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Latest Angle between Vectors MCQ Objective Questions

Top Angle between Vectors MCQ Objective Questions

Angle between Vectors Question 1:

If a and b are vectors such that |a|=13,|b|=2 and a×b=3i^+4j^+12k^ then what is the acute angle between a and b?

  1. 60°
  2. 45°
  3. 30°
  4. 90°

Answer (Detailed Solution Below)

Option 3 : 30°

Angle between Vectors Question 1 Detailed Solution

Concept:

If two vector a and b then |a×b|=|a||b|sinθ

Where θ is the angle between vector  a and b.

Consider vector  A=xi^+yj^+zk^ then magnitude of A =  |A| = x2+y2+z2

 

Calculation:

Given that,

|a|=13,|b|=2

a×b=3i^+4j^+12k^

Taking Magnitude both sides, we get

⇒ |a×b|=|3i^+4j^+12k^|

⇒ |a||b|sinθ=32+42+122

⇒ 13 × 2 × sin θ = 169

⇒ 13 × 2 × sin θ = 13

⇒ sin θ = 12

∴ θ = 30°

Angle between Vectors Question 2:

If a and b are unit vectors and θ is angle between them, then |ab|2 is:

  1. sin θ
  2. sin 2θ
  3. sinθ2
  4. sin2 θ

Answer (Detailed Solution Below)

Option 3 : sinθ2

Angle between Vectors Question 2 Detailed Solution

Concept:

Let a=xi+yj+zk

The magnitude of the vector of a = |a|=x2+y2+z2

Unit vector = a^=a|a|

Let the two vectors be aandb then dot Product of two vectors: ab=|a||b|cosθ

Where, |a| = Magnitude of vectors a and |b| = Magnitude of vectors b and θ is the angle between a and b

Calculation:

Given: aandb are unit vectors

|a|=1and|b|=1

Now,

|ab|2=(ab)(ab)=|a|2+|b|22(ab)

|ab|2=|a|2+|b|22|a||b|cosθ

 = 1 + 1 - 2 × 1 × 1 cos θ

= 2 - 2 cos θ

= 2 (1 - cos θ)

= 2 × 2 sin2 (θ/2)

|ab|=2sin(θ2)

|ab2|=sin(θ2)

Angle between Vectors Question 3:

If |a|=1,|b|=3 and |ab|=7, then the angle between a and b is

  1. 0
  2. π6
  3. π4
  4. π3

Answer (Detailed Solution Below)

Option 4 : π3

Angle between Vectors Question 3 Detailed Solution

Solution:

Given that, |a|=1,|b|=3 and |ab|=7

⇒ |ab|2=(7)2

⇒ |a|2+|b|22a.b=7

⇒ (1)2+(3)22a.b=7

⇒ 1+92a.b=7

⇒ 2a.b=107

⇒ a.b=32

⇒ |a||b|cosθ=32

⇒ 1×3×cosθ=32

⇒ cosθ=12

⇒ θ=π3

Hence, The angle between a and b  is π3

∴ The correct option is (4)

Angle between Vectors Question 4:

If a¯ and b¯ are unit vectors and θ is the angle between them then |a¯b¯2| is

  1. sinθ2
  2. sin θ
  3. 2 sin θ
  4. sin 2θ 

Answer (Detailed Solution Below)

Option 1 : sinθ2

Angle between Vectors Question 4 Detailed Solution

Concept:

Angle between vectors aandb is θ

then a.b=|a||b|cosθ

If aandb are unit vectors, then we have 

a.b=cosθ (|a|=|b|=1)

Calculation:

Given: that aandb are two unit vectors

|ab|2=|a|2+|b|22a.b

=1+12|a||b|cosθ

|ab|2=22cosθ

We know that cosθ=12sin2(θ2)

1cosθ=2sin2(θ2)

|ab|2=2(1cosθ)

|ab|2=2×2sin2θ2

Taking square root on both sides.

|ab2|=sinθ2

Angle between Vectors Question 5:

The position vectors of vertices A, B and C of triangle ABC are respectively j^+k^,3i^+j+5k^^ and 3j^+3k^. What is angle C equal to?

  1. π6
  2. π4
  3. π3
  4. π2

Answer (Detailed Solution Below)

Option 4 : π2

Angle between Vectors Question 5 Detailed Solution

Concept:

  • The angle between two vectors  a and b is given by, cosθ=a.b|a||b|​ 
  • If a = a1i^ + a2j^ + a3k^ and  b = b1i^ + b2j^ + b3k^, then  a.b=a1b1+a2b2+a3b3
  • If position vectors of A and B are a1i^ + a2j^ + a3k^ and b1i^ + b2j^ + b3k^ respectively, then the vector AB = (b1 - a1)i^ + (b2 - a2)j^ + (b3 - a3)k^

F1 Madhuri Defence 20.09.2022 D13

Calculation:

Given: The position vectors of vertices A, B and C of triangle ABC are respectively j^+k^,3i^+j+5k^^ and 3j^+3k^.

⇒ AB = (30)i^+(11)j+(51)k^^=3i^+4k^

 BC = (03)i^+(31)j+(35)k^^=3i^+2j^2k^

 CA = (00)i^+(31)j+(31)k^^=2j^+2k^

The angle between the vectors BC and CA is angle C

cosC=BC.CA|BC||CA|

⇒ cosC=(3i^+2j^2k^).(2j^+2k^)|3i^+2j^2k^||2j^+2k^|

⇒ cosC=3(0)+2(2)2(2)32+22+(2)222+22 = 0

⇒ cos C = 0

⇒ C = π2

∴ The correct option is (4).

Angle between Vectors Question 6:

The angle between the vectors î − ĵ and ĵ − k̂ is

  1. π3
  2. 2π3
  3. π3
  4. 5π6

Answer (Detailed Solution Below)

Option 2 : 2π3

Angle between Vectors Question 6 Detailed Solution

Concept:

1) Let the two vectors be a and b then the angle between a and b is given by cosθ=ab|a||b|

2) Magnitude of vector  aî + bĵ + ck̂ is a2+b2+c2

Calculation:

Let a=i^j^ and  b=j^k^

ab=1×01×10×1=1

|a|=12+(1)2+02=2  

|b|=02+12+(1)2=2

The angle between a and b is given by cosθ=ab|a||b| :

⇒ θ = cos1(1)2×2 = cos1(1)2

⇒ θ = cos1(cos2π3) = 2π3

The angle between the vectors î − ĵ and ĵ − k̂ is 2π/3.

The correct answer is option 2.

Angle between Vectors Question 7:

If a and b are vectors such that |a|=13,|b|=2 and a×b=3i^+4j^+12k^ then what is the acute angle between a and b?

  1. 60°
  2. 45°
  3. 30°
  4. 90°
  5. 120° 

Answer (Detailed Solution Below)

Option 3 : 30°

Angle between Vectors Question 7 Detailed Solution

Concept:

If two vector a and b then |a×b|=|a||b|sinθ

Where θ is the angle between vector  a and b.

Consider vector  A=xi^+yj^+zk^ then magnitude of A =  |A| = x2+y2+z2

Calculation:

Given that,

|a|=13,|b|=2

a×b=3i^+4j^+12k^

Taking Magnitude both sides, we get

⇒ |a×b|=|3i^+4j^+12k^|

⇒ |a||b|sinθ=32+42+122

⇒ 13 × 2 × sin θ = 169

⇒ 13 × 2 × sin θ = 13

⇒ sin θ = 12

∴ θ = 30°

Angle between Vectors Question 8:

If  a=2i^+3j^+4k^ and b=3i^+2j^λk^  perpendicular, then what is the value of λ

  1. 2
  2. 3
  3. 4
  4. 5

Answer (Detailed Solution Below)

Option 2 : 3

Angle between Vectors Question 8 Detailed Solution

Concept:

  • aandb are two vectors perpendicular to each other  ab=0

 

Calculation:

Given a=2i^+3j^+4k^ 1) and b=3i^+2j^λk^ perpendicular

 ab=0

(2i^+3j^+4k^)(3i^+2j^λk^)=0 

6 + 6 – 4λ = 0

12 – 4λ = 0

λ = 3

Angle between Vectors Question 9:

If the sum of two unit vectors is a unit vector, then the magnitude of their difference will be :

  1. 1
  2. 2
  3. 3
  4. 0

Answer (Detailed Solution Below)

Option 3 : 3

Angle between Vectors Question 9 Detailed Solution

Concept Used:

If the angle between two vector a and b is θ.

then |a+b|=|a|2+|b|2+2|a||b|cosθ and 

|ab|=|a|2+|b|22|a||b|cosθ

where |a| represents the magnitude of the vector.

Calculation:

Let three unit vectors be ab and c such that a+b=c

|a+b|=|a|2+|b|2+2|a||b|cosθ=|c|=1

⇒ 1+1+211cosθ=1

⇒ 2+2cosθ=1

⇒ 2cosθ=1

⇒ cosθ=12 ⇒ θ = ππ3

⇒ θ = 2π3

​Hence, the required value is:

|ab|=|a|2+|b|22|a||b|cosθ

⇒ |ab|=1+1211cos2π3

⇒ |ab|=1+1212

⇒ |ab|=1+1+1=3

Angle between Vectors Question 10:

If A and B are two vectors, then angle between vectors A+B and A×B is

  1. 60°
  2. 90°
  3. 120°

Answer (Detailed Solution Below)

Option 3 : 90°

Angle between Vectors Question 10 Detailed Solution

Concept:

For calculating the angle between two vectors, we generally use the dot product method. The dot product of two vectors a and b is defined as:

a.b=|a|.|b|cos θ

where θ is the angle between vectors a and b.

Scalar Triple Product:

 [a b c]=a.(b×c)=(a×b).c

If any two of the are same then [a b c] = 0

Calculation:

We know that,

A×B is perpendicular to both A and B.

⇒ (A + B) (A × B) cos α = (A + B).(A × B)

⇒ A.(A × B) + B.(A × B)

⇒ B.(A × A) + A.(B × B)

⇒ [B A  A] + [A  B B]

⇒ 0 + 0 =0

∴ cos α = 0

⇒ α = 90° 

Hence, The angle between vectors A and B is 90°. 

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