Vectors

Scalers:

Those physical quantities which have magnitude but no direction is called scalars. They follow ordinary algebraic rules for the operation of scalars. Eg. speed, mass, length, time, work done, etc.

Vectors:

Those physical quantities which have both magnitude and direction and follow the rules of vector rules of operation. For Eg. velocity, displacement, force, linear momentum, acceleration, etc.

Vectors are represented by the straight lines with arrow heads. Length of the line gives the magnitude of that vector and arrow heads gives the direction.

Laws of addition of vector:

1) Head tail combination:

a) [latex]\vec{P} + \vec{Q}[/latex]

b) [latex]\vec{Q} + \vec{P}[/latex]

Triangle law of vector addition:

It states that if two vectors acting at a point simultaneously be represented in magnitude and direction by two sides of a triangle taken in the same order then third side of the triangle in the reverse order represents magnitude and direction of the resultant vector of these two vectors.

Suppose two vector P and Q acting on a point simultaneously making an angle [latex]\theta[/latex] shown in figure (a). These two vectors are represented in magnitude and direction by two sides [latex]\vec{AB}[/latex] and [latex]\vec{B}[/latex] of a triangle [latex]\Delta[/latex]ABC in the same order then third side of the triangle in a reversed order i.e. [latex]\vec{AC}[/latex] represents magnitude and direction of the resultant vectors as shown in the figure (b).

In [latex]\Delta[/latex]ABC,

[latex]\vec{AC}=\vec{AB}+\vec{BC}[/latex] 

[latex]\vec{R}=\vec{P}+\vec{Q}[/latex]………………….. (i)

Draw CD ⊥ AB produced.

In [latex]\Delta[/latex]BDC,

Cos[latex]\theta = \frac{BD}{BC}[/latex]

Or, BD = QCos[latex]\theta[/latex]

Sin[latex]\theta = \frac{CD}{BC}[/latex]

Or, CD = QSin[latex]\theta[/latex]

Then,

AD = AB + BD

              = P + QCos[latex]\theta[/latex]

Now, using Pythagoras theorem in [latex]\Delta[/latex]ADC, we get,

AC² = AD² + CD²

R² = (P + QCos[latex]\theta[/latex])² + (QSin[latex]\theta[/latex])²

R² = P² + 2PQCos[latex]\theta[/latex] + Q²Cos²[latex]\theta[/latex] + Q²Sin²[latex]\theta[/latex]

R² = P² + 2PQCos[latex]\theta[/latex] + Q²(Cos²[latex]\theta[/latex] + Sin²[latex]\theta[/latex])

R² = P² + 2PQCos[latex]\theta[/latex] + Q²

R = [latex]\sqrt{P^2+Q^2+2PQCos\theta}[/latex] ………………… (ii)

Let, β be the angle made by [latex]\vec{R}[/latex] with [latex]\vec{P}[/latex], then,

Tanβ = [latex]\frac{CD}{AD}[/latex]

Or, Tanβ = [latex]\frac{QSin\theta}{P+QCos\theta}[/latex] ………………. (iii)

Specific cases:

Case I:

If [latex]\theta[/latex] = 0^o, then,

R = [latex]\sqrt{P^2+Q^2+2PQCos0^o}[/latex]

Or, R = [latex]\sqrt{P^2+Q^2+2PQ}[/latex]

Or, R = P + Q

∴ Tanβ = [latex]\frac{QSin0^o}{P+QCos0^o}[/latex] = 0

∴ β = 0^o

Case II:

If [latex]\theta[/latex] = 90^o, then,

R = [latex]\sqrt{P^2+Q^2+2PQCos90^o}[/latex]

Or, R = [latex]\sqrt{P^2+Q^2}[/latex]

∴ Tanβ = [latex]\frac{QSin90^o}{P+QCos90^o} = \frac{Q}{P}[/latex]

∴ β = Tan^-1 [latex](\frac{Q}{P})[/latex]

Case III:

If [latex]\theta[/latex] = 180^o, then,

R = [latex]\sqrt{P^2+Q^2+2PQCos180^o}[/latex]

              = [latex]\sqrt{P^2+Q^2-2PQ}[/latex]

              = P – Q if P > Q

              = Q – P if Q > P

Tanβ = [latex]\frac{QSin180^o}{P+QCos180^o} = 0[/latex]

∴ β = 0^o.

Parallelogram law of vector addition:

It states that, if two vectors acting at a point simultaneously be represented in magnitude and direction by two adjacent sides of a parallelogram then the diagonal of it passing through the point of intersection of these adjacent sides represents magnitude and direction of the resultant vector of these two vectors.

Suppose, two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are acting at a point O simultaneously making an angle [latex]\theta[/latex] as shown in fig. (a). These two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are represented in magnitude and direction by two adjacent sides [latex]\vec{OP}[/latex] and [latex]\vec{OS}[/latex]. On parallelogram OPQS, then, the diagonal [latex]\vec{OQ}[/latex] represents the magnitude and direction of the resultant vector [latex]\vec{R}[/latex].

So, [latex]\vec{OQ}=\vec{OP}+\vec{PQ}[/latex] 

 [latex]\vec{R}=\vec{A}+\vec{B}[/latex]…….…………… (A)

Now, drop a perpendicular from Q on OM so they meet at M.

From right angled triangle [latex]\Delta[/latex]OQM,

OQ² = OM² + QM²

Or, R² = (OP + PM)² + QM²

Or, R² = OP² + 2.OP.PM + PM² + QM² ……………. (i)

In right angled [latex]\Delta[/latex]PQM,

Sin[latex]\theta = \frac{QM}{PQ} = \frac{QM}{B}[/latex]

Or, QM = BSin[latex]\theta[/latex]

And, Cos[latex]\theta = \frac{PM}{PQ} = \frac{PM}{B}[/latex]

Or, PM = BCos[latex]\theta[/latex]

Then, by substituting the values of OP, QM and PM, we get,

R² = A² + 2ABCos[latex]\theta[/latex] + B²Cos²[latex]\theta[/latex] + B²Sin²[latex]\theta[/latex]

Or, R² = A² + 2ABCos[latex]\theta[/latex] + B² (Cos²[latex]\theta[/latex] + Sin²[latex]\theta[/latex])

Or, R² = A² + 2ABCos[latex]\theta[/latex] + B² [Since, Sin²[latex]\theta[/latex] + Cos²[latex]\theta[/latex] = 1]

Or, R = [latex]\sqrt{A^2+2ABCos\theta + B^2}[/latex]

∴ R = [latex]\sqrt{A^2+B^2+2ABCos\theta}[/latex]

Which is the magnitude of resultant of [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex].

Now, for direction of resultant [latex]\vec{R}[/latex], let us suppose, ‘β’ is the angle made by resultant with the direction of [latex]\vec{A}[/latex]

In [latex]\Delta[/latex]OQM,

tanβ = [latex]\frac{QM}{OM} = \frac{BSin\theta}{OM} = \frac{BSin\theta}{A+BCos\theta}[/latex]

Or, β = tan^-1 [latex](\frac{BSin\theta}{A+BCos\theta})[/latex]

This gives the direction of resultant vector.

Special Cases:

1) When two vectors are acting in the same direction, then [latex]\theta[/latex] = 0^o.

∴ R = [latex]\sqrt{A^2+B^2+2ABCos0^o}[/latex]

= [latex]\sqrt{A^2+B^2+2AB}[/latex] [Since, Cos0^o = 1]

Or, R = [latex]\sqrt{(A+B)^2}[/latex]

Or, R = (A + B)

i.e., resultant is algebraic sum of magnitude of two vectors.

And, β = tan^-1 [latex](\frac{BSin0^o}{A+BCos0^o})[/latex]

= tan^-1 ([latex]\frac{B\times 0}{A+B\times 1}[/latex])

Or, β = tan^-1(0) = 0

i.e., the resultant coincides with given two vectors.

2) When two vectors are acting at right angles, then [latex]\theta[/latex] = 90^o.

∴ R = [latex]\sqrt{A^2+B^2+2ABCos90^o}[/latex]

= [latex]\sqrt{A^2+B^2+0}[/latex] [Since, Cos90^o = 0]

Or, R = [latex]\sqrt{A^2+B^2}[/latex]

And, β = tan^-1 [latex](\frac{B}{A})[/latex]

3) When two vectors are acting in the opposite direction, then, [latex]\theta[/latex] = 180^o.

∴ R = [latex]\sqrt{A^2+B^2+2ABCos180^o}[/latex]

= [latex]\sqrt{A^2+B^2-2AB}[/latex] [Since, Cos180^o = -1]

Or, R = [latex]\sqrt{(A-B)^2}[/latex]

Or, R = A – B

i.e., Resultant is the difference of magnitude of two vectors.

And, β = tan^-1 [latex](\frac{BSin180^o}{A+BCos180^o})[/latex]

Or, β = tan^-1(0) = 0

i.e., resultant is directed towards the bigger one.

Polygonal law of vector addition:

It states that, if many vectors acting at a point simultaneously be represented in magnitude and direction by sides of a polygon taken in same order then, the last side of it taken in the reverse order gives magnitude and direction of the resultant vector of these vectors.

Suppose, a number of vectors P, Q, R, S, T and U are acting at a point simultaneously as shown in figure (a). These vectors are represented in magnitude and direction by sides [latex]\vec{AB}[/latex], [latex]\vec{BC}[/latex], [latex]\vec{CD}[/latex], [latex]\vec{DE}[/latex], [latex]\vec{EF}[/latex] and [latex]\vec{FA}[/latex]. A polygon ABCDEF in the same order, the last side of a polygon [latex]\vec{FA}[/latex]  i.e. [latex]\vec{AF}[/latex] in the reverse order gives magnitude and direction of the resultant vector.

Join AC, AD and AE.

In [latex]\Delta[/latex]ABC,

             [latex]\vec{AC}[/latex] = [latex]\vec{AB}[/latex] + [latex]\vec{BC}[/latex]

= [latex]\vec{P}+\vec{Q}[/latex]

In [latex]\vec{\Delta}[/latex]ACD,

             [latex]\vec{AD}[/latex] = [latex]\vec{AC}[/latex] + [latex]\vec{CD} = \vec{P}+\vec{Q}+\vec{S}[/latex]

In [latex]\Delta[/latex]ADE,

             [latex]\vec{AE}=\vec{AD}+\vec{DE}[/latex]

= [latex]\vec{P}+\vec{Q}+\vec{S}+\vec{T}[/latex]

In [latex]\Delta[/latex]ADE,

[latex]\vec{AF}=\vec{AE}+\vec{EF}[/latex]

  [latex]\vec{R}=\vec{P}+\vec{Q}+\vec{S}+\vec{T}+\vec{U}[/latex]

Resolution of a vector: rectangular components:

The process of splitting of a vector into its parts is known as resolution and split parts are called component. If they are perpendicular to each other then they are called rectangular components.

Let, q be the angle made by [latex]\vec{OP}[/latex] = [latex]\vec{F}[/latex] with x – axis. Draw PN perpendicular to x – axis and PM ⊥ y axis.

Then, ON = F_x and OM = F_y are rectangular components of [latex]\vec{F}[/latex].

In [latex]\Delta[/latex]ONP,

Cos[latex]\theta = \frac{ON}{OP}[/latex]

Or, ON = OPCos[latex]\theta[/latex]

Or, F_x = FCos[latex]\theta[/latex] ……………. (i)

Again,

Sin[latex]\theta[/latex] = [latex]\frac{PN}{OP}[/latex]

Or, PN = OPSin[latex]\theta[/latex]

Or, OM = PN = OPSin[latex]\theta[/latex]

Or, F_y = FSin[latex]\theta[/latex] ………………. (ii)

Squatting and adding eqⁿ. (i) and (ii), we get,

F_x² + F_y² = F²Sin²[latex]\theta[/latex] + F²Cos²[latex]\theta[/latex]

              = F² x 1

F_x² + F_y²= F

Or, F = [latex]\sqrt{F_x^2+F_y^2}[/latex]

Dividing eq^n. (ii) by (i), we get,

[latex]\frac{F_y}{F_x} = Tan\theta[/latex]

Or, Tan[latex]\theta = \frac{F_y}{F_x}[/latex]

Dot or Scalar Product:

Let, [latex]\theta[/latex] be the angle between two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex]. The dot or scalar product of these two vectors is defined as:

[latex]\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|Cos\theta[/latex]

              = ABCos[latex]\theta[/latex]  ………………. (i)

Where, [latex]\vec{i}, \vec{j}[/latex] and [latex]\vec{k}[/latex] are unit vectors along x – axis, y – axis and z – axis respectively.

[latex]\hat{i}.\hat{i}=|\hat{i}||\hat{i}|Cos0^o[/latex]

              = 1 x 1 x 1 = 1

∴ [latex]\hat{i}.\hat{i} = \hat{j}.\hat{j}=\hat{k}.\hat{k} = 1[/latex]

∴ [latex]\hat{i}.\hat{j}=|\hat{i}||\hat{j}|Cos90^o[/latex]

              = 1 x 1 x 0 = 0

∴ [latex]\hat{i}.\hat{j}=\hat{j}.\hat{k}=\hat{k}.\hat{i}[/latex] = 0

If [latex]\vec{A}[/latex] = A_x [latex]\hat{i}[/latex] + A_y [latex]\hat{j}[/latex] and

 [latex]\vec{B}[/latex] = B_x [latex]\hat{i}[/latex] + B_y [latex]\hat{j}[/latex], then,

             [latex]\vec{A}.\vec{B} = (A_x\hat{i}+A_y\hat{j}+A_z\hat{k}).(B_x\hat{i}+B_y\hat{j}+B_z\hat{k})[/latex]

= A_xB_x + A_yB_y + A_zB_z ………………….. (ii)

Which is required expression for dot or scalar product.

1. [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are two non-zero vectors. If |[latex]\vec{A}\times \vec{B}[/latex]| = [latex]\vec{A}.\vec{B}[/latex], what is the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] ?

   OR, If the scalar product of two vectors in equal magnitude of their product, find the angle between them.
   Solution:
   Let [latex]\theta[/latex] be the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex]. The scalar product of [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] is given by

   [latex]\vec{A}.\vec{B}=ABCos\theta[/latex]            ……………………… (i)
   The magnitude of the vector product of  and  is given by,
   |[latex]\vec{A}\times \vec{B}[/latex]| = ABsin[latex]\theta[/latex]  ……………….. (ii)
   According to the question, we have,
   [latex]\vec{A}.\vec{B}=|\vec{A}\times \vec{B}|[/latex]

   Or, ABcos[latex]\theta = ABSin\theta[/latex]

   Or, [latex]tan\theta = 1[/latex]

   [latex]∴ \theta = 45^o[/latex]

   Hence, required angle is 45^o.

2. If [latex]\vec{A}.\vec{B}=0,[/latex] what is the value of angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex]?
   Solution: 
   [latex]\vec{A}.\vec{B}=|\vec{A}||\vec{B}|Cos\theta = AB Cos\theta[/latex], where [latex]\theta[/latex] be the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex].

   When two vectors [latex]\vec{A}[/latex] & [latex]\vec{B}[/latex] are perpendicular to each other, i.e.

   When [latex]\theta = 90^o[/latex], then, [latex]\vec{B}=AB Cos90^o[/latex] = 0              

   Hence, for [latex]\vec{A}.\vec{B}=0[/latex], the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] must be 90^o.

3. Resultant of two equal forces may have the magnitude equal to one of the forces. At what angle between the two equal forces this is possible? Justify your answer.
   OR, Two vectors have equal magnitudes and their resultant also has the same magnitude. What is the angle between the two vectors?
   Ans: [latex]\theta[/latex] Let  be the angle between two equal vectors, each vector [latex]\vec{A}[/latex], so that resultant (sum) vector [latex]\vec{R}[/latex] has also magnitude A.
   We have, R² = A² + B² + 2ABcos[latex]\theta[/latex]
   ∴ [latex]A^2=A^2+A^2+2AACos\theta[/latex]      [∵ R = a and A = B]
   Or, A² = 2A² + 2A²cos[latex]\theta[/latex]
   Or, 1 = 2 + 2cos[latex]\theta[/latex]
   Or, 2cos[latex]\theta = -1[/latex]
   Or, cos[latex]\theta = -\frac{1}{2}[/latex]
   ∴ [latex]\theta = 120^o[/latex]
   Hence, required angle is 120^o.

4. Two vectors are given as [latex]\vec{V_1} = 2\hat{i}+3\vec{j}+4\vec{k}[/latex] and [latex]\vec{V_2}=3\hat{i}-2\hat{j}-4\hat{k}[/latex]. Which one of the two is larger in magnitude? Justify your answer.
   Ans: [latex]\vec{V_1}=2\hat{i}+3\hat{j}+4\hat{k}[/latex]
   [latex]\vec{V_2}=3\hat{i}-2\hat{j}-4\hat{k}[/latex]
   Magnitude of V₁ = [latex]\sqrt{2^2+3^2+4^2 = \sqrt{29}}[/latex]
   Magnitude of V₂ = [latex]\sqrt{3^2+(-2)^2+(-4)^2} = \sqrt{29}[/latex]             
   Both vectors have same magnitude and are equal.

5. A vector is defined as [latex]\vec{E}=2\hat{i}+3\hat{j}-4\hat{k}[/latex]. What is the magnitude of y-component of [latex]\vec{E}[/latex]?
   Ans: Every vector [latex]\hat{A}[/latex] can be represented in three rectangular components of the vector as   
   [latex]\vec{A}=\vec{A_x}+\vec{A_y}+\vec{A_z}[/latex]
   Or, [latex]\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}[/latex]
   Here, the given vector is [latex]\vec{E}=2\hat{i}+3\hat{j}-4\hat{k}[/latex]
   Hence, x-component = 2
   y-component = 3                             
   z-component = – 4

6. Given two vectors [latex]\vec{A}=4.00\hat{i}+3.00\hat{j}[/latex] and [latex]\vec{B}=5.00\hat{i}-2.00\hat{j}[/latex]. Find the magnitude of each vector.
   Ans: Here given
   [latex]\vec{A}=4.00\hat{i}+3.00\hat{j}[/latex]
   [latex]\vec{B}=5.00\hat{i}-2.00\hat{j}[/latex]
   Magnitude of each vector =?
   Here, A_x = 4.00 and A_y = 3.00
   B_x = 5.00 and B_y = – 2.00
   Now, magnitude of [latex]\vec{A}, |\vec{A}|=\sqrt{A_x^2+A_y^2}=\sqrt{4.00^2+3.00^2}=5.00[/latex]
   Magnitude of [latex]\vec{B}, |\vec{B}|=\sqrt{B_x^2+B_y^2}=\sqrt{5.00^2+(-2.00)^2}=5.39[/latex]

7. A vector [latex]\vec{F}=\hat{i}+2\hat{j}-3\hat{k}[/latex] is given. What is the magnitude of the y-component of the vector?
   Ans: Every vector [latex]\vec{A}[/latex] can be represented in three rectangular components of the vector as
   [latex]\vec{A}=\vec{A_x}+\vec{A_y}+\vec{A_z}[/latex]   
   Or, [latex]\vec{A} = A_x\hat{i}+A_y\hat{j}+A_z\hat{k}[/latex]
   Here, given vector [latex]\vec{F} = \hat{i}+2\hat{j}-3\hat{k}[/latex]
   Hence, x – component = 1
   y – component = 2
   z – component = – 3

8. Is a physical quantity having magnitude and direction necessarily a vector quantity? Explain.
   Ans: No. A physical quantity having magnitude and direction shouldn’t be necessarily a vector. The quantity should have one additional property: it should obey the laws of vectors (for example, triangle law of vector addition, parallelogram law of vector addition). Consider the physical quantity pressure which has both magnitude and direction, still it isn’t a vector. This is because it doesn’t obey laws of vectors instead it behaves like a scalar quantity.

9. A force (in Newton) expressed in vector notation as [latex]\vec{F}=2\hat{i}+\hat{j}-3\hat{k}[/latex] is applied on a body so that the displacement produced in meter is given by [latex]\vec{D}=\hat{i}-2\hat{j}-3\hat{k}[/latex]. Express the result and nature of the work done.
   Ans: Here,
   Now,
   W = [latex]\vec{F}.\vec{D} = (2\hat{i}+\hat{j}-3\hat{k})(\hat{i}-2\hat{j}-3\hat{k})[/latex] = 2 – 2 + 9
                                             = 9 J (positive work)

10. A force (in Newton) expressed in vector notation as [latex]\vec{F}=4\hat{i}+7\hat{j}-3\hat{k}[/latex] is applied on a body and produces a displacement (in meter), [latex]\vec{D}=3\hat{i}-2\hat{j}-5\hat{k}[/latex] in 4 seconds. Estimate the power.
    Ans: Here,
    [latex]\vec{F}=4\hat{i}+7\hat{j}-3\hat{k}[/latex]
    [latex]\vec{D}=3\hat{i}-2\hat{j}-5\hat{k}[/latex]
    t = 4 seconds
    Now,
    W = [latex]\vec{F}.\vec{D} = (4\hat{i}+7\hat{j}-3\hat{k}).(3\hat{i}-2\hat{j}-5\hat{k})[/latex] 
    = 12 – 14 + 15
    = 13 J
    ∴ [latex]P=\frac{W}{t}=\frac{13}{4}=3.25 Watt[/latex]

11. What does [latex]\vec{A}.\vec{A}[/latex], the scalar product of a vector with itself give? What about [latex]\vec{A}\times \vec{A}[/latex] the vector product of a vector with itself?
    Ans: The scalar product of a vector with itself gives the square of its magnitude.
    i.e. [latex]\vec{A}.\vec{A}=|A||A|cos0^o[/latex]        (∵[latex]\theta = 0^o[/latex])
    Or, [latex]\vec{A}.\vec{A}=A^2[/latex]
    But, the vector product of a vector with itself is zero.
    i.e. [latex]\vec{A}\times \vec{A}=|A||A|sin0^o[/latex]
    ∴ [latex]\vec{A}\times \vec{A}=0[/latex]

12. If [latex]\hat{i}, \hat{j}[/latex] and [latex]\hat{k}[/latex] are unit vectors along x, y and z – axis respectively, find [latex]\hat{i}.(\hat{k}\times \hat{j})[/latex].
    Ans: Here,
    [latex]\hat{i}, \hat{j}[/latex] and [latex]\hat{k}[/latex] are the unit vectors along x, y and z – axis respectively.
    Then, [latex]\hat{i}.(\hat{k}\times \hat{j})[/latex]
    = [latex]\hat{i}.-\hat{i}[/latex] [∵ [latex]\hat{k}\times \hat{j}=-\hat{i}[/latex]]   
    = – 1 [∵ [latex]\hat{i}.\hat{i} = 1][/latex]       

13. The angle between two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] is [latex]\theta[/latex]. Find the magnitude and direction of [latex]\vec{A}\times \vec{B}[/latex] and [latex]\vec{A}.\vec{B}[/latex].
    Ans: The scalar product of two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] is the product of the magnitude of one of the vectors and the projection of the second vector on the first.
    ∴ [latex]\vec{A}.\vec{B}=ABcos\theta[/latex]
    
    The vector product of two vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] is a vector which is perpendicular to the plane of the vectors [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] and whose magnitude is ABSin[latex]\theta[/latex]. Where A and B are the magnitude of [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] and [latex]\theta[/latex] is the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex].
    Symbolically,
    [latex]\vec{A}\times \vec{B}=ABsin\theta \hat{n}[/latex]
    Where,[latex]\hat{n}[/latex]  is the unit vector normal to both [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex].

14. [latex]\hat{i}, \hat{j}[/latex] and [latex]\hat{k}[/latex] are the unit vectors of a force along X, Y and Z axes respectively. Find the magnitude and direction of the vector product of two forces [latex]\vec{F_1}[/latex] and [latex]\vec{F_2}[/latex]. If [latex]\vec{F_1} = 3\hat{i}[/latex] and [latex]\vec{F_2} = – 2\hat{k}[/latex].
    Solution:
    Here, [latex]\vec{F_1} = 3\hat{i}[/latex] and [latex]\vec{F_2} = – 2\hat{k}[/latex]

    [latex]\therefore \vec{F_1} \times \vec{F_2} = 3\hat{i} \times (- 2\hat{k}) = 6\hat{j}[/latex]

    The plane of [latex]\vec{F_1}[/latex] and [latex]\vec{F_2}[/latex] (the [latex]\hat{i} – \hat{k}[/latex] plane) is perpendicular to the direction of [latex]\vec{F_1} \times \vec{F_2}[/latex] (the [latex]\hat{j}[/latex] direction).

15. If [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are non zero vectors, is it possible for [latex]\vec{A} \times \vec{B}[/latex] and [latex]\vec{A}.\vec{B}[/latex] both to be zero? Explain.
    Ans:
    We know,
    [latex]\vec{A} \times \vec{B} = ABsin\theta \hat{n}[/latex]
    Or, 0 = [latex]ABsin\theta \hat{n}[/latex]
    i.e. either A = 0 or B = 0 or Sin[latex]\theta = 0[/latex]  ………………….. (i)
    Also, [latex]\vec{A}.\vec{B} = ABcos\theta[/latex]
    Or, 0 = ABCos[latex]\theta[/latex]
    i.e. either A = 0 or B = 0 or cos[latex]\theta[/latex] = 0  ……………………….. (ii)
    If [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] are non zero vectors, from equations (i) & (ii), it follows that [latex]\sin\theta[/latex] and [latex]\cos\theta[/latex] both should be zero simultaneously, which is impossible.
16. [latex]\vec{C}[/latex] is the vector sum of [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex] i.e. [latex]\vec{C} = \vec{A} + \vec{B}[/latex] for C = A + B to be true, what is the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex]?
    Ans: Let, [latex]\theta[/latex] be the angle between [latex]\vec{A}[/latex] and [latex]\vec{B}[/latex].
    We know,
    C² = [latex]\vec{C}.\vec{C}[/latex]
    = [latex](\vec{A} + \vec{B})(\vec{A} + \vec{B})[/latex]
    = [latex]\vec{A}.\vec{A} + \vec{A}.\vec{B} + \vec{B}.\vec{A} + \vec{B}.\vec{B}[/latex]
    = A² + ABcos[latex]\theta + BAcos\theta + B^2[/latex]
    Or, C² = A² + B² + 2ABCos[latex]\theta[/latex]  ………………… (i)
    Again, we have,
    C = A + B
    Or, C² = (A + B)²
    Or, C² = A² + B² + 2AB …………….. (ii)
    Comparing equations (i) and (ii), we get,
    cos[latex]\theta = 1[/latex]
    Or, [latex]\theta = cos^{-1}(1) = 0^o[/latex]
    [latex]\therefore[/latex] Angle between two vectors is 0^o, i.e. they are acting in the same direction.
17. The magnitude of two vectors are 3 and 4, and their product is 6. What is the angle between them?
    Ans: Let, a and b be the magnitude of two vectors [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] respectively. According to the question, we have a = 3, b = 4 and [latex]\vec{a}.\vec{b} = 6[/latex] (Note that the product of [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex] can’t be vector product i.e. [latex]\vec{a}\times \vec{b}[/latex]. This is because [latex]\vec{a}\times \vec{b}[/latex] is a vector but in accordance to the question the product is a scalar i.e. 6).
    [latex]\therefore abcos\theta[/latex] = 6, [latex]\theta[/latex] is the angle between [latex]\vec{a}[/latex] and [latex]\vec{b}[/latex].
    Or, 3 x 4 cos[latex]\theta = 6[/latex]
    Or, cos[latex]\theta = \frac{1}{2}[/latex]
    [latex]\therefore \theta = 60^o[/latex]             
    Hence, the required angle is 60^o.
18. What is the difference between scalar and vector product of two vectors? Explain.
    Ans: Following are differences between scalar and vector products of two vectors.