5. The area of the triangle, two whose sides
are the vectors 𝑎⃗ = i + j and 𝑏⃗⃗ = i – j inclined
VECTORS
1. If ABCDEF is a regular hexagon and if
⃗⃗⃗⃗⃗⃗ and 𝑏⃗⃗ = 𝐵𝐶
⃗⃗⃗⃗⃗⃗ then 𝐹𝐷
⃗⃗⃗⃗⃗⃗ =
𝑎⃗ = 𝐴𝐵
(1) 𝑎⃗ + 𝑏⃗⃗
(3) 𝑏⃗⃗ − 2𝑏⃗⃗
at an angle 30° is
(1) 0.50 (2) 0.30
(2) 𝑎⃗ + 2𝑏⃗⃗
(4) 2𝑎⃗ − 𝑏⃗⃗
position vectors ai – 52j, 60i + 3j and 40i – 8j
are collinear. The value of ‘a’ is
Solution: Ans(2).
⇒
60 − 𝑎
− 20
=
55
− 11
=
40 − 60
60 − 𝑎
⇒
(4) − 20
(3) 20
60 − 𝑎
− 20
3 –( −52 )
=−5
(1) 0.50
(2) 0.30
(3) 0.25 (4) 0.40
6
2
2
(2i – 2j + k) with the positive direction of the
3
y-axis is
−2
)
−1
2
3
3
,
−2
(4) cos- 1(
3
cosines of the given vector are
2
3
,
2
and
3
and magnitude of the vector is 6
(2) ( − 1, 2, 1)
(4) (2, − 4, 4 )
3
,
2
3
2
, ) = ( − 2, 4, 4)
3
8. If ai + j + k, i + bj + k and i + j + ck are
coplanar then a + b + c =
(2) abc – 2 (3) 0
(4) abc + 2
a 1 1
Solution: Ans(4).|1 b 1| = 0
1 1 c
⇒ a( bc – 1) – 1( c – 1 ) + 1( 1 – b ) = 0
)
2i – 2j + k is √4 + 4 + 1 = 3, the given
vector is a unit vector. ∴ The direction
−2
3
1
and .
∴ The angle made by the given vector with
the positive direction of the y-axis is
3
⃗⃗⃗⃗⃗⃗ .
⃗⃗⃗⃗⃗⃗ = 2𝑃𝐶
= ⃗⃗⃗⃗⃗⃗
𝑃𝐶 . ∴ ⃗⃗⃗⃗⃗⃗
𝑃𝐴 + 𝑃𝐵
2
(1) abc
2
(2) cos- 1( )
3
Solution: Ans(4). Since the magnitude of
).
⃗⃗⃗⃗⃗⃗
𝑃𝐴 + ⃗⃗⃗⃗⃗⃗
𝑃𝐵
−1
1
−2
Solution: Ans(2). By the midpoint formula
6(
4. The angle made by the vector
cos- 1(
2
Solution: Ans(3) Required vector is
parallel to 𝑐⃗ )
(3)
2
units. The vector is
(1) ( 1, 2, 3)
(3) ( − 2, 4, 4)
1
∴ [a⃗⃗ ⃗⃗
b c⃗] = . ( Note: Here 𝑎⃗ × 𝑏⃗⃗ is
4
3
1
cos- 1( )
3
2
7. Direction cosines of a vector are
Solution: Ans(3). [a⃗⃗ ⃗⃗
b c⃗] = (𝑎⃗ × 𝑏⃗⃗) . 𝑐⃗
𝜋
1
= |𝑎⃗ × 𝑏⃗⃗||𝑐⃗|cos0° = |𝑎⃗||𝑏⃗⃗|sin = 1.1.
(1) sin- 1(
1
6. If C is the midpoint of the line joining the
points A and B and if P is any other point non
collinear with A and B then
always
⃗⃗⃗⃗⃗⃗ + 𝑃𝐵
⃗⃗⃗⃗⃗⃗ = 𝑜⃗
⃗⃗⃗⃗⃗⃗ + 𝑃𝐶
(1) 𝑃𝐴
2
then [a⃗⃗ ⃗⃗
b c⃗] =
6
2
(4) ⃗⃗⃗⃗⃗⃗
𝑃𝐴 + ⃗⃗⃗⃗⃗⃗
𝑃𝐵 = ⃗⃗⃗⃗⃗⃗
𝑃𝐶
3. If 𝑐⃗ is a unit vector perpendicular to the
unit vectors 𝑎⃗ and 𝑏⃗⃗ and if the angle between
𝜋
1
⃗⃗⃗⃗⃗⃗
(2) ⃗⃗⃗⃗⃗⃗
𝑃𝐴 + ⃗⃗⃗⃗⃗⃗
𝑃𝐵 = 2𝑃𝐶
⃗⃗⃗⃗⃗⃗ + 𝑃𝐵
⃗⃗⃗⃗⃗⃗ = 𝑜⃗
(3) 𝑃𝐴
−8−3
⇒ 60 – 𝑎 = 100 ⇒ 𝑎 = −40.
𝑎⃗ and 𝑏⃗⃗ is
2
1
= (√2 . √2) = .
2. The points with the
(2) – 40
(4) 0.40
Solution: Ans(1) Area of the triangle
1
1
= |𝑎⃗ × 𝑏⃗⃗| = |𝑎⃗||𝑏⃗⃗|sin30°
Solution: Ans(1). ⃗⃗⃗⃗⃗⃗
𝐹𝐷 =
⃗⃗⃗⃗⃗⃗ = 𝐴𝐵
⃗⃗⃗⃗⃗⃗ + 𝐵𝐶
⃗⃗⃗⃗⃗⃗ = 𝑎⃗ + 𝑏⃗⃗
𝐴𝐶
(1) 40
(3) 0.25
3
⇒ abc – a – c + 1 + 1 – b = 0
⇒ a + b + c = abc + 2.
9. If 𝑎⃗ and 𝑏⃗⃗ are unit vectors and if
|𝑎⃗ + 𝑏⃗⃗| = √3 then the angle between 𝑎⃗ and 𝑏⃗⃗
is
(1) 60°
(2) 30°
(3) 45°
(4) 90°
Solution: Ans(1).
2
2
|𝑎⃗ + 𝑏⃗⃗| = |𝑎⃗|2 + |𝑏⃗⃗| + 2|𝑎⃗||𝑏⃗⃗| cos 𝜃
2
⇒ √3 = 12 + 12 + 2.1.1. cos 𝜃
⇒ cos 𝜃 =
1
2
⇒ 𝜃 = 60°.
Solution: Ans(2). Dot product
10. The unit vector perpendicular to i + j and
j + k forming a right handed system is
(1)
(3)
𝑖+𝑗+𝑘
√3
𝑖−2𝑗 + 𝑘
√3
(2)
(4)
𝑖−𝑗 + 𝑘
√3
2𝑖−𝑗−𝑘
√3
√3
11. If 𝑎⃗ = 2i + j – 2k and 𝑏⃗⃗ = i – 2j + 2k then
the angle between 𝑎⃗ + 𝑏⃗⃗ and 𝑎⃗ − 𝑏⃗⃗ is
(2) 30°
(3) 90°
(4) 45°
⃗⃗| = 3. Since
Solution: Ans(3). Here |𝑎⃗| = |b
⃗⃗|, vectors 𝑎⃗ + 𝑏⃗⃗ and 𝑎⃗ − 𝑏⃗⃗ are
|𝑎⃗| = |b
orthogonal. Hence angle between 𝑎⃗ + 𝑏⃗⃗ and
12. If 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are unit vectors such that
𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = 𝑜⃗ then 𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ =
(2) 1
(3)
−2
(4)
3
Solution: Ans(4). |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗|
−3
2
2
= |𝑎|2 + |𝑏|2 + |𝑐|2 + 2(𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ )
⇒ 0 = 1 + 1 + 1 + 2(𝑎⃗.𝑏⃗⃗ + 𝑏⃗⃗.𝑐⃗ + 𝑐⃗.𝑎⃗ )
−3
∴ 𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ = .
2
(2) i
(3)
𝑖+𝑗
(4)
√2
𝑖−𝑗
√2
Solution: Ans(3). Now
i × ( j × ( i + j ) = i × ( ( j × i) + ( j × j ))
= i × ( − k + 𝑜⃗) = j. This is the required
unit vector. (concept :
(1) 30°
(2) 60°
(3) 45°
⃗⃗ × 𝑐⃗)
𝑎⃗⃗ × (𝑏
⃗⃗ × 𝑐⃗)|
|𝑎⃗⃗ × (𝑏
)
(4) 90°
Solution: Ans(3). |𝑎⃗. 𝑏⃗⃗| = |𝑎⃗||𝑏⃗⃗|cos𝜃 and
|𝑎⃗ × 𝑏⃗⃗| = |𝑎⃗||𝑏⃗⃗|sin𝜃.
∴ |𝑎⃗. 𝑏⃗⃗| = |𝑎⃗ × 𝑏⃗⃗| ⇒ cos𝜃 = sin𝜃
⇒ 𝜃 = 45°.
⃗⃗⃗⃗⃗⃗ = j – k then unit
16. If ⃗⃗⃗⃗⃗⃗
𝑂𝐴 = i – j and 𝑂𝐵
vector perpendicular to the plane of ∆ABC is
(1)
(3)
𝑖–𝑗+𝑘
(2)
√3
𝑖 + 2𝑗 + 2𝑘
(4)
√3
2𝑖 − 𝑗 + 2 𝑘
√3
𝑖+𝑗+𝑘
√3
⃗⃗⃗⃗⃗⃗⃗
𝑂𝐴× ⃗⃗⃗⃗⃗⃗⃗
𝑂𝐵
⃗⃗⃗⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗⃗⃗
|𝑂𝐴
𝑂𝐵 |
.
𝑖
𝑗
𝑘
⃗⃗⃗⃗⃗⃗ × 𝑂𝐵
⃗⃗⃗⃗⃗⃗ = |1 − 1 0 |
𝑂𝐴
0 1 −1
= 𝑖. 1 − 𝑗(− 1) + 𝑘(1) = 𝑖 + 𝑗 + 𝑘.
∴ Required is
𝑖+𝑗+𝑘
√3
17. If A, B, C and D are the vertices of a
parallelogram, P is the point of intersection of
the two diagonals and O is the origin then
⃗⃗⃗⃗⃗⃗ + 𝑂𝐵
⃗⃗⃗⃗⃗⃗ + 𝑂𝐶
⃗⃗⃗⃗⃗⃗ + 𝑂𝐷
⃗⃗⃗⃗⃗⃗⃗ =
𝑂𝐴
⃗⃗⃗⃗⃗⃗
(1) 4𝑂𝑃
13. A unit vector perpendicular to i and
coplanar with j and i + j is
(1) j
⇒ 5 – p = 0 ⇒ p = 5.
Solution: Ans(4). Required is
𝑎⃗ − 𝑏⃗⃗ is 90°.
(1) 3
(1 – p).3 +2(1+ p).1+ ( 3 + p).0 = 0
15. If |𝑎⃗. 𝑏⃗⃗| = |𝑎⃗ × 𝑏⃗⃗| then the angle
between 𝑎⃗ and 𝑏⃗⃗ is
Solution: Ans(2). Cross product of the given
𝑖 𝑗 𝑘
two vectors is |1 1 0| = 𝑖 − 𝑗 + 𝑘.
0 1 1
𝑖−𝑗+𝑘
Required unit vector is
.
(1) 60°
14. If (1 – p)i + 2( 1 + p )j + (3 + p)k and
3i +j are at right angles to each other then
p=
(1) 2.5
(2) 5
(3) 3
(4) 4
⃗⃗⃗⃗⃗⃗
(2) 2𝑂𝑃
⃗⃗⃗⃗⃗⃗
(3) 𝑂𝑃
Solution: Ans(1). By the
midpoint formula, ⃗⃗⃗⃗⃗⃗
𝑂𝐴 +
⃗⃗⃗⃗⃗⃗ = 2𝑂𝑃
⃗⃗⃗⃗⃗⃗ and 𝑂𝐵
⃗⃗⃗⃗⃗⃗ +
𝑂𝐶
⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑂𝐷 = 2𝑂𝑃
∴
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗.
𝑂𝐴 + 𝑂𝐵
𝑂𝐶 + ⃗⃗⃗⃗⃗⃗⃗
𝑂𝐷 = 4𝑂𝑃
⃗⃗⃗⃗⃗⃗
(4)− 𝑂𝑃
18. If 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are the three coterminous
edges of a parallelepiped having volume
3cubic units then
[𝑎⃗ × 𝑏⃗⃗ 𝑏⃗⃗ × 𝑐⃗ 𝑐⃗ × 𝑎⃗] =
(1) 6
(2) 9
(3) 3
(4) 27
⇒ 3( − 5 ) + 𝛼( − 5) + 5( − 5 ) = 0
⇒ 𝛼 = − 8.
22. The projection of 𝑎⃗ = 2i + 3j on
𝑏⃗⃗ = 3j – 2k is
(1)
13
√15
(2)
9
√13
(3)
12
(4)
√5
9
√5
Solution: Ans(2). [𝑎⃗ × 𝑏⃗⃗ 𝑏⃗⃗ × 𝑐⃗ 𝑐⃗ × 𝑎⃗]
= [𝑎⃗ 𝑏⃗⃗ 𝑐⃗] 2 = 32 = 9.
Solution: Ans(2). Required =
19. If 𝑎⃗ × 𝑏⃗⃗ = 𝑎⃗ × 𝑐⃗, 𝑎⃗ ≠ 0 then for a
constant 𝑘,
(1) 𝑏⃗⃗ = 𝑎⃗ + 𝑘𝑐⃗
(2) 𝑎⃗ = 𝑏⃗⃗ + 𝑘𝑐⃗
(3) 𝑏⃗⃗ = 𝑐⃗ + 𝑘𝑎⃗
(4) 𝑎⃗ = 𝑐⃗ + 𝑘𝑏⃗⃗
5i + 3j – 2k and 12i – 8j − k form the sides
of
(1) an isosceles ∆
(2) an equilateral ∆
(3) a right angled ∆ (4) a scalene ∆
Solution: Ans(3). 𝑎⃗ × 𝑏⃗⃗ = 𝑎⃗ × 𝑐⃗
⇒ 𝑎⃗ × 𝑏⃗⃗ − 𝑎⃗ × 𝑐⃗ = 𝑜⃗
⇒ 𝑎⃗ × ( 𝑏⃗⃗ − 𝑐⃗ ) = 𝑜⃗
Solution: Ans(3). Magnitudes of the three
⇒ 𝑎⃗ and 𝑏⃗⃗ − 𝑐⃗ are parallel
⇒ 𝑏⃗⃗ − 𝑐⃗ = 𝑘𝑎⃗ ⇒ 𝑏⃗⃗ = 𝑐⃗ + 𝑘𝑎⃗.
20. If 𝑎⃗ and 𝑏⃗⃗ are two unit vectors inclined at
an angle 𝜃 to each other then |𝑎⃗ + 𝑏⃗⃗| will be
less than 1if 𝜃 is greater than
(1) 30°
(2)75°
(3)135°
(4)120°
Solution: Ans(4).
2
2
|𝑎⃗ + 𝑏⃗⃗| = |𝑎⃗|2 + |𝑏⃗⃗| + 2|𝑎⃗||𝑏⃗⃗|cos𝜃
= 1 + 1 + 2cos𝜃 = 2 + 2cos𝜃
2
𝜃
= 2( 1 + cos𝜃) = 4cos ( ).
2
𝜃
∴ |𝑎⃗ + 𝑏⃗⃗| = 2cos( ).
2
𝜃
𝜃
2
=
9
.
√13
23. The three vectors 7i – 11j + k,
vectors are √49 + 121 + 1 = √171,
√25 + 9 + 4 = √38 and
√144 + 64 + 1 = √209.
Since 171 + 38 = 209 the triangle formed by
the vectors is right angled.
24. The unit vectors 𝑎⃗ and 𝑏⃗⃗ are inclined at an
angle 90°. The magnitude of
(𝑎⃗ + 3𝑏⃗⃗) × (3𝑎⃗ − 𝑏⃗⃗) is
(1) 24
(2) 20
(3) 5
(4) 10
Solution: Ans(4). (𝑎⃗ + 3𝑏⃗⃗) × (3𝑎⃗ − 𝑏⃗⃗)
= 𝑎⃗ × 3𝑎⃗ + 3𝑏⃗⃗ × 3𝑎⃗ − 𝑎⃗ × 𝑏⃗⃗ − 3𝑏⃗⃗ × 𝑏⃗⃗
= 9𝑏⃗⃗ × 𝑎⃗ + 𝑏⃗⃗ × 𝑎⃗ = 10𝑏⃗⃗ × 𝑎⃗.
∴ |(𝑎⃗ + 3𝑏⃗⃗) × (3𝑎⃗ − 𝑏⃗⃗)| = |10𝑏⃗⃗ × 𝑎⃗|
= 10|𝑏⃗⃗||𝑎⃗|sin90° = 10.
1
This is less than one when cos( ) <
2
2
or
⃗⃗
𝑎⃗⃗.𝑏
⃗⃗|
|𝑏
> 60° or 𝜃 > 120°.
21. For what value of 𝛼 the vector
3i – 𝛼j + 5k lie on the plane determined by
the vectors (2, 1, − 1) and (1, − 2, − 3) ?
(1) − 8
(2) 8
(3) 6
(4) − 6
Solution: Ans(1). Three vectors are coplanar
3 −𝛼
5
⇒ |2
1
− 1| = 0
1 −2 −3
25. The points A, B and C having the position
vectors i + 2j + 3k, 𝛼i + 4j + 7k and
– 3i – 2j – 5k respectively are collinear iff
𝛼=
(1) 3
(2) 2
(3) 1
(4) 4
Solution: Ans(1). ⃗⃗⃗⃗⃗⃗
𝐴𝐵 = ( 𝛼 – 1, 2, 4 ) and
⃗⃗⃗⃗⃗⃗ = ( − 4, − 4, − 8). Now ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ are
𝐴𝐶
𝐴𝐵 and 𝐴𝐶
parallel. ∴ 𝛼 – 1 = 2. ∴ 𝛼 = 3.
26. If 𝑎⃗ is a non zero vector of magnitude 'a'
units and 𝑚 is a scalar then m𝑎⃗ is a unit
vector if a =
2
1
(1) |𝑚|
(2) |𝑚|
(3)
1
2|𝑚|
1
(4) |𝑚2|
Solution: Ans(2). |𝑚𝑎⃗| = 1 ⇒ |𝑚||𝑎⃗| = 1
1
⇒ |𝑚|a = 1 ⇒ a = |𝑚|.
27. If 𝛼, 𝛽 and 𝛾 are the angles made by the
vector 𝑎⃗ = 2i + 3j + 4k with the positive
direction of the x-, y- and z-axes then
sin2𝛼 + sin2𝛽 + sin2𝛾 =
(1) 1
(2) 4
(3) 2
(4)3
Solution: Ans(3).
Formula: cos2𝛼 + cos2𝛽 + cos2𝛾 =1.
∴ sin2𝛼 + sin2𝛽 + sin2𝛾
= 1 − cos2𝛼 + 1− cos2𝛽 + 1− cos2𝛾
2
2
2
= 3 – (cos 𝛼 + cos 𝛽 + cos 𝛾) = 3 – 1 = 2.
28. The position vector of the centroid of the
triangle formed by the points having position
vectors 𝑎⃗ = i +2j + 3k, 𝑏⃗⃗ = 2i – 5j + 3k and
𝑐⃗ = 3i + 6j – 12k is
(1) 3i + 2j – k
(3) 2i – j + 3k
(2) i – j + k
(4) 2i + j – 2k
Solution: Ans(4). The position vector of the
centroid is =
⃗⃗ + 𝑐⃗
𝑎⃗⃗ + 𝑏
3
=
6𝑖 + 3𝑗 −6𝑘
3
= 2𝑖 + 𝑗 − 2𝑘.
29. If A, B, C and D are the four points such
⃗⃗⃗⃗⃗⃗ + 𝐵𝐷
⃗⃗⃗⃗⃗⃗⃗ =
that ⃗⃗⃗⃗⃗⃗
𝐴𝐵 = ⃗⃗⃗⃗⃗⃗
𝐷𝐶 then 𝐴𝐶
⃗⃗⃗⃗⃗⃗ (2) 2𝐷𝐴
⃗⃗⃗⃗⃗⃗ (3) 2𝐶𝐷
⃗⃗⃗⃗⃗⃗ (4) 2𝐴𝐵
⃗⃗⃗⃗⃗⃗
(1) 2𝐵𝐶
Solution: Ans(1). By the triangle law of
⃗⃗⃗⃗⃗⃗ + 𝐵𝐷
⃗⃗⃗⃗⃗⃗⃗
addition of the vectors, 𝐴𝐶
= ⃗⃗⃗⃗⃗⃗
𝐴𝐵 +
= ⃗⃗⃗⃗⃗⃗
𝐴𝐵 +
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗
𝐵𝐶 +𝐵𝐶
𝐶𝐷
⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ ∵ ⃗⃗⃗⃗⃗⃗
2𝐵𝐶
𝐷𝐶 = 2𝐵𝐶
𝐴𝐵 = ⃗⃗⃗⃗⃗⃗
𝐷𝐶 .
30. Which of the following is FALSE?
(Here 𝑖, 𝑗 and 𝑘 are unit vectors with the
usual meaning)
(1) ∑ 𝑖 × (𝑗 × 𝑘) = 𝑜⃗ (2) ∑ 𝑖. ( 𝑗 × 𝑘 ) = 0
(3) ∑ 𝑖. (𝑗 + 𝑘) = 0 (4) ∑ 𝑖 × (𝑗 + 𝑘) = 𝑜⃗
Solution: Ans(2).
∑ 𝑖 × (𝑗 × 𝑘) = ∑ 𝑖 × 𝑖 = 𝑜⃗.
∑ 𝑖. ( 𝑗 × 𝑘 ) = ∑ 𝑖. 𝑖 = 3.
∑ 𝑖. (𝑗 + 𝑘) = ∑( 𝑖. 𝑗 + 𝑖. 𝑘) = ∑ 0 = 0.
∑ 𝑖 × (𝑗 + 𝑘) = ∑( 𝑖 × 𝑗 + 𝑖 × 𝑘)
= ∑( 𝑘 − 𝑗) = 𝑜⃗.
31. If 𝑥⃗ = 𝑎⃗ − 3𝑏⃗⃗ and 𝑦⃗ = 3𝑎⃗ − 2𝑏⃗⃗ and if
|𝑎⃗| = |𝑏⃗⃗| = 1 and 𝑥⃗ ⊥ 𝑦⃗ then angle
between 𝑎⃗ and 𝑏⃗⃗ is
7
(2) cos – 1 ( )
13
5
9
(4) cos – 1 ( )
5
(1) cos – 1 ( )
12
2
(3) cos – 1 ( )
11
Solution: Ans(3). (𝑎⃗ − 3𝑏⃗⃗). (3𝑎⃗ − 2𝑏⃗⃗) = 0
⇒ 𝑎⃗. 3𝑎⃗ − 3𝑏⃗⃗. 3𝑎⃗ − 𝑎⃗. 2𝑏⃗⃗ + 3𝑏⃗⃗. 2𝑏⃗⃗ = 0.
2
⇒ 3|𝑎⃗|2 – 11𝑎⃗. 𝑏⃗⃗ + 6|𝑏⃗⃗| = 0 ⇒ 𝑎⃗. 𝑏⃗⃗ =
⇒ |𝑎⃗||𝑏⃗⃗|cos𝜃 =
9
11
⇒ cos𝜃 =
9
11
9
11
9
⇒ 𝜃 = cos – 1 ( ).
11
32. If 𝑎⃗, 𝑏⃗⃗, 𝑐⃗ and 𝑑⃗ are the position vectors
of the four points A, B, C and D such that
(𝑑⃗ −
(𝑑⃗ −
⃗⃗ + 𝑐⃗
𝑏
) . (𝑏⃗⃗ − 𝑐⃗) = 0 and
2
𝑎⃗⃗ + 𝑐⃗
2
) . (𝑎⃗ − 𝑐⃗) = 0 then for the ∆ABC
the point D is
(1) centroid
(3) orthocenter
(2) incenter
(4) circumcenter
Solution: Ans(4). Let E
and F be the midpoints of
BC and AC. By the given
⃗⃗⃗⃗⃗⃗. ⃗⃗⃗⃗⃗⃗
condition 𝐸𝐷
𝐶𝐵 = 0 and
⃗⃗⃗⃗⃗⃗ = 0.
⃗⃗⃗⃗⃗⃗
𝐹𝐷. 𝐶𝐴
⃗⃗⃗⃗⃗⃗ and 𝐹𝐷
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ ⊥ 𝐶𝐵
⃗⃗⃗⃗⃗⃗ ⊥ 𝐶𝐴
∴ 𝐸𝐷
This shows that D is the circumcenter of the
∆ABC.
33. If 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are three vectors such that
each one is perpendicular to the sum of the
other two and if |𝑎⃗| = 3, |𝑏⃗⃗| = 4 and |𝑐⃗| = 5
then |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| =
(1) √50
(2) √35
(3) √54
(4) √15
Solution: Ans(1). Given 𝑎⃗ ⊥ ( 𝑏⃗⃗ + 𝑐⃗ ), 𝑏⃗⃗ ⊥
( 𝑐⃗ + 𝑎⃗ ) and 𝑐⃗ ⊥ ( 𝑎⃗ + 𝑏⃗⃗ ).
∴ 𝑎⃗. ( 𝑏⃗⃗ + 𝑐⃗ ) = 0, 𝑏⃗⃗. (𝑐⃗ + 𝑎⃗ ) = 0 and
𝑐⃗. ( 𝑎⃗ + 𝑏⃗⃗ ) = 0.
∴ 𝑎⃗. (𝑏⃗⃗ + 𝑐⃗) + 𝑏⃗⃗. (𝑐⃗ + 𝑎⃗) + 𝑐⃗. (𝑎⃗ + 𝑏⃗⃗) = 0
∴ 2( 𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ ) = 0.
2
2
∴ |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = |𝑎⃗|2 + |𝑏⃗⃗| + |𝑐⃗|2
= 9 + 16 +25 = 50.
∴ |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = √50.
34. If p, q, r are direction cosines of a vector
perpendicular to the vector 2i – 3j + 4k then
(1) p + q + r = 0
(2) 2p – 3q + 4r = 0
(3)
𝑝
2
+
𝑞
−3
+
𝑟
4
= 0 (4)
𝑝
=
2
𝑞
−3
𝑟
=
4
Solution: Ans(2). The unit vector pi + qj + rk
is perpendicular to the vector 2i – 3j + 4k.
∴ the dot product of the vectors is zero.
35. If 𝑎⃗ and 𝑏⃗⃗ are two vectors of magnitudes
2
2
2 units each then |𝑎⃗ + 𝑏⃗⃗| + |𝑎⃗ − 𝑏⃗⃗| =
(1) 0
(2) 4
(3) 16
(4) 8
2
Solution: Ans(3). |𝑎⃗ + 𝑏⃗⃗| + |𝑎⃗ − 𝑏⃗⃗|
2
2
= 2( |𝑎⃗|2 + |𝑏⃗⃗| ) = 2( 4 + 4) = 16.
36. If the angle made by 𝑎⃗ = 2i + 2j – k with
the positive direction of x-axis is 60° then
|𝑎⃗ × 𝑖| =
(1)
√3
4
(2)
2√2
3
(3)
3
2
(4)
3√3
2
Solution: Ans(4). |𝑎⃗ × 𝑖| = |𝑎⃗||𝑖|sin60°
√3
2
= 3.1.
=
3√3
.
2
37. If 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are three mutually
perpendicular vectors each of magnitudes 3
units the [𝑎⃗ 𝑏⃗⃗ 𝑐⃗] =
(1) 27
(2) 18
(3) 9
(4) 6
Solution: Ans(1). [𝑎⃗ 𝑏⃗⃗ 𝑐⃗] = 𝑎⃗. (𝑏⃗⃗ × 𝑐⃗)
= |𝑎⃗||𝑏⃗⃗ × 𝑐⃗|cos0° = |𝑎⃗||𝑏⃗⃗||𝑐⃗|sin90°
= |𝑎⃗||𝑏⃗⃗||𝑐⃗| =3.3.3 = 27. ( Here 𝑎⃗ is
parallel to 𝑏⃗⃗ × 𝑐⃗ )
38.The vertices A, B and D of a
parallelogram ABCD are
i +2j, i + 2k and k + 2i. The point C is
(1) i – 3j + k
(2) 2i – 2j + 2k
(3) 3i – 2j + 3k
(4) i – j + k
Solution: Ans(2). Let C ≡ (x, y, z ). Midpoint
of AC is same as the midpoint of BD.
i +2j + xi + yj + zk = i + 2k + k + 2i
⇒ xi + yj + zk = 2i – 2j + 3k
39. 𝑎⃗ × {𝑎⃗ × (𝑎⃗ × 𝑏⃗⃗)} =
(1) (𝑎⃗. 𝑎⃗)( 𝑎⃗ × 𝑏⃗⃗ )
(2) (𝑎⃗. 𝑏)( 𝑏⃗⃗ × 𝑎⃗ ).
(3) (𝑎⃗. 𝑎⃗)( 𝑏⃗⃗ × 𝑎⃗ )
(4) (𝑎⃗. 𝑏)( 𝑎⃗ × 𝑏⃗⃗ ).
Solution: Ans(3). 𝑎⃗ × (𝑎⃗ × 𝑏⃗⃗)
= (𝑎⃗. 𝑏⃗⃗)𝑎⃗ − (𝑎⃗. 𝑎⃗)𝑏⃗⃗
∴ required = 𝑎⃗ × ((𝑎⃗. 𝑏⃗⃗)𝑎⃗ − (𝑎⃗. 𝑎⃗)𝑏⃗⃗)
= 𝑎⃗ × ((𝑎⃗. 𝑏⃗⃗)𝑎⃗ ) − 𝑎⃗ × ((𝑎⃗. 𝑎⃗)𝑏⃗⃗)
= (𝑎⃗. 𝑏⃗⃗)( 𝑎⃗ × 𝑎⃗ ) − (𝑎⃗. 𝑎⃗)( 𝑎⃗ × 𝑏⃗⃗ )
= (𝑎⃗. 𝑏⃗⃗)( 𝑜⃗ ) − (𝑎⃗. 𝑎⃗)( 𝑎⃗ × 𝑏⃗⃗ )
= (𝑎⃗. 𝑎⃗)( 𝑏⃗⃗ × 𝑎⃗ ).
40. If 𝑟⃗ × i = 2j – 3k then 𝑟⃗ =
(1) xi + j + k
(2) xi − 3j + 2k
(3) xi − j + k
(4) xi + 3j + 2k
Solution: Ans(4). Let 𝑟⃗ = xi + yj + zk. Then
𝑖 𝑗 𝑘
𝑟⃗ × i = |𝑥 𝑦 𝑧 |
1 0 0
= 𝑖( 0 )– 𝑗( – 𝑧) + 𝑘( −𝑦 ) = 𝑧𝑗 – 𝑦𝑘.
∴ y = 3 and z = 2. ∴ 𝑟⃗ = xi + 3j + 2k.