos p 3 =
a1
1
\a1 = 1 2
a
cos b = r2
a
cos p 4 =
\a2 = 1
a2
1
2
a
cos g = r3
a
cosq = a3
a3 = cosq
r 1
1
\a = i +
j + cosq k
2
2
2
2
r
æ1ö æ 1 ö
2
a = ç ÷ +ç
÷ + (cosq )
è2ø è 2 ø
1 1
+ + cos 2 q
4 2
3
\1 = + cos 2 q
4
2
cos q = 1 4
cosq = ± 1 2
1=
q = 600 or 1200
r 1
1
1
\a = i +
j+ k
2
2
2
or
r 1
1
1
a = i+
j- k
2
2
2
11
12) Find the area of triangle with vertices A(1, 1, 2), B(2, 3, 5), C (1, 5, 5)
A º (1, 1, 2)
OA = i + j + 2k
B º (2, 3, 5)
OB = 2i + 3 j + 5k
C º (1, 5, 5)
OC = i + 5 j + 5k
AB = OB - OA
= 2i + 3 j + 5k - i - j - 2k
AB = i + 2 j + 3k
AC = OC - OA
= i + 5 j + 5k - i - j - 2k
AC = 4 j + 3k
i j k
AB ´ AC = 1 2 3
0 4 3
= i(6 - 12 ) - j (3 - 0) + k (4 - 0) = -6i - 3 j + 4k
AB ´ AC =
(- 6 )2 + (- 3)2 + 4 2
Area of triangle ABC =
= 36 + 9 + 16 = 51
1
1
AB ´ AC =
51 sq units
2
2
12
13) Find the area of parallelog ram whose adjecent sides are 2i - 4j + 5k and
i - 2j - 3k. Also find unit vector parallel to its diagonal.
r
r
a = 2i - 4j + 5k
b = i - 2j - 3k
i
j
r r
a´b = 2 -4
k
5 = i{12 + 10} - j{- 6 - 5}+ k {- 4 + 4 } = 22i + 11 j
1 -2 -3
r r
a´b =
(22)2 + (11)2 + 0 =
242 + 121 + 0 = 363
r r
Area of parallelog ram ABCD = a ´ b = 363 sq.units
r r
Diagonal AC = a + b
= 2i - 4j + 5k + i - 2j - 3k
AC = 3i - 6 j + 2k
AC = 32 + (- 6 ) + 2 2 = 9 + 36 + 4 = 49 = 7
2
^
Unit vector along AC,
AC =
AC
AC
=
3i - 6 j + 2k
7
Unit vector parallel to AC = Unit vector along AC =
3i - 6 j + 2k
7
r
r
14) Show that the points A, B, C with position v ectors a = 3i - 4j - 4k, b = 2i - j + k
r
& c = i - 3j - 5k respectively form the vertices of a right angled triangle.
Given A º (3, - 4, - 4) B º (2, - 1,1)
C º (1, - 3, - 5)
Given OA = 3i - 4 j - 4k
13
OB = 2i - j + k
OC = i - 3 j - 5k
AB = OB - OA
= 2i - j + k - 3i + 4 j + 4k
\ AB = -i + 3 j + 5k
BC = OC - OB
= i - 3 j - 5k - 2i + j - k
\BC = -i - 2 j - 6k
CA = OA - OC
= 3i - 4 j - 4k - i + 3 j + 5k = 2i - j + k
AB + BC + CA = -i + 3 j + 5k - i - 2 j - 6k + 2i - j + k
r
\ AB + BC + CA = O
\Dle law is satisfied
r r r
\ a , b , c form a Dle
AB =
(- 1)2 + 32 + 5 2
BC =
(- 1)2 + (2)2 + 6 2
= 35
= 41
2
AB = 35
2
BC = 41
CA = 2 2 + (- 1) + (- 1) = 6
2
2
We can see that AB + AC
2
\ AB + AC
2
2
CA = 6
2
2
= BC
= 35 + 6 = 41 = BC
2
\Pythagrous theorem is satified.
r r r
i.e. a, b, c form a right angled triangle
14
2
[
] [ ]
) {( )
}
){
}
){
}
) ( )
( ) (
)
) ( ) ( ) [ ]
r r r r r r
rrr
14 ) Prove that a + b , b + c , c + a = 2 a b c
r r
r r
r r
LHS = a + b . b + c ´ (c + a )
r r r r r r r r r r
= a + b . b´c + b´a +c´c + c´a
r r r r r r r r
= a + b . b´c- a´b+ c´a
r r r
r r r
r r r
r r r
r r r
r r r
= a . b ´ c - a . a ´ b + a .(c ´ a ) + b b ´ c - b a ´ b + b .(c ´ a )
r r r
r r r
= a . b ´ c - 0 + 0 + 0 - 0 + b .(c ´ a )
r r r
r r r
r r r
r r r
= a. b ´ c + a b ´ c = 2a. b ´ c = 2 a , b , c
(
(
(
(
(
(
)
14) Show that the vectors 4i - j + k, 3i - 2j - k and i + j + 2k are coplanar
r
Let a = 4i - j + k
r
b = 3i - 2j - k
r
c = i + j + 2k
4 -1 1
r r r
consider a. b ´ c = 3 - 2 - 1
(
)
1
1
2
= 4{- 4 + 1} + 1{ 6 + 1 }+ 1{3 + 2}
= -12 + 7 + 5
r
r
r
\a. b ´ c = 0
\vectors are coplanar
(
)
Scalar Triple Product of vectors
(
) (
)
r r r
r r r
r r r
If a, b, c are non zero vectors then a. b ´ c or a ´ b .c
is called scalar triple product of vectors.
r r r
r r r
r r r
The scalar triple product of a, b, c is denoted by a b c or a, b, c
[
a1
r r r
Then a. b ´ c = b1
c1
(
)
a2
b2
c2
] [
a3
b3
c3
since scalar triple product is a determinen t , all determinen t
properties are satisfied.
Properties
15
]
( )
( )
[ ][ ] [
( )
[ ]
( ) (
[ ] [ ]
r r r
1) a. b ´ c is a scalar.
r r r r r r r r r
2) a. b ´ c = b.(c ´ a ) = c . a ´ b
rrr r r r
rrr
abc = b c a = c ab
r r r
r r r
3) a. a ´ b = 0 ,
a. b ´ b = 0
rr r
r r r
aa b = 0
a b b =0
r r r
r r r
4 ) a. b ´ c = - a . c ´ b
rrr
rr r
a b c = - a c b etc.
5) Dot and cross can be interchanged
r r r
r r r
a. b ´ c = a ´ b . c
r r r r r r
b ´ c . a = b . (c . a ) etc.
(
(
)
) (
(
[
)
(
]
)
)
]
)
Coplanar vector:
The vectors are said to be coplanar if they lie on same plane
or parallel planes.
( )
r r r
The condition for the vectors to be coplanar is a. b ´ c = 0
Problems:
1) Find the scalar triple product of vectors i + 2j + 3k, - i - j + k and i + j + k
1 2 3
Scalar triple product = - 1 - 1 1
1
1
1
= 1 (- 1 - 1) - 2 (- 1 - 1) + 3(- 1 + 1)
=- 2+ 4+0 = 2
16
[
] [ ]
r r r r r r
rrr
2) Prove that a + b, b + c, c + a = 2 a b c
( ) {( )
}
( ){
}
( ){
}
( ) ( )
( ) ( )
( )
( ) ( ) ( ) [ ]
r r r r
r r
LHS = a + b . b + c ´ (c + a )
r r r r r r r r r r
= a + b . b´ c + b´a + c ´ c + c ´a
r r r r r r r r
= a + b . b´ c - a ´ b + c´a
r r r r r r r r r rr r rr r r r r
= a. b ´ c - a. a ´ b + a.(c ´ a ) + b b ´ c - b a ´ b + b.(c ´ a )
r r r
r r r
= a. b ´ c - 0 + 0 + 0 - 0 + b.(c ´ a )
r r r rr r
r r r
r r r
= a. b ´ c + a b ´ c = 2a. b ´ c = 2 a, b , c
3 ) Show that the vectors 4i - j + k, 3i - 2j - k
and i + j + 2k are coplanar
r
Let
a = 4i - j + k
r
b = 3i - 2j - k
r
c = i + j + 2k
4
r
r
r
consider a . b ´ c = 3
(
)
-1
-2
1
-1
1
2
1
(
)
= 4{- 4 + 1} + 1{ 6 + 1 }+ 1{3 + 2}
= - 12 + 7 + 5
r r r
\ a. b ´ c = 0
\ vectors are coplanar
4 ) Find l if the vectors i + j + 2k, l i - j + k & 3i - 2j - k are coplanar.
17
Given that vectors are coplanar
1
\ l
1
2
-1
1 =0
3 - 2 -1
= 1(1 + 2 ) - 1(- l - 3) + 2(- 2l + 3) = 0
= 3 + l + 3 - 4l + 6 = 0
= -3l + 12 = 0
\l = 4
18