Magnitude and Directions of a Vector MCQ Quiz - Objective Question with Answer for Magnitude and Directions of a Vector - Download Free PDF

Last updated on May 1, 2025

Latest Magnitude and Directions of a Vector MCQ Objective Questions

Magnitude and Directions of a Vector Question 1:

Find the magnitude of vector 5a, where a=2i^+3j^+7k^?

  1. 562
  2. 560
  3. 62
  4. 40

Answer (Detailed Solution Below)

Option 1 : 562

Magnitude and Directions of a Vector Question 1 Detailed Solution

Concept:

Magnitude of vector z=ai^+bj^+ck^ then magnitude of vector is given by |z|=(a2+b2+c2)

Calculation:

Given: Let z=5awhere a=2i^+3j^+7k^

⇒ z=5a=10i^+15j^+35k^

As we know that, if z=ai^+bj^+ck^ then |z|=(a2+b2+c2)

⇒ |z|=102+152+352=1550

⇒  |z|=562

Hence, option 1 is correct.

Magnitude and Directions of a Vector Question 2:

The magnitude of a given vector with end points (5, –5, 0) and (2, 3, 0) must be ______.

  1. 83
  2. 65
  3. 97
  4. 73
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : 73

Magnitude and Directions of a Vector Question 2 Detailed Solution

Given:

The vector with end points (5, –5, 0) and (2, 3, 0) 

Concept:

Let a and b be the position vector of the points A and B respectively,

then the vector equation of the line passing through A and B is given by I r I = I(ba)I

Calculation:

Let the position vector of points A (-1, 0, 2) and B (3, 4, 6) be a and b respectively.

Then, a=5^i5j^+0k^

And, b=2i^+3j^+0k^

⇒  ba=3i^+8j^+0k^

⇒ I r I = I3i^+8j^+0k^I

⇒ I r I (3)2(8)2+(0)2

⇒ I r I ​√​73

∴ Correct answer is √​73.

Magnitude and Directions of a Vector Question 3:

The magnitude of a given vector with end points (5, –5, 0) and (2, 3, 0) must be ______.

  1. 83
  2. 65
  3. 97
  4. 73

Answer (Detailed Solution Below)

Option 4 : 73

Magnitude and Directions of a Vector Question 3 Detailed Solution

To determine the magnitude of a vector with given endpoints:

Magnitude=(x2x1)2+(y2y1)2+(z2z1)2.

Calculation:

(x1,y1,z1)=(5,5,0) and (x2,y2,z2)=(2,3,0).

Step 1: Subtract the coordinates
x2x1=25=3,y2y1=3(5)=8,z2z1=00=0.

Step 2: Square the differencesStep 3: Sum the squares

(3)2=9,82=64,02=0.

⇒ 9 + 64 + 0 = 73

Step 4: Take the square rootFinal Answer:

Magnitude=73.

Hence, The Correct Answer is Option 4.

Magnitude and Directions of a Vector Question 4:

Given that A+B+C=0, out of these three vectors two are equal in magnitude and the magnitude of the third vector 2 times as that of either of the two having equal magnitude. Then the angles between vectors are given by :-

  1. 30,60,90
  2. 45,45,90
  3. 45,60,90
  4. 90,135,135

Answer (Detailed Solution Below)

Option 4 : 90,135,135

Magnitude and Directions of a Vector Question 4 Detailed Solution

A+B+C=0A+B=C

A2+B2+2ABcosθ1=C2(A=B)

B2+B2+2ABcosθ1=2B2(C=2B)

2ABcosθ1=0cosθ1=0θ1=90

B+C=AB2+C2+2BCcosθ2=A2

B2+2B2+2B2Bcosθ2=B2

2B2(1+2cosθ2)=12θ2=135

Thus, angles are 90,135,135

Magnitude and Directions of a Vector Question 5:

Scalar projection of the line segment joining the points A(-2, 0,3), B(1, 4, 2) on the line whose direction ratios are 6, -2, 3 is

  1. 237
  2. 1
  3. 7
  4. 17

Answer (Detailed Solution Below)

Option 2 : 1

Magnitude and Directions of a Vector Question 5 Detailed Solution

Answer : 2

Solution :

Let a̅ be the vector joining A(-2, 0, 3) and B(1, 4, 2).

∴ a=(1(2))i^+(40)j^+(23)k^

3i^+4j^k^

and b = b¯=6i^2j^+3k^

∴ Projection = a b| b|=3×6+4×(2)1×362+(2)2+32

188349

77 

= 1

Top Magnitude and Directions of a Vector MCQ Objective Questions

What is the value of p for which the vector p(2î - ĵ + 2k̂) is of 3 units length?

  1. 1
  2. 2
  3. 3
  4. 6

Answer (Detailed Solution Below)

Option 1 : 1

Magnitude and Directions of a Vector Question 6 Detailed Solution

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Concept:

Let a=xi+yj+zk then magnitude of the vector of a = |a|=x2+y2+z2

Calculation:

Let a =  p(2î - ĵ + 2k̂)

Given, |a|=3

⇒ 4p2+p2+4p2=3

⇒ 9p2=3

⇒ 3p = 3

∴ p = 1

If A = 5i^2j^+4k^ and B = i^+3j^7k^ , then what is the value of |AB|?

  1. 6√2
  2. 7√2
  3. 8√2
  4. 9√2

Answer (Detailed Solution Below)

Option 4 : 9√2

Magnitude and Directions of a Vector Question 7 Detailed Solution

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Concept:

If A=xi^yj^+zk^, then |A|=x2+y2+z2

Calculation:

Given A = 5i^2j^+4k^ and B = i^+3j^7k^

AB=BA

AB = i^+3j^7k^(5i^2j^+4k^)

AB = 4i^+5j^11k^

Now |AB|=(4)2+52+(11)2

|AB|=16+25+121

|AB|=162 = 9√2

If a = 2î + ĵ + k̂ and b = î + 2ĵ + k̂, then the magnitude of their resultant is:

  1. 2√5
  2. 2√6
  3. √22
  4. None of these.

Answer (Detailed Solution Below)

Option 3 : √22

Magnitude and Directions of a Vector Question 8 Detailed Solution

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Concept:

The magnitude of a vector A = a1î + a2ĵ + a3k̂ is given as |A|=a12+a22+a32.

The magnitude of the sum of vectors A and B can also be calculated as |A+B|=A2+B2+2AB.

The resultant of a set of vectors acting at a point is simply the algebraic sum of the vectors.

Calculation:

The resultant of the vectors a = 2î + ĵ + k̂ and ​b​ = î + 2ĵ + k̂ is:

r=a+b = (2î + ĵ + k̂) + (î + 2ĵ + k̂) = 3î + 3ĵ + 2k̂

Now, |a+b|=32+32+22=22.

What is the value of k for which the vector k(2î -  ĵ -  2k̂) is of 6 units length?

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

Answer (Detailed Solution Below)

Option 2 : 2

Magnitude and Directions of a Vector Question 9 Detailed Solution

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Concept:

Length of the vector ai^+bj^+ck^ from origin is a2+b2+c2

Calculation:

Length of the vector k(2î -  ĵ -  2k̂) from origin is 

(2k)2+(k)2+(2k)2

4k2+k2+4k2 

9k2 

= 3k

Length is 6 units given

3k = 6

k = 6/3

k = 2

Hence option 2 is correct.

If a=3i^+2j^andb=2i^+3j^ are two vectors then find the value of |ab| ?

  1. 7
  2. 5
  3. 3
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Magnitude and Directions of a Vector Question 10 Detailed Solution

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CONCEPT:

If a=a1i^+a2j^+a3k^ is a vector then magnitude of a is given by: |a|=a12+a22+a32

CALCULATION:

Given: a=3i^+2j^andb=2i^+3j^

Here, we have to find the value of |ab|

⇒ ab=i^j^

As we know that, if a=a1i^+a2j^+a3k^  then magnitude of a is given by: |a|=a12+a22+a32

⇒ |ab|=12+(1)2+02=2

Hence, option D is the correct answer.

Find the direction cosines of the vector 7î + 4ĵ - 3k̂.

  1. 774474374
  2. 774474374
  3. Both options 1 and 2
  4. None of these

Answer (Detailed Solution Below)

Option 3 : Both options 1 and 2

Magnitude and Directions of a Vector Question 11 Detailed Solution

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Concept:

The direction cosines of the vector aî + bĵ + ck̂ are given by α = ±aa2+b2+c2, β = ±ba2+b2+c2 and γ = ±ca2+b2+c2.

Calculation:

For the given vector 7î + 4ĵ - 3k̂, a = 7, b = 4 and c = -3.

The direction cosines of the vector are:

α = ±772+42+(3)2, β = ±472+42+(3)2 and γ = ±372+42+(3)2

⇒ α = ±774, β = ±474 and γ = 374 

∴ (α , β , γ ) = (774,474,374) or (774,474,374​)

The vector (cosαcosβ)i^+(cosαsinβ)j^+(sinα)k^ is

  1. Parallel to (i^+j^+k^)
  2. Null vector
  3. Unit vector
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : Unit vector

Magnitude and Directions of a Vector Question 12 Detailed Solution

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Concept:

Unit vector: a vector that has a magnitude of one.

  • Let a=xi+yj+zk
  • Magnitude of vector of a = |a|=x2+y2+z2
  • Unit vector = a^=a|a|

 

Calculation:

Given Vector is (cosαcosβ)i^+(cosαsinβ)j^+(sinα)k^

LetA=(cosαcosβ)i^+(cosαsinβ)j^+(sinα)k^

Now calculate the magnitude of Vector A,

|A|=(cos αcos β)2 + (cos αsin β)2 + (sin α)2

|A|=cos2α[cos2β+sin2β]+(sinα)2

|A|=cos2α+sin2α=1

So, A is a unit vector.

If the position vectors of points A and B are i^+2j^k^ and 3i^2j^k^ respectively, then what is the length of AB?

  1. √5
  2. 3√5
  3. 2√5 
  4. 4

Answer (Detailed Solution Below)

Option 3 : 2√5 

Magnitude and Directions of a Vector Question 13 Detailed Solution

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Concept:

If A and B are points with position vectors aandb respectively then AB=ba

If a=a1i^+a2j^+a3k^ is a vector then the magnitude of the vector is given by

|a|=a12+a22+a32

 

Calculation:

Given: The position vectors of points A and B are i^+2j^k^ and 3i^2j^k^ respectively

As we know, If A and B are points with position vectors aandb respectively then AB=ba

AB=(3i^2j^k^)(i^+2j^k^)=2i^4j^

As we know that, If a=a1i^+a2j^+a3k^ is a vector then the magnitude of the vector is given by

|a|=a12+a22+a32

|AB|=(2)2+(4)2+(0)2=20=25

What are the values of x for which the angle between the vectors 2x2i^ + 3xj^ + k^ and i^ −2j^ + x2k^ is obtuse ?

  1. 0 < x < 2
  2. x < 0
  3. x > 2
  4. 0 ≤ x ≤ 2

Answer (Detailed Solution Below)

Option 1 : 0 < x < 2

Magnitude and Directions of a Vector Question 14 Detailed Solution

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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

Calculation:

Given: The angle between the vectors 2x2i^ + 3xj^ + k^ and i^ −2j^ + x2k^ is obtuse

The angle between the vectors 2x2i^ + 3xj^ + k^ and i^ −2j^ + x2k^ is given by

cosθ=(2x2i^+3xj^+k^).(i^2j^+x2k^)|2x2i^+3xj^+k^||i^2j^+x2k^|

⇒ cosθ=(2x2(1)+3x(2)+1(x2))(2x2)2+(3x)2+(1)2(1)2+(2)2+(x2)2

⇒ cosθ=3x26x4x4+9x2+15+x4

Since θ is obtuse, 

⇒ cos θ < 0

⇒ 3x2 - 6x < 0

⇒ x(x - 2) < 0

⇒ 0 < x < 2

∴ The correct option is (1).

Determine |a|and|b|,if(a+b).(ab)=8 and|a|=8|b|?

  1. 16237,2237
  2. 8237,2237
  3. 16637,2237
  4. 8337,2337

Answer (Detailed Solution Below)

Option 1 : 16237,2237

Magnitude and Directions of a Vector Question 15 Detailed Solution

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Given:

(a+b).(ab)=8 

|a|=8|b|

Calculation:

We have,

(a+b).(ab)=8 

⇒ a.a a.b+b.ab.b=8

⇒  |a|2a.b+a.b|b|2=8        [∵ b.a=a.b]

⇒ |a|2|b|2=8

⇒ (8|b|)2|b|2=8       [|a|=8|b|]

⇒ 64|b|2|b|2=8

⇒ 63|b|2=8

⇒ |b|2=863

⇒ |b|=863

⇒ |b|=2237

⇒ |a|=8|b|

Put the value of b

⇒ |a|=8×2237

⇒ |a|=16237 

∴ |a|=16237  and |b|=2237

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