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Vector -02
1. 𝐚⃗, 𝐛, 𝐜 are mutually perpendicular unit vectors, a) −40 b) −20
then |𝐚⃗ + 𝐛 + 𝐜| is equal to c) 20 d) 40
10. The unit vector perpendicular to 𝐢̇̂ − 𝐣̇̂ and
a) √3 b) 3
c) 1 d) 0 coplanar with 𝐢̇̂ + 2𝐣̇̂ and 𝐢̇̂ + 3𝐣̇̂ is
2𝐢̇̂−5𝐣̇̂
2. If 𝐚⃗, 𝐛, 𝐜 are three non-coplanar vectors and a) b) 2𝐢̇̂ + 5𝐣̇̂
√29
1
⃗ ,𝐪
𝐩 ⃗ , 𝐫, are reciprocal vectors, then (𝑙𝐚⃗ + 𝑚𝐛 + c) (𝐢̇̂ + ̂𝐣̇) d) 𝐢̇̂ + ̂𝐣̇
√2
𝑛𝐜) ∙ (𝑙𝐩 ⃗ + 𝑚𝐪 ⃗ + 𝑛𝐫) is
11. The vectors 2𝑖̂ + 3𝑗̂ − 4𝑘̂ and 𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂ are
a) 𝑙 + 𝑚 + 𝑛 b) 𝑙 3 + 𝑚3 + 𝑛3
2 2 2
perpendicular when
c) 𝑙 + 𝑚 + 𝑛 d) None of these a) 𝑎 = 2, 𝑏 = 3, 𝑐 = −4 b) 𝑎 = 4, 𝑏 = 4, 𝑐 = 5
3. A constant force 𝐅 = 2𝐢̇̂ − 3𝐣̇̂ + 2𝐤̂ is acting on a
c) 𝑎 = 4, 𝑏 = 4, 𝑐 = −2 d) None of these
particle such that the particle is displaced from ̂ and 𝑎𝐢̇̂ + 𝑏𝐣̇̂ − 15𝐤
12. If vectors 3𝐢̇̂ + 𝐣̇̂ − 5𝐤 ̂ are
the point(1,2,3) to the point (3,4,5). The work collinear, then
done by the force is a) 𝑎 = 3, 𝑏 = 1 b) 𝑎 = 9, 𝑏 = 1
a) 2 b) 3
c) 𝑎 = 3, 𝑏 = 3 d) 𝑎 = 9, 𝑏 = 3
c) 4 d) 5
13. The point collinear with (1, −2, −3)and (2,0,0)
4. If (𝑎 × 𝑏⃗) × 𝑐 = 𝑎 × (𝑏⃗ × 𝑐), then among the following is
a) 𝑏⃗ × (𝑐 × 𝑎) = ⃗0 b) 𝑎 × (𝑏⃗ × 𝑐) = ⃗0 a) (0,4,6) b) (0, −4, −5)
c) 𝑐 × 𝑎 = 𝑎 × 𝑏⃗ d) 𝑐 × 𝑏⃗ = 𝑏⃗ × 𝑎 c) (0, −4, −6) d) (0, −4,6)
̂ is acting at the point
5. If a force 𝐅 = 3𝐢̇̂ + 2𝐣̇̂ − 4𝐤 14. If θ is the angle between the lines 𝐴𝐵 and 𝐴𝐶
where 𝐴, 𝐵 and 𝐶 are the three points with
𝑃(1, −1,2) then the magnitude of moment of 𝐅
coordinates (1,2, −1), (2,0,3), (3, −1,2)
about the point 𝑄(2, −1,3) is
respectively, then √462 cos θ is equal to
a) √57 b) √39
a) 20 b) 10
c) 12 d) 17
c) 30 d) 40
6. If 𝑎̂, 𝑏, 𝑐̂ are three unit vectors such that 𝑏̂ and 𝑐̂
̂
1 15. Each of the angle between vectors 𝑎, 𝑏⃗ and 𝑐 is
are non-parallel and 𝑎̂ × (𝑏̂ × 𝑐̂ ) = 𝑏̂, then the2 equal to 60°. If |𝑎| = 4, |𝑏⃗| = 2 and |𝑐| = 6, then
angle between 𝑎̂ and 𝑐̂ is
a) 30° b) 45° the modulus of 𝑎 + 𝑏⃗ + 𝑐 , is
a) 10 b) 15
c) 60° d) 90°
c) 12 d) None of these
7. The equation of the plane perpendicular to the
16. The equation of the plane passing through three
line
𝑥−1 𝑦−2 non-collinear points 𝐚⃗, 𝐛, 𝐜 is
=
1 −1 a) 𝐫 ∙ (𝐛 × 𝐜 + 𝐜 × 𝐚⃗ + 𝐚⃗ × 𝐛) = 0
𝑧+1
= and passing through the point(2,3,1) is b) 𝐫 ∙ (𝐛 × 𝐜 + 𝐜 × 𝐚⃗ + 𝐚⃗ × 𝐛) = [𝐚⃗ 𝐛 𝐜]
2
a) 𝐫 ∙ (𝐢̇̂ + 𝐣̇̂ + 2𝐤̂ ) = 1 b) 𝐫 ∙ (𝐢̇̂ − 𝐣̇̂ + 2𝐤
̂)=1 c) 𝐫 ∙ (𝐚⃗ × (𝐛 × 𝐜)) = [𝐚⃗ 𝐛 𝐜]
̂ ) = 7 d) 𝐫 ∙ (𝐢̇̂ + ̂𝐣̇ − 2𝐤
c) 𝐫 ∙ (𝐢̇̂ − ̂𝐣̇ + 2𝐤 ̂ ) = 10 d) 𝐫 ∙ (𝐚⃗ + 𝐛 + 𝐜) = 0
8. If 𝑛̂1 , 𝑛̂2 are two unit vectors and 𝜃 is the angle 17. If the point whose position vectors are 2𝐢̇̂ + 𝐣̇̂ + 𝐤 ̂,
between them, then cos 𝜃/2 = ̂ and 14𝐢̇̂ − 5𝐣̇̂ + 𝑝𝐤
6𝐢̇̂ − 𝐣̇̂ + 2𝐤 ̂ are collinear, then
1 1
a) |𝑛̂1 + 𝑛̂2 | b) |𝑛̂1 − 𝑛̂2 | the value of 𝑝 is
2 2
1 |𝑛̂ ×𝑛̂ | a) 2 b) 4
c) (𝑛̂1 . 𝑛̂2 ) d) 2|𝑛̂1 | |𝑛̂2 |
2 1 2 c) 6 d) 8
9. If the points with position vectors 60𝐢̇̂ + 3𝐣̇̂, 40𝐢̇̂ −
8𝐣̇̂ and 𝑎𝐢̇̂ − 52𝐣̇̂ are collinear, then 𝑎 is equal to

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18. Let 𝐴𝐵𝐶 be a triangle, the position vectors of ⃗
a) 𝟎 b) 0
whose vertices are respectively 7𝐢̇̂ + 10𝐤 ̂ , −𝐢̇̂ + c) 𝐚⃗ × 𝐛 d) 𝐛 × 𝐚⃗
̂ and −4𝐢̇̂ + 9𝐣̇̂ + 6𝐤
6𝐣̇̂ + 6𝐤 ̂ Then, the ∆𝐴𝐵𝐶 is 20. If the position vectors of 𝐴, 𝐵 and 𝐶are
a) Isosceles ̂ and 3𝐢̇̂ − 4𝐉̂̇ −
̂ , 𝐢̇̂ − 3𝐣̇̂ − 5𝐤
respectively 2𝐢̇̂ − 𝐣̇̂ + 𝐤
b) Equilateral ̂ then cos 2 𝐴 is equal to
4𝐤
c) Right angled isosceles 6
a) 0 b) 41
d) None of these
35
19. If 𝐚⃗, 𝐛 are any two vwctors, then (2𝒂
⃗ + 3𝐛) × c) 41
d) 1
(5𝐚⃗ + 7𝐛) + 𝐚⃗ × 𝐛 is equal to

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Solution
1 (a) 6 (c)
⃗ , c are mutually perpendicular unit vectors.
Since, a⃗, b We have,
1
⇒ |𝐚⃗| = | 𝐛| = |𝐜| = 1 𝑎̂ × (𝑏̂ × 𝑐̂ ) = 𝑏̂
2
and 𝐚⃗ ∙ 𝐛 = 𝐛 ∙ 𝐜 = 𝐜 ∙ 𝐚⃗ = 0 …..(i) 1
2 ⇒ (𝑎̂ ∙ 𝑐̂ )𝑏̂ − (𝑎̂ ∙ 𝑏̂)𝑐̂ = 𝑏̂
Now, |𝐚⃗ + 𝐛 + 𝐜| = ( 𝐚⃗ + 𝐛 + 𝐜) ∙ ( 𝐚⃗ + 𝐛 + 𝐜) 2
2 1
= |𝐚⃗|2 + |𝐛| + |𝐜|2 + 2(𝐚⃗ ∙ 𝐛 + 𝐛 ∙ 𝐜 + 𝐜 ∙ 𝐚⃗ ) ⇒ {(𝑎̂ ∙ 𝑐̂ ) − } 𝑏̂ − (𝑎̂ ∙ 𝑏̂)𝑐̂ = 0
2
= 1 + 1 + 1 + 0 = 3 [from Eq. (i) ] 1
⇒ 𝑎̂ ∙ 𝑐̂ − = 0 and 𝑎̂ ∙ 𝑏̂
⇒ |𝐚⃗ + 𝐛 + 𝐜| = √3 2
=0 [ ∵ 𝑏̂, 𝑐̂ ]
2 (c) are non − collinear vectors
1
∵ 𝐩⃗ ,𝐪
⃗ , 𝐫 are reciprocal vectors 𝐚⃗, 𝐛, 𝐜 respectively. ⇒ cos 𝜃 = 2, where 𝜃 is the angle between 𝑎̂ and 𝑐̂
∴𝐩⃗ ∙ 𝐚⃗ = 1, 𝐩⃗ ∙ 𝐛 = 0, 𝐩
⃗ ∙ 𝐜 etc. ⇒ 𝜃 = 𝜋/3
∴ (𝑙𝐚⃗ + 𝑚𝐛 + 𝑛𝐜) ∙ (𝑙𝐩 ⃗ + 𝑚𝐪 ⃗ + 𝑛𝐫) = 𝑙 2 + 𝑚2 + 𝑛2
7 (b)
The given line is parallel to the vector 𝐧 ̂.
⃗ = 𝐢̇̂ − 𝐣̇̂ + 2𝐤
3 (a)
The required plane passing through the point (2, 3, 1)
̂ and ⃗⃗⃗⃗⃗⃗
Let 𝐎𝐀 = 𝐢̇̂ + 2𝐣̇̂ + 3𝐤
⃗⃗⃗⃗⃗⃗ ̂
𝐎𝐁 = 3𝐢̇̂ + 4𝐣̇̂ + 5𝐤
𝑖. 𝑒. , 2𝐢̇̂ + 3𝐣̇̂ + 𝐤̂ and is perpendicular to the vector
𝐀𝐁 = 2𝐢̇̂ + 2𝐣̇̂ + 2𝐤
∴ ⃗⃗⃗⃗⃗ ̂
⃗ = 𝐢̇̂ − 𝐣̇̂ + 2𝐤
𝐧 ̂
∴ work don, 𝑊 = 𝐅 ∙ ⃗⃗⃗⃗⃗𝐀𝐁 ∴ Its equation is
̂ ̂ ̂ ̂)
= (2𝐢̇ − 3𝐣̇ + 2𝐤) ∙ (2𝐢̇̂ + 2𝐣̇̂ + 2𝐤 ̂ )] ∙ (𝐢̇̂ − ̂𝐣̇ + 2𝐤
̂)=0
[(𝐫 − (2𝐢̇̂ + 3𝐣̇̂ + 𝐤
=4−6+4=2 ⟹ 𝐫 ∙ (𝐢̇̂ − ̂𝐣̇ + 2𝐤 ̂)=1

4 (a)
8 (a)
We have,
We have,
(𝑎 × 𝑏⃗) × 𝑐 = 𝑎 × (𝑏⃗ × 𝑐) |𝑛̂1 + 𝑛̂2 |2 = |𝑛̂1 | + |𝑛̂2 | + 2𝑛̂1 ∙ 𝑛̂2
⇔ −𝑐 × (𝑎 × 𝑏⃗) = 𝑎 × (𝑏⃗ × 𝑐) ⇒ |𝑛̂1 + 𝑛̂2 |2 = |𝑛̂1 |2 + |𝑛̂2 |2 + 2|𝑛̂1 | + |𝑛̂2 | cos 𝜃
⇔ −{(𝑐 ∙ 𝑏⃗)𝑎 − (𝑐 ∙ 𝑎)𝑏⃗} = (𝑎 ∙ 𝑐 )𝑏⃗ − (𝑎 ∙ 𝑏⃗)𝑐 𝜃
⇒ |𝑛̂1 + 𝑛̂2 |2 = 1 + 1 + 2 cos 𝜃 = 4 cos2
⇔ (𝑎 ∙ 𝑏⃗)𝑐 − (𝑐 ∙ 𝑏⃗)𝑎 = 0 2
𝜃 1
⇔ (𝑏⃗ ∙ 𝑎)𝑐 − (𝑏⃗ ∙ 𝑐)𝑎 = 0 ∴ cos = |𝑛̂1 + 𝑛̂2 |
2 2
⇔ 𝑏⃗ × (𝑐 × 𝑎) = 0
9 (a)
5 (a) Let 𝑃(60𝐢̇̂ + 3𝐣̇̂), 𝑄(40𝐢̇̂ − 8𝐣̇̂)and 𝑅(𝑎𝐢̇̂ − 52𝐣̇̂) be the
𝐏𝐐 = (2𝐢̇̂ − ̂𝐣̇ + 3𝐤
⃗⃗⃗⃗⃗ ̂ ) − (𝐢̇̂ − ̂𝐣̇ + 2𝐤
̂) collinear points. Then 𝐏𝐐 ⃗⃗⃗⃗⃗ = 𝜆𝐐𝐑
⃗⃗⃗⃗⃗⃗
= 𝐢̇̂ + 𝐤 ̂ for some scalar λ
and 𝐅 = 3𝐢̇̂ + 2𝐣̇̂ − 4𝐤 ̂ ⟹ (−20𝐢̇̂ − 11𝐣̇̂) = 𝜆[(𝑎 − 40)𝐢̇̂ − 44𝐣̇̂]
∴ Moment = |𝐏𝐐 ⃗⃗⃗⃗⃗ × 𝐅| ⟹ 𝜆(𝑎 − 40) = −20, −44𝜆 = −11
̂ 1
𝐢̇̂ 𝐣̇̂ 𝐤 ⟹ 𝜆(𝑎 − 40) = −20, 𝜆 =
= |1 0 1 | 4
∴ 𝑎 − 40 = −20 × 4 ⟹ 𝑎 = −40
3 2 −4
= −2𝐢̇̂ + 7𝐣̇̂ + 2𝐤 ̂
10 (c)
∴Magnitude of moment= √4 + 49 + 4 = √57

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𝐢̇̂ + ̂𝐣̇ 15 (a)
Let the unit vector is perpendicular to 𝐢̇̂
√2
̂
− 𝐣̇, then we get 16 (b)
(𝐢̇̂ + 𝐣̇̂) ∙ (𝐢̇̂ − 𝐣̇̂) 1 − 1 We know that the equation of the plane passing
= =0 through three non-collinear points 𝐚⃗, 𝐛, 𝐜 is
√2 √2
𝐢̇̂ + ̂𝐣̇ 𝐫 ∙ (𝐛 × 𝐜 + 𝐜 × 𝐚⃗ + 𝐚⃗ × 𝐛) = [𝐚⃗ 𝐛 𝐜]
∴ is the unit vector
√2
17 (b)
11 (b) 2 1 1
Given vectors 2𝑖̂ + 3𝑗̂ − 4𝑘̂ and 𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂ will be Given vectors are collinear, if | 6 −1 2| = 0
14 −5 p
perpendicular, if
⟹ 2[−𝑝 + 10] − 1[6𝑝 − 28] + 1[−30 + 14] = 0
(2𝑖̂ + 3𝑗̂ − 4𝑘̂ ). (𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂ ) = 0 ⇒ 2𝑎 + 3𝑏 − 𝑐
⟹ −8𝑝 + 32 = 0
=0 ⟹𝑝=4
Clearly, 𝑎 = 4, 𝑏 = 4, 𝑐 = 5 satisfy the above equation
18 (c)
12 (d) ̂, 𝐁 ̂ and 𝐂 = −4𝐢̇̂ +
Let 𝐀 ⃗ = 7𝐣̇̂ + 10𝐤 ⃗ = −𝐢̇̂ + 6𝐣̇̂ + 6𝐤
̂ and 𝐝 = 𝑎𝐢̇̂ + 𝑏𝐣̇̂ − 15𝐤
Let 𝐜 = 3𝐢̇̂ + 𝐣̇̂ − 5𝐤 ̂
̂
9𝐣̇̂ + 6𝐤
̂𝐢̇ ̂𝐣̇ ̂𝐤
𝐀𝐁 = −𝐢̇̂ − ̂𝐣̇ − 4𝐤
Now, ⃗⃗⃗⃗⃗ ̂, 𝐁𝐂 = −3𝐢̇̂ + 3𝐣̇̂
⃗⃗⃗⃗⃗
For collinears, 𝐜 × 𝐝 = |3 1 −5 | = ⃗𝟎
𝐂𝐀 = 4𝐢̇̂ − 2𝐣̇̂ + 4𝐤
and ⃗⃗⃗⃗⃗ ̂
𝑎 𝑏 −15
⟹ 𝐢̇̂(−15 + 5𝑏) − 𝐣̇̂(−45 + 5𝑎) + 𝐤 ̂ (3𝑏 − 𝑎) = 𝟎
⃗ Here, |𝐀𝐁⃗⃗⃗⃗⃗ | = |𝐁𝐂
⃗⃗⃗⃗⃗ | = 3√2 and |𝐂𝐀 ⃗⃗⃗⃗⃗ | = 6
⟹ −15 + 5𝑏 = 0, −45 + 5𝑎 = 0, 3𝑏 − 𝑎 = 0 Now, |𝐀𝐁 ⃗⃗⃗⃗⃗ |2 + |𝐁𝐂
⃗⃗⃗⃗⃗ |2 = |𝐀𝐂
⃗⃗⃗⃗⃗ |2
⟹ 𝑏 = 3, 𝑎 = 9 Hence, the triangle is right angled isosceles triangle.

13 (c) 19 (a)
̂ , 𝐛 = 2𝐢̇̂ + 0𝐣̇̂ + 0𝐤
Let 𝐚⃗ = 𝐢̇̂ − 2𝐣̇̂ − 3𝐤 ̂ (2𝐚⃗ + 3𝐛) × (5𝐚⃗ + 7𝐛) + 𝐚⃗ × 𝐛
Now take option (c). = ⃗𝟎 + 14(𝐚⃗ × 𝐛) − 15(𝐚⃗ × 𝐛) + ⃗𝟎 + 𝐚⃗ × 𝐛
Let 𝐜 = 0𝐢̇̂ − 4𝐣̇̂ − 6𝐤 ̂
⃗
=𝟎
1 −2 −3
Now, 𝐚⃗ ∙ (𝐛 × 𝐜) = |2 0 0|
0 −4 −6 20 (c)
= 1(0) + 2(−12) − 3(−8) = 0 𝐎𝐀 = 2𝐢̇̂ − 𝐣̇̂ + 𝐤
Let ⃗⃗⃗⃗⃗⃗ ̂ , 𝐎𝐁 ̂
⃗⃗⃗⃗⃗⃗ = 𝐢̇̂ − 3𝐣̇̂ − 5𝐤
and 𝐎𝐂⃗⃗⃗⃗⃗ = 3𝐢̇̂ − 4𝐣̇̂ − 4𝐤 ̂
14 (a) ⃗⃗⃗⃗⃗⃗ | = √6, 𝑏 = |𝐎𝐁 ⃗⃗⃗⃗⃗⃗ | = √35
∴ 𝑎 = |𝐎𝐀
𝐀𝐁 = (2 − 1)𝐢̇̂ + (0 − 2)𝐣̇̂ + (3 + 1)𝐤
⃗⃗⃗⃗⃗ ̂
⃗⃗⃗⃗⃗ | = √41
and 𝐜|𝐎𝐂
= 𝐢̇̂ − 2𝐣̇̂ + 4𝐤 ̂
𝑏 2 + 𝑐 2 + 𝑎2
and ∴ cos 𝐴 =
⃗⃗⃗⃗⃗ = (3 − 1)𝐢̇̂ + (−1 − 2)𝐣̇̂ + (2 + 1)𝐤 ̂ 2𝑏𝑐
𝐀𝐂 2 2 2
̂ ̂ ̂ (√35) + (√41) − (√6)
= 2𝐢̇ − 3𝐣̇ + 3𝐤 =
̂ ) ∙ (2𝐢̇̂ − 3𝐣̇̂ + 3𝐤
(𝐢̇̂ − 2𝐣̇̂ + 4𝐤 ̂) 2√35 √41
cos θ =
√1 + 4 + 16√4 + 9 + 9 35
2 + 6 + 12 20 ⟹ cos 𝐴 = √
41
= =
√21√22 √462 35
⟹ √462 cos θ = 20 ⟹ sin2 𝐴 =
41

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