ANSWER SHEET
SELF PRACTICE
|THEORY 1 |
2. (b)
26. Let A and B be the points which divides the
1
3
→
Explanation: Consider a = (ˆ
i + ˆj + ˆ
k)
→
So unit vector a =
\ The value of p is
1 ˆ ˆ ˆ
(ˆ
i + ˆj + ˆ
k)
(i + j + k )
=
3
1+1+1
1
.
3
→
Let R divide AB externally in the ratio 2 : 1.
1
A
→
At λ = –3,

λa = |–3|4 = 12
At λ = 0,

λa = 0 × 4 = 0
At λ = 2,

λa = 2 × 4 = 8

So, the range λa of is [0, 12].
→
|a| =
\
62 + 22 + 32
= 36 + 4 + 9
=
49
= 7
25. Two vectors are equal, if coefficients of their
components are equal.
→
Given, a = b
xiˆ + 2ˆj − zkˆ = 3ˆ
i − yˆj + kˆ
On comparing the coefficients both sides, we
get
x = 3, y = – 2, z = – 1
90
→
→
→
→
→
→
ˆ
a = 3ˆ
i + 2ˆj + 9ˆ
k and b =−
i 2pjˆ + 3kˆ
The given two vectors will be parallel if their
direction ratios are proportional.
2
9
3
=
\
=
− 2p 3
1
⇒
→
→
→
i + (7 – 5) ĵ + (4 – 0) k̂
= (–3 – 2) ˆ
Explanation: Here, a = 6ˆ
i + 2ˆj + 3kˆ
→
32. Given vectors are
⇒
13. (b) 7
→
= 3a + 4 b
Explanation: Required vector
i + 2 ĵ + 4k̂
= –5ˆ
→
= 4a + 2b − a + 2b
⇒
7. (c) –5iˆ+2jˆ+ 4kˆ
R
= 2 × (2 a × b ) − 1 × ( a − 2 b )
2 −1


λa = λ a
Now,
\
B
2
\ Position vector of R

Explanation: Here, a = 4 and –3 ≤ λ ≤ 2
⇒
→
2a + b.
5. (c) [0, 12]
→
→
→
join of points with position vectors a − 2 b and
2
3
=
− 2p
1
– 6p = 2
−1
p=
3
Hence, the value of p for which the two vectors
−1
will be parallel is
⋅
3

35. Since, r is equally indined to the three axes, so
its direction cosines will be same i.e. l = m = n.
Since, l2 + m2 + n2 = 1
⇒
l2 + l2 + l2 = 1
⇒
l2 =
1
3
1
3
⇒
l=±
So,
r̂ = ±
Now,
→

r = r̂ | r |
1 ˆ
1 ˆ 1 ˆ
i±
j±
k
3
3
3
x+y+z=3–2–1=0
Mathematics Class XII
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Also, K divides AB in the ratio 1 : 3. Therefore,
1 ˆ 1 ˆ
 1
= ±  ˆ
i+
j+
k ×2 3
3
3 
 3
→
= ± 2ˆ
i ± 2ˆj ± 2kˆ

→
b + 3a
the position vector of K is
4
→
→
→
= ±2 (ˆ
i + ˆj + ˆ
k)
Now, AL + BM + CN
Concept Applied
→
 → →
  →

4 b + c →   3a + 2c → 
= 
−a+
−b
5
5

 


If a vector a is inclined at equal angles to the three

→
 →

7a + 3b → 

+
−c
10


axes, then, direction cosines of the vector a will be
same.
→ →
→
43. Let, a , b and c be the position vectors of
the vertices A, B and C of DABC, respectively.
Then, the position vectors of L, M and N are
→
→
→
→
→
→
→
→
→
→
 → →
4 b + c − 5a + 3a + 2c − 5b 
= 

5


→
→
 →
7 a + 3 b − 10 c 
+ 

10


4 b + c 3a + 2c
7a + 3b
,
and
respectively.
5
5
10
A( a )
→ →
→

→  →
→
 − 2 a − b + 3 c   7 a + 3 b − 10 c 
+
= 
 

5
10

 

2
3
(7 a 3 b )
N
10
(3 a 2 c )
M
5
3
7
→
4
1
L
B( b )
=
C( c )
→
→
3a + b − 4 c
=
10
(4 b c )
5
4 →
CK
10
→
→
→
→
Hence, AL + BM + CN is parallel to CK .
|THEORY 2 |
3. (d) 16



Explanation: Here, a = 10, b = 2 and a.b = 12.
We know,
⇒
⇒
∴
Now,
π
2
→ →
4
5
 
 
a × b = a b sin θ
= 16
 
Explanation: a . b ≥ 0
1 – cos2 θ =
= 10 × 2 ×

 
7. (b) 0 ≤ θ ≤
12 3
=
20 5
sin θ = ±
 
Since, a × b = a b sin θ and a.b = a b cos θ, so
we shall use these formulae to get the value a × b .
12 = 10 × 2 cos θ
sin θ =

 

 
a.b = a b cos θ
cos θ =
Caution
4
5
9
1–
25
→
⇒
→
| a || b | cos θ ≥ 0
⇒
cos θ ≥ 0
⇒
0≤θ≤
9. (c)
π
2
15
sq. units
4
Explanation: We know that the area of the
→
→
triangle formed by two vectors a and b is
1 → →
| a × b | sq. units.
2
Vector Algebra
91
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1 → →
| a× b |
2
Required area =
1 → →
= | a || b | sin θ
2
= 1 × 12 + 22 + 22 × 5 × sin p
2
6
1
1
⇒
→
→
→
→
Projection of b on a
→
→ →
→
a . b/ | a |
|a|
|b|
6 – 4 –1
1
=
6 × 26
2 39
→
→
→ →
24. We have, | a × b |2 + ( a . b )2
52 + ( − 3)2 + ( − 4)2
→
→
→
→
→ →
= | a |2 | b |2 sin2 θ + ( a . b )2
9
3
=
50 5 2
=
2 + (–1)2 + 12 32 + 42 + (–1)2
Hence, the angle between the given vectors is
 1 
cos–1 
.
 2 39 
→
22 + ( − 1)2 + 22
=
(2ˆ
i – ˆj + kˆ).(3ˆ
i + 4ˆj – kˆ)
2
 1 
θ = cos–1 

 2 39 
⇒
→
=
cos θ =
=
Projection of a on b
a . b/ | b |
→

We know, the angle between two vectors a

and b is given by

a .b
cos θ =  
a b
3
13. (c)
5 2
=
→


22. Let a = 2ˆ
i – ˆj + kˆ and b = 3ˆ
i + 4ˆj – ˆ
k.
15
=
sq. units
4
→ →
→
(D) (a) BC and ED
= ×3×5×
2
2
Explanation:
→
19. (B) (b) AB and EF
→ →
= | a |2 | b |2 (1 − cos2 θ ) + ( a . b )2
17. The movement of the girl is shown here:
→
→
→
→
→ →
= | a |2 | b |2 − | a |2| b |2 cos2 θ + ( a . b )2
B
North
→
→
= | a |2 | b |2
4 km
O
30°
A
6 km
→
East
→
(C) (a) OA + AB
 
 
27. We will find cross product of ( a – d ) and ( b – c ) .
 
 
       
(a – d ) × (b – c ) = a × b – a × c – d × b + d × c
       
= c × d – b × d + b × d – c × d
(E) (a) 2 7
→ → → →
 a× b = c × d 
→ → → →
  a× c = b× d 


 → → → →
 -d× b = b × d 






⇒
(a – d ) × (b – c ) = 0
 
 
Hence, ( a – d ) is parallel to ( b – c ) .
Explanation: From part (D), we have
→
\


OB = 4i + 2 3 j
→
2
2
| OB | = ( 4 ) + (2 3 )
= 16 + 12 = 2 7
→
18. (A) We have,
A = PV of P2 – PV of P1
= (21 – 6, 8 – 8, 4 – 4)
= (15, 0, 0)
Concept Applied
→
B = PV of P4 – PV of P1
= (6 – 6, 16 – 8, 10 – 4)
= (0, 8, 6)
92

Two non–zero vectors are parallel, if their cross product
is zero.
Mathematics Class XII
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→
→
Here,
→
→
30. Given, (2 a + 3 b ) × (5 a + 7 b )
→
→
→
→
→
→
= 10( a × a ) + 14( a × b ) + 15( b × a )
→ →
+ 21( b × b )
→
→
→
→
= 0 + 14( a × b ) − 15( a × b ) + 0
 → → → →

 a× a = b× b =0 

→ →
→ →
 and a × b =− b × a 


→
→
= −(a × b)
→
33. Given,
→
→
→
→
→
and b = 3ˆ
i − 2ˆj + ˆ
k
→2
( a + b )2 =
(− c )
→
→
→
→
→
→
⇒ ( a + b ).( a + b ) =
( − c ).( − c )
→ →
→ →
→ →
→ →
→ →
→
→
2
→
→
→
2
→
Here,
|a| =
→
→
25 + 60 cos q + 36 = 81
⇒
60 cos q = 81 – 61 = 20
cos q =
⇒
12 + ( − 2)2 + 32
32 + ( − 2)2 + 12
→
→
ˆj kˆ
−2 3
−2 1
=ˆ
i ( − 2 + 6) − ˆj(1 − 9) + kˆ( − 2 + 6)
1
3
Hence, the angle between
→
ˆ
i
a× b = 1
3
→
= 4ˆ
i + 8ˆj + 4ˆ
k
1
q = cos −1  
3
⇒
→
= 9 + 4 + 1 =14
=
[ a | 5,=
| b | 6,=
| c | 9]
⇒
→
| a || b |
|b| =
52 + 2 × 5 × 6 cos q + 62 = 92
→
→
| a× b |
→
2
⇒ | a | + 2 | a || b | cos θ + | b | = | c |
⇒
sin q =
= 1 + 4 + 9 =14
2
2
| c |2
⇒ | a | +2a .b + | b | =
→
Then,
→ →
⇒ a .a + a .b + b .a + b .b =
c .c
→
[Using triangle law of vector addition]
 

= a × ( b + ka )
 
 
= a × b
0]
[ a × a =
→
39. Let, a =ˆ
i − 2ˆj + 3ˆ
k
On squaring both sides, we get,
→
 
∴ Area of parallelogram ABCD = a × b
 
Now, area of parallelogram ABEF = AB × AE
  
= AB × (AD + DE)
Hence, proved.
→
a + b =− c
→
Let
= Area of parallelogram ABCD
a+ b+ c =
0
⇒
AB || CD and AE || BF




AB = a and AD = b
→
a
and
= 4(ˆ
i + 2ˆj + kˆ)
→
b
→
is
1
cos–1   .
3
35. Consider ABCD and ABFE as parallelograms
on the same base AB and between the same
parallel lines AB and DF.
→
| a × b | = 4 12 + 22 + 12 =
4 6
\
sin q =
4 6
4 6 2 6
= =
14
7
14 14


43. Here, a = ˆ
k.
i + ˆj + ˆ
k and b = ˆj – ˆ

i + y ĵ + z k̂
Let
c = xˆ

 
Now,
a×c = b
[Given]
ˆ
i ˆj kˆ
⇒
1 1 1 = ˆj – kˆ
x y z
i (z – y) – ĵ (z – x) + k̂(y – x)
⇒ˆ
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93
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= ˆj – kˆ
∴
Also,
⇒
And
Also,
z–y
–(z – x)
x–z
x–y
 
a .c
=0
=1
=1
=1
…(i)
…(ii)
...(iii)
=3
[Given]
⇒ (ˆ
i + ˆj + kˆ) .( xiˆ + yjˆ + zkˆ) = 3
x+y+z=3
On adding equations (ii) and (iii), we get
2x – y – z = 2
On adding equations (iv) and (v), we get
3x = 5
5
⇒
x=
3
…(iv)
...(v)
= 2 – 2 cos q
= 2(1 – cos q)
θ
= 2 × 2sin2
2
θ
= 4sin2
2
⇒
sin2
θ
1
ˆ |2
ˆ−b
= |a
4
2
⇒
sin
θ
1
ˆ|
ˆ−b
= |a
2
2
Hence, proved.
→ →
→ →
...(i)
→ →
a . b 0,
=
b . c 0 and=
c .a 0
and=
→
→
→ 2
→ 2
→
...(ii)
→ 2
Now, | a + b + c | = | a | + | b |2 + | c |
From (i), we have,
z–y=0
→ →
→ →
→ →
+ 2( a . b + b . c + c . a )
= l2 + l2 + l2 + 2(0 + 0 + 0)
2
3
⇒
y=z=
Now,

i + y ĵ + z k̂
c = xˆ
= 3l2
→
→
→
| a+ b+ c | =
⇒
5
2
2
= ˆ
i + ˆj + kˆ
3
3
3
3λ
[Q length cannot be negative]
→
→
→
Suppose ( a + b + c ) is inclined at angles q1, q2
1
= (5ˆ
i + 2ˆj + 2kˆ)
3
→ →
→
and q3 respectively with vectors a , b and c .
Then
Caution
 
 Remember that cross product of vectors a × b =
ˆ
ˆj kˆ
i
a1 a2 a3 and their dot product
b1 b2 b3

a.b = a1b1 + a2 b2 + a3 b3.
→
→
→ →
→ →
→ →
→
→
→
→
( a + b + c ). a = | a + b + c || a | cos θ1
→
⇒ | a |2 + b . a + c . a =
λ2 + 0 + 0 =
⇒
3λ × λ cos θ1
3λ2 cos θ1
[Using equations (i) and (ii)]
→
47. Given,=
| a | 1,=
|b| 1
1
cos θ1 =
3
\
θ 1
ˆ|
ˆ−b
|a
To prove: sin
=
2 2
Similarly,
ˆ |2 = ( a
ˆ).( a
ˆ)
Proof: | a
ˆ−b
ˆ−b
ˆ−b
→
= a
ˆ.a
ˆ+ b
ˆ.b
ˆ
ˆ. a
ˆ− b
ˆ−a
ˆ.b
ˆ+ | b
ˆ |2
= |a
ˆ |2 − 2a
ˆ.b
ˆ=b
ˆ.a
ˆ.b
ˆ]
[a
94
→
50. Given, | =
a | |=
b | |=
c | λ (say)
5
–z =1
3
5
2
z = –1=
3
3
→
→
→
∴ From (ii), we get
⇒
ˆ | cos θ + | b
ˆ |2
ˆ |2 − 2 | a
ˆ || b
=|a
= 1 – 2.1. cos q + 1
→
→ →
→
→
→
→
( a + b + c ). b = | a + b + c || b | cos θ2
→ →
→
→ →
⇒ a . b + | b |2 + c . b=
⇒
0 + λ2 + 0=
3λ × λ × cos θ2
3λ2 cos θ2
Mathematics Class XII
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1
cos θ2 =
3
⇒
And,
→
→
→ →
→
→
→
( a + b + c ). c = | a + b + c || c | cos θ3
→ →
→ →
→
2
⇒ a . c + b . c + | c |=
⇒
→
0 + 0 + λ2=
1
cos θ3 =
3
⇒
Thus, cos θ1= cos θ2= cos θ3=
→
→
3λ × .λ cos θ3
3λ2 cos θ3
1
3
→
Hence, ( a + b + c ) is equally inclined with the
→ →
→
vectors a , b and c .
Hence, proved.
Vector Algebra
95
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