Magnitude and Directions of a Vector MCQ Quiz in বাংলা - Objective Question with Answer for Magnitude and Directions of a Vector - বিনামূল্যে ডাউনলোড করুন [PDF]

Let

For two vectors to be collinear,

So,

On comparing,

3 = 2λb, 4 = -3λ and -a = 5λ

λ = (-4)/3

a = 20/3

And b = (-9)/8

a – b = 20/3 - ((-9)/8)

a – b = 187/24

Calculation:

Here, 

∴

Hence, option (1) is correct. 

Concept:

Calculation:

Here,  and 

⇒

⇒ 3² = 5² + 7² + 2(5)(7) cos θ 

⇒ 9 - 25 - 49 = 70 cos θ 

⇒ cos θ = (-65/70) = (-13/14)

Hence, option (4) is correct. 

Concept:

If  is unit vector, then || = 1

For any vector , . = ||²

For any vector ,  ×  = 0

For any two vectors  and , 

For any two vectors  and , 

If  ,  and  are three coplanar vectors, then their scalar triple product is 0

If any two vectors in scalar triple product are equal, then scalar triple product is 0.

Calculation:

Given,  ,  and  be unit vectors lying on the same plane. 

⇒  ,  and  are coplanar vectors. 

that is 

and  || = 1,  || = 1 and  || = 1

Now, 

Using distributive property,

⇒ 

⇒      (∵   ×  = 0)

⇒   

Again using distributive property,

⇒ \(\{ 10(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{b}} − 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}- 8(\vec{\text{b}}×\vec{\text{c}})⋅\vec{\text{b}}\} + \{ 20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}} − 24(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{c}}- 16(\vec{\text{b}}×\vec{\text{c}})⋅\vec{\text{c}}\}\) 

As, if any two vectors in scalar triple product are equal, then scalar triple product is 0.

⇒ \(\{ 10(0) − 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}- 8(0)\} + \{ 20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}} − 24(0)- 16(0)\}\) 

⇒ \(20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}}− 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}\) 

As,  ,  and  are coplanar vectors. 

⇒ 20(0) - 12(0) = 0

∴ The correct option is (4).

Concept:

Magnitude of vector  then magnitude of vector is given by 

Calculation:

Given: Let  and 

As we know that, if  then 

⇒ 

⇒ 

By squaring both the sides, we get

⇒ 36 = 26 + λ²

⇒ λ2 = 10 

Hence, option 1 is correct.

Concept:

If vector v = ai + bj + ck, then magnitude of vector v is .

Two vectors a and b are said to be parallel vectors if one is a scalar multiple of the other. i.e., a = λb, where 'λ' is a scalar.

Formulae

(a × b) = -(b × a)

Calculation:

We have,

a × (3i + 2j + 4k) = (3i + 2j + 4k) × b

⇒ a × (3i + 2j + 4k) - (3i + 2j + 4k) × b = 0

⇒ a × (3i + 2j + 4k) + b × (3i + 2j + 4k) = 0             [(a × b) = -(b × a)]

⇒ (a + b) × (3i + 2j + 4k) = 0

So, (a + b) and (3i + 2j + 4k) are parallel vectors.

⇒ (a + b) = λ(3i + 2j + 4k)

⇒  = 

⇒ 

⇒ 

⇒ λ = ± 1

a + b = ± (3i + 2j + 4k)

Now,

(a + b).(2i - 7j + 3k) =  ± (3i + 2j + 4k).(2i - 7j + 3k)

 = ± (6 - 14 + 12)

 = ± 4

∴ The possible value of (a + b).(2i - 7j + 3k) is 4.

Given:

Formula:

The magnitude of vector aî + bĵ + ck̂ is given by √(a² + b² +c²)

Calculation:

Magnitude of  

= √[(1/√3)2 + (1/√3)2 + (1/√3)2]

= 1

Concept:

The unit vector in the direction of a vector  is given by .

Calculation:

Let  î − 2ĵ + 2k̂

The unit vector in the direction of a vector  is given by .

= 

= 

∴ Vector in the direction of the vector î − 2ĵ + 2k̂ that has magnitude 9 is

9 ×  

= 3(î − 2ĵ + 2k̂)

The vector in the direction of the vector î − 2ĵ + 2k̂ that has magnitude 9 is  3(î − 2ĵ + 2k̂).

The correct answer is option 3.

Concept:

The direction cosines of the vector aî + bĵ + ck̂ are given by α = , β =  and γ = .

Calculation:

For the given vector î + 2ĵ - k̂, a = 1, b = 2 and  = -1.

The direction cosines of the vector are:

α = , β =  and γ = 

⇒ α = , β =  and γ = 

CONCEPT:

If  is a vector then magnitude of  is given by: 

CALCULATION:

Given: 

Here, we have to find the value of 

⇒ 

⇒