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Class 11 Vectors Notes

On the basis of magnitude and direction, physical quantities are classified into two classes:

Scalar Quantity: Those physical quantities which have only magnitude but no direction are called scalar quantities. For example, mass, length, etc.

Vector Quantity: Those physical quantities which have both magnitude and direction are called vector quantities. For example, velocity, force, etc.

Note:

  • Necessary condition: direction
  • Sufficient condition: follow vector algebra

Electric current having both magnitude and direction is a scalar quantity because it does not follow vector algebra.

Graphical Representation of Vector:

Graphically, a vector can be represented by a straight line with an arrowhead. The length of the line represents the magnitude, and the arrow gives the direction.

Symbolic Representation of Vector:

Normally, a vector can be represented by a small or capital English alphabet over an arrowhead. For example: [katex] ( \vec{A}, \vec{B}, \vec{a}, \vec{b} ) [/katex].

Types of Vector

  1. Unit Vector:
    A vector is said to be a unit vector if it has a magnitude of one. If [katex]( \vec{A} )[/katex] is a vector having magnitude [katex] ( \left| \vec{A} \right| ) [/katex], then its unit vector is given by:
    [katex] \hat{A} = \frac{\vec{A}}{\left| \vec{A} \right|} [/katex]
    Unit vectors along the x-axis, y-axis, and z-axis are represented as [katex] ( \hat{i}, \hat{j}, \text{and} \hat{k} ) [/katex] respectively.
  2. Null Vector (Zero Vector):
    A vector is said to be a null vector if it has zero magnitude and no particular direction.
  3. Negative Vector:
    A vector is said to be the negative of another if they have the same magnitude but opposite directions.
  4. Collinear Vector:
    Two vectors are said to be collinear if they lie on the same line.
  5. Coplanar Vector:
    Two vectors are said to be coplanar if they lie on the same plane.

Resolution of Vector

The phenomenon of splitting a vector into two or more vectors is called the resolution of a vector. The resulting vectors are called components of the vector. If the components are mutually perpendicular to each other, they are rectangular components, and this process is known as the rectangular resolution of a vector.

Suppose a vector [katex] ( P ) [/katex] is represented by a line [katex] ( OA ) [/katex], making an angle [katex] ( \theta ) [/katex] with the x-axis. We have to find the components of vector [katex] ( P ) [/katex] along the x-axis and y-axis. For this, draw normals [katex] ( AN ) [/katex] and [katex] ( AM ) [/katex] from [katex] ( A ) [/katex] to the x-axis and y-axis, respectively.

In [katex] ( \triangle ONA ) [/katex]:

[katex] \cos \theta = \frac{ON}{OA} \Rightarrow ON = OA \cos \theta \Rightarrow P_x = P \cos \theta [/katex]

In [katex] ( \triangle OMA [/katex]):

[katex] \sin \theta = \frac{OM}{OA} \Rightarrow OM = OA \sin \theta \Rightarrow P_y = P \sin \theta [/katex]

Equation (i) and (ii) are the components of a vector. Here [katex] ( P_x ) [/katex] and [katex] ( P_y ) [/katex] are mutually perpendicular to each other, so they are rectangular components.

By squaring and adding equations (i) and (ii), we get:

[katex] P_x^2 + P_y^2 = P^2 \cos^2 \theta + P^2 \sin^2 \theta \Rightarrow P_x^2 + P_y^2 = P^2 [/katex]

Triangle Law of Vector

It states that “Two vectors acting at a point, represented in magnitude and direction by two sides of a triangle taken in order, have a resultant vector represented by the closing side of the triangle in the opposite order.”

Suppose two vectors [katex] ( P ) [/katex] and [katex] ( Q ) [/katex] acting at a point [katex] ( O ) [/katex] are represented in magnitude and direction by two sides [katex] ( AB ) [/katex] and [katex] ( BC ) [/katex] of a triangle [katex] ( \triangle ABC ). ( \theta ) [/katex] is the angle between the vectors. According to the triangle law of vectors, the closing side [katex] ( AC ) [/katex] in the opposite order represents the magnitude and direction of the resultant vector.

In [katex] ( \triangle ANC ) [/katex]:

[katex] AC^2 = AN^2 + CN^2 = (AB + BN)^2 + CN^2 [/katex]

In [katex] ( \triangle BNC ) [/katex]:

[katex] CN = BC \sin \theta = Q \sin \theta [/katex]

Again:

[katex] BN = BC \cos \theta = Q \cos \theta [/katex]

Thus:
[katex] AC^2 = (P + Q \cos \theta)^2 + Q^2 \sin^2 \theta [/katex]
[katex] \Rightarrow R^2 = P^2 + 2PQ \cos \theta + Q^2 [/katex]

[katex] \Rightarrow R = \sqrt{P^2 + 2PQ \cos \theta + Q^2} [/katex]

This is the magnitude of the resultant vector. Let [katex] ( \beta ) [/katex] be the angle made by the resultant vector with the initial vector [katex] ( P ) [/katex].

In [katex] ( \triangle ANC ) [/katex]:
[katex] \tan \beta = \frac{CN}{AN} = \frac{Q \sin \theta}{P + Q \cos \theta} [/katex]

[katex] \Rightarrow \beta = \tan^{-1} \left( \frac{Q \sin \theta}{P + Q \cos \theta} \right) [/katex]

Parallelogram Law of Vectors

It states that “Two vectors acting at a point, represented in magnitude and direction by two adjacent sides of a parallelogram, have a resultant vector represented by the diagonal passing through the same point.”

Suppose two vectors [katex] ( P ) [/katex] and [katex] ( Q ) [/katex] acting at a point [katex] ( O ) [/katex] are represented in magnitude and direction by two adjacent sides [katex] ( AB ) [/katex] and [katex] ( AD ) [/katex] of a parallelogram [katex] ( ABCD ) [/katex]. [katex] ( \theta ) [/katex] is the angle between the vectors. According to the parallelogram law of vectors, the diagonal [katex] ( AC ) [/katex] passing through the same point [katex] ( A ) [/katex] represents the magnitude and direction of the resultant vector.

In [katex] ( \triangle ANC ) [/katex]:

[katex] AC^2 = AN^2 + CN^2 = (AB + BN)^2 + CN^2 [/katex]

In [katex] ( \triangle BNC ) [/katex]:

[katex] CN = BC \sin \theta = Q \sin \theta [/katex]

Again:

[katex] BN = BC \cos \theta = Q \cos \theta [/katex]

Thus:
[katex] AC^2 = (P + Q \cos \theta)^2 + Q^2 \sin^2 \theta [/katex]
[katex] \Rightarrow R^2 = P^2 + 2PQ \cos \theta + Q^2 [/katex]

[katex] \Rightarrow R = \sqrt{P^2 + 2PQ \cos \theta + Q^2} [/katex]

This is the magnitude of the resultant vector. Let [katex] ( \beta ) [/katex] be the angle made by the resultant vector with the initial vector [katex] ( P ) [/katex].

In [katex] ( \triangle ANC ) [/katex]:
[katex] \tan \beta = \frac{CN}{AN} = \frac{Q \sin \theta}{P + Q \cos \theta} [/katex]

[katex] \Rightarrow \beta = \tan^{-1} \left( \frac{Q \sin \theta}{P + Q \cos \theta} \right) [/katex]

Polygon Law of Vector

It states that “If a number of vectors acting at a point are represented in magnitude and direction by the regular sides of a polygon taken in order, then the closing side in opposite order represents the magnitude and direction of the resultant vector.”

Suppose given vectors [katex] (\vec{P}), (\vec{Q}), (\vec{R }) [/katex] and [katex] ( \vec{S} ) [/katex] acting at a point [katex] ( O ) [/katex] are represented in magnitude and direction by the regular sides [katex] ( AB, BC, CD ) [/katex] and [katex] ( DE ) [/katex] of a polygon [katex] ( ABCDE ) [/katex]. According to the polygon law of vectors, the closing side in opposite order [katex] ( AE ) [/katex] represents the magnitude and direction of the resultant vector.

To find the resultant of [katex] ( AB ) [/katex] and [katex] ( BC ) [/katex], join [katex] ( A ) [/katex] with [katex] ( C ) [/katex].

[katex] AC = AB + BC \Rightarrow AC = P + Q [/katex]

Similarly, to find the resultant of [katex] ( AC ) [/katex] and [katex] ( CD ) [/katex], join [katex] ( A ) [/katex] with [katex] ( D ) [/katex].

[katex] AD = AC + CD \Rightarrow AD = P + Q + R [/katex]

To find the resultant of [katex] ( AD ) [/katex] and [katex] ( DE ) [/katex], join [katex] ( A ) [/katex] with [katex] ( E ) [/katex].

[katex] AE = AD + DE \Rightarrow AE = P + Q + R + \vec{S} [/katex]

This is the required expression for the resultant of given vectors.

Subtraction of a Vector

Subtraction of a vector is the sum of a vector with the negative of another vector. If [katex] ( P ) [/katex] and [katex] ( Q ) [/katex] are two vectors, then the subtraction of [katex] ( Q ) [/katex] from [katex] ( P ) [/katex] is the sum of [katex] ( P ) [/katex] vector with the negative of [katex] ( Q ) [/katex].

[katex] P – Q = P + (-Q) [/katex]

Using the parallelogram law of vectors:

[katex] P – Q = \sqrt{P^2 + Q^2 + 2PQ \cos (\pi – \theta)} [/katex]

If [katex] ( \beta ) [/katex] is the angle made by the resultant with the [katex] ( P ) [/katex] vector:
[katex] \tan \beta = \frac{Q \sin (\pi – \theta)}{P + Q \cos (\pi – \theta)} [/katex]

[katex] \beta = \tan^{-1} \left( \frac{Q \sin (\pi – \theta)}{P + Q \cos (\pi – \theta)} \right) [/katex]

Multiplication of a Vector

When a vector is multiplied with another vector, it either gives a scalar or a vector. So, the multiplication of a vector is of two types:

  1. Scalar Multiplication (Dot Product):
    It is defined as the product of the magnitudes of the given vectors with the cosine of the angle between them.

[katex] P \cdot Q = PQ \cos \theta [/katex]

Properties:

  • The magnitude of the dot product depends on the angle between the vectors.
    • If two vectors lie on the same line [katex] (( \theta = 0^\circ )) [/katex]:
      [katex] P \cdot Q = PQ \cos 0^\circ = PQ [/katex]
    • If two vectors are perpendicular to each other [katex] (( \theta = 90^\circ )) [/katex]:
      [katex] P \cdot Q = PQ \cos 90^\circ = 0 [/katex]
  • It follows the commutative law:
    [katex] P \cdot Q = Q \cdot P [/katex]
  • It follows the distributive law:
    [katex] P \cdot (Q + R) = P \cdot Q + P \cdot R [/katex]
  • The dot product of a vector with itself gives the square of its magnitude:
    [katex] P \cdot P = P^2 [/katex]
  • For unit vectors along the x, y, and z axes:
    [katex] \hat{i} \cdot \hat{i} = 1 [/katex]
    [katex] \hat{i} \cdot \hat{j} = 0 [/katex]
  1. Vector Multiplication (Cross Product):
    It is defined as the product of the magnitudes of the given vectors with the sine of the angle between them.
    [katex] P \times Q = PQ \sin \theta \hat{n} [/katex]
    where [katex] ( \hat{n} ) [/katex] gives the direction of the cross product, always perpendicular to the plane containing vectors [katex] ( P ) [/katex] and [katex] ( Q ) [/katex].

Properties:

  • The magnitude of the cross product depends on the angle between the vectors.
    • If two vectors lie on the same line [katex] (( \theta = 0^\circ )) [/katex]:
      [katex] P \times Q = PQ \sin 0^\circ = 0 [/katex]
    • If two vectors are perpendicular to each other [katex] (( \theta = 90^\circ )) [/katex]:
      [katex] P \times Q = PQ \sin 90^\circ = PQ [/katex]
  • It does not follow the commutative law:

[katex] P \times Q = -Q \times P [/katex]

It follows the distributive law:

[katex] P \times (Q + R) = P \times Q + P \times R [/katex]

The cross product of a vector with itself gives the null vector:

[katex] P \times P = 0 [/katex]

For unit vectors along the [katex] x, y [/katex], and [katex] z [/katex] axes:
[katex] \hat{i} \times \hat{i} = 0 [/katex]

[katex] \hat{i} \times \hat{j} = \hat{k} [/katex]

If the given vectors represent the adjacent sides of a parallelogram, the magnitude of the cross product gives the area of the parallelogram:
[katex] P \times Q = PQ \sin \theta [/katex]

[katex] \text{Area of parallelogram} = \text{base} \times \text{height} [/katex]