Vector Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Vector Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:  

Calculation:

Given

Additional Information

Properties of Scalar Product

 (Scalar product is commutative)

 (Distributive of scalar product over addition)

In terms of orthogonal coordinates for mutually perpendicular vectors, it is seen that 

Properties of Vector Product

Concept:

•  and  are two vectors parallel to each other ⇔  
• Cross product of parallel vectors are zero ⇔  
• A cross or vector product is not commutative ⇔ 

Calculation:

We have to find the value of 

We know that 

∴ Option 3 is correct.

Concept:

Dot Product: it is also called the inner product or scalar product

• Let the two vectors be  then dot Product of two vector:   Where, || = Magnitude of vectors a and || = Magnitude of vectors b and θ is angle between a and b 

Calculation:

Let

⇒ 1 + 0 + 0 = √3 × 1 × cos θ

⇒ cos θ = 1/√3

∴ θ = cos⁻¹ (1/√3)

Concept:

Length of the vector  from origin is 

Calculation:

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

=  

=  

= 3k

Length is 6 units given

3k = 6

k = 6/3

k = 2

Hence option 2 is correct.

Concept:

If θ is the angle between x and y then x.y = xycosθ 

Calculation:

GIven, a + b + c = 0

⇒ a + b = -c, squaring both sides we get

⇒  (a + b)² = (-c)²

⇒ |a + b|2 = |c|2

⇒ |a|² + |b|² +2|a||b|cosθ = |c|², where θ = angle betwee a and b

Putting the values we get,

3² + 5² + 2× 3× 5 cosθ = 7²

⇒ 9 + 25 + 30cosθ = 49

⇒ 30cosθ = 49 - 34 = 15

⇒ cosθ = 

∴ θ = 60° 

Concept:

Dot Product: it is also called the inner product or scalar product

Let the two vectors are  and 

Dot Product of two vectors is given by:   = |a||b| cos θ

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

Formulas of Dot Product:

Calculation:

Given that,

(â + b̂ + ĉ) = 0    ----(1)

We know that,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

⇒ (â + b̂ + ĉ)² = â ⋅ â + b̂ ⋅ b̂ + ĉ ⋅ ĉ + 2(â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) 

From equation (1), we get

⇒ (1 + 1 + 1) + 2 (â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) = 0

∴ (â ⋅ b̂ + b̂ ⋅ ĉ + ĉ ⋅ â) = - 3/2        

Concept:

For unit vectors:

When two vectors and  are perpendicular to each other then the dot product is zero i.e.

Calculation:

Given:

, 

  [∵ perpendicular to each other]

∴ θ = π/3

Concept:

If vectors  are perpendicular then 

Calculation:

Given:  and  are perpendicular

Let  and 

We know that, If vectors  are perpendicular then 

⇒ -2 - 20 - λ = 0 

⇒ -22 - λ = 0 

∴ λ = -22

Concept:

If  are two vectors parallel to each other then  or 

Calculation:

Given:

  and  are parallel vectors,

Therefore, 

Equating the coefficient of 

⇒ x = 2λ                    .... (1)

⇒ -2 = -4λ 

∴ λ = 1/2

Put the value of λ in equation (1), we get

x = 2 × (1/2)

So, x = 1

Concept:

Scaler triple product of the vectors:

The scaler triple product of the vectors  and  is given by:

Coplaner vectors:

Three vectors  and  are said to be coplaner if the scaler triple product

Solution:

Let the given vectors be  and .

It is given that the vectors are coplaner therefore the scaler triple product

Therefore,

Perform the coloumn operation C₁ – C₂ as follows:

Now perform another coloum operation C₂ – C₃ as follows:

Take (1 - a)(1 - b)(1 - c) common. Note that it is given that a,b,c ≠ 0 therefore this action is jstified.

Since a, b, c ≠ 0 therefore (1 - a)(1 - b)(1 - c) ≠ 0.Therefore the determinant has to be zero.

Simplify the above expression as follows:

Therefore, .