CBSE Class 11 Physics Vectors Notes Set 01

Important Notes and Formulas

1. Scalar

In physics we deal with two types of physical quantities: one is scalar and the other is vector. Each scalar quantity has a magnitude and a unit.
For example, mass = 4 kg
Magnitude of mass = 4
and unit of mass = kg
Examples of scalar quantities: mass, speed, distance etc.
Scalar quantities can be added, subtracted and multiplied by simple laws of algebra.

2. Vector

Vectors are the physical quantities having magnitude as well as a specified direction.
For example:
Speed = 4 m/s (is a scalar)
Velocity = 4 m/s toward north (is a vector)
If someone wants to reach some location then it is not sufficient to provide information about the distance of that location; it is also essential to tell him the proper direction from the initial location to the destination.
The magnitude of a vector \( (\vec{A}) \) is the absolute value of the vector and is indicated by \( |\vec{A}| \) or A.
Examples of vector quantities: displacement, velocity, acceleration, force etc.

Knowledge of direction

3. General Points Regarding Vectors

3.1 Representation of vector: Geometrically, the vector is represented by a line with an arrow indicating the direction of the vector.
Mathematically, a vector is represented by \( \vec{A} \). Sometimes it is represented by the bold letter A.
Thus, the arrow in the figure below represents a vector \( \vec{A} \) in the xy-plane making an angle θ with the x-axis.
A representation of a vector will be complete if it gives us direction and magnitude.

Symbolic form: \( \vec{v}, \vec{a}, \vec{F}, \vec{s} \) are used to separate a vector quantity from scalar quantities (u, i, m).
Graphical form: A vector is represented by a directed straight line, having the magnitude and direction of the quantity represented by it.
For example, if we want to represent a force of 5 N acting 45° N of E:
(i) We choose direction co-ordinates.
(ii) We choose a convenient scale like 1 cm = 1 N.
(iii) We draw a line of length equal in magnitude and in the direction of the vector to the chosen scale.
(iv) We put an arrow in the direction of the vector.
The vector is \( \overrightarrow{AB} \) and its magnitude is \( |\overrightarrow{AB}| = 5 \) N.

3.2 Angle between two vectors (θ): The angle between two vectors means the smaller of the two angles between the vectors when they are placed tail to tail by displacing either of the vectors parallel to itself (i.e. 0 ≤ θ ≤ π).

3.3 Negative of vector: It means a vector of the same magnitude but opposite in direction.

3.4 Equality of vectors: Vectors having equal magnitude and the same direction are called equal vectors.
If \( |\vec{A}| = |\vec{B}| = |\vec{C}| \) and \( \hat{A} = \hat{B} = \hat{C} \), then \( \vec{A} = \vec{B} = \vec{C} \).

3.5 Collinear vectors: If two vectors are collinear then one can be expressed in terms of the other: \( \vec{a} = \lambda \vec{b} \) (where λ is a constant).

3.6 Co-initial vectors: If two or more vectors start from the same point then they are called co-initial vectors. In the figure, \( \vec{A}, \vec{B}, \vec{C}, \vec{D} \) are co-initial.

3.7 Coplanar vectors: Three (or more) vectors are called coplanar vectors if they lie in the same plane or are parallel to the same plane. Two (free) vectors are always coplanar.

Important points
(1) If the frame of reference is translated or rotated, the vector does not change (though its components may change).
(2) Two vectors are called equal if their magnitudes and directions are the same, and they represent values of the same physical quantity.

3.8 Multiplication and division of a vector by a scalar: Multiplying a vector \( \vec{A} \) by a positive number λ gives a vector \( \vec{B} = \lambda \vec{A} \) whose magnitude becomes λ times but the direction is the same as that of \( \vec{A} \). Multiplying a vector \( \vec{A} \) by a negative number λ gives a vector \( \vec{B} \) whose direction is opposite to the direction of \( \vec{A} \) and whose magnitude is −λ times \( |\vec{A}| \).
The division of vector \( \vec{A} \) by a non-zero scalar m is defined as multiplication of \( \vec{A} \) by \( \frac{1}{m} \).
Here \( \vec{A} \) and \( \vec{B} \) are collinear vectors.

4. Laws of Addition of Vectors

4.1 Triangle rule of addition: Steps for adding two vectors representing the same physical quantity by the triangle law:
(i) Keep the vectors such that the tail of one vector coincides with the head of the other.
(ii) Join the tail of the first to the head of the other by a line with an arrow at the head of the second.
(iii) This new vector is the sum of the two vectors (also called the resultant).

4.2 Polygon law of addition: This law is used for adding more than two vectors. This is an extension of the triangle law of addition. We keep on arranging vectors such that the tail of the next vector lies on the head of the former. When we connect the tail of the first vector to the head of the last, we get the resultant of all the vectors.
Note: \( \vec{P} = (\vec{a} + \vec{b} + \vec{c}) + \vec{d} = (\vec{c} + \vec{a} + \vec{d}) + \vec{b} \) [Associative law]

4.3 Parallelogram law of addition:
Steps:
(i) Keep the two vectors such that their tails coincide.
(ii) Draw parallel vectors to both of them, considering both of them as sides of a parallelogram.
(iii) Then the diagonal drawn from the point where the tails coincide represents the sum of the two vectors, with its tail at the point of coincidence of the two vectors.

Note: \( \overrightarrow{AC} = \vec{a} + \vec{b} \) and \( \overrightarrow{AC} = \vec{b} + \vec{a} \); thus \( \vec{a} + \vec{b} = \vec{b} + \vec{a} \) [Commutative law]
Note: The angle between 2 vectors is the angle between their positive directions.
Suppose the angle between these two vectors is θ, and \( |\vec{a}| = a,\ |\vec{b}| = b \). In the figure below, AB = a, BD = b and DE ⊥ AE:
(AD)² = (AE)² + (DE)²
= (AB + BE)² + (DE)²
= (a + b cos θ)² + (b sin θ)²
= a² + b² cos²θ + 2ab cos θ + b² sin²θ
= a² + b² + 2ab cos θ
Thus, \( AD = \sqrt{a^2 + b^2 + 2ab\cos\theta} \) or \( |\vec{a} + \vec{b}| = \sqrt{a^2 + b^2 + 2ab\cos\theta} \)
The angle α of the resultant with vector \( \vec{a} \) is given by \( \tan\alpha = \frac{DE}{AE} = \frac{b\sin\theta}{a + b\cos\theta} \)

Important points:
(1) To a vector, only a vector of the same type can be added that represents the same physical quantity, and the resultant is also a vector of the same type.
(2) As R = [A² + B² + 2AB cos θ]^1/2, R will be maximum when cos θ = max = 1, i.e. θ = 0°, i.e. the vectors are like or parallel and R_max = A + B.
(3) If \( |\vec{A}| = |\vec{B}| \) and the angle between them is θ, then R = 2A cos (θ/2).
(4) If \( |\vec{A}| = |\vec{B}| \) and the angle between them is π − θ, then R = 2A sin (θ/2).
(5) The resultant will be minimum if cos θ = min = −1, i.e. θ = 180°, i.e. the vectors are antiparallel and R_min = A − B.
(6) If the vectors A and B are orthogonal, i.e. θ = 90°, \( R = \sqrt{A^2 + B^2} \).
(7) The resultant of two vectors can have any value from (A − B) to (A + B) depending on the angle between them, and the magnitude of the resultant decreases as θ increases from 0° to 180°.
(8) The minimum number of unequal coplanar vectors whose sum can be zero is three.
(9) The resultant of three non-coplanar vectors can never be zero, i.e. the minimum number of non-coplanar vectors whose sum can be zero is four.

5. Subtraction of Vectors

The negative of a vector, say \( -\vec{A} \), is a vector of the same magnitude as vector \( \vec{A} \) but pointing in a direction opposite to that of \( \vec{A} \).
Thus, \( \vec{A} - \vec{B} \) can be written as \( \vec{A} + (-\vec{B}) \), i.e. \( \vec{A} - \vec{B} \) is really the vector addition of \( \vec{A} \) and \( -\vec{B} \).
Suppose the angle between two vectors \( \vec{A} \) and \( \vec{B} \) is θ. Then the angle between \( \vec{A} \) and \( -\vec{B} \) will be 180° − θ, as shown in the figure.

The magnitude of \( \vec{S} = \vec{A} - \vec{B} \) will thus be given by
\( S = |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos(180^\circ - \theta)} \)
or \( S = \sqrt{A^2 + B^2 - 2AB\cos\theta} \)    …(i)
For the direction of \( \vec{S} \) we calculate either angle α (with \( \vec{A} \)) or β (with \( -\vec{B} \)), where
\( \tan\alpha = \frac{B\sin(180^\circ - \theta)}{A + B\cos(180^\circ - \theta)} = \frac{B\sin\theta}{A - B\cos\theta} \)    …(ii)
or \( \tan\beta = \frac{A\sin(180^\circ - \theta)}{B + A\cos(180^\circ - \theta)} = \frac{A\sin\theta}{B - A\cos\theta} \)    …(iii)
Note: \( \vec{A} - \vec{B} \) or \( \vec{B} - \vec{A} \) can also be found by making triangles as shown in figures (a) and (b).

6. Unit Vector and Zero Vector

A unit vector is a vector which has unit magnitude and points in a particular direction. Any vector \( (\vec{A}) \) can be written as the product of the unit vector \( (\hat{A}) \) in that direction and the magnitude of the given vector:
\( \vec{A} = A\hat{A} \) or \( \hat{A} = \frac{\vec{A}}{A} \)
A unit vector has no dimensions and no unit. Unit vectors along the positive x-, y- and z-axes of a rectangular coordinate system are denoted by \( \hat{i}, \hat{j} \) and \( \hat{k} \) respectively, such that \( |\hat{i}| = |\hat{j}| = |\hat{k}| = 1 \).

A vector of zero magnitude is called a zero or null vector. Its direction is arbitrary.

7. Resolution of Vectors

If \( \vec{a} \) and \( \vec{b} \) are any two non-zero vectors in a plane with different directions and \( \vec{A} \) is another vector in the same plane, \( \vec{A} \) can be expressed as a sum of two vectors: one obtained by multiplying \( \vec{a} \) by a real number and the other obtained by multiplying \( \vec{b} \) by another real number:
\( \vec{A} = \lambda\vec{a} + \mu\vec{b} \) (where λ and μ are real numbers)
We say that \( \vec{A} \) has been resolved into two component vectors, namely \( \lambda\vec{a} \) and \( \mu\vec{b} \), along \( \vec{a} \) and \( \vec{b} \) respectively. Hence one can resolve a given vector into two component vectors along a set of two vectors, all three lying in the same plane.

7.1 Resolution along rectangular components: It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude, which we call unit vectors. \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors along the x, y and z-axis as shown in the figure above.

7.2 Resolution in two dimensions: Consider a vector \( \vec{A} \) that lies in the xy plane as shown in the figure:
\( \vec{A} = \vec{A}_1 + \vec{A}_2 \), where \( \vec{A}_1 = A_x\hat{i},\ \vec{A}_2 = A_y\hat{j} \) ⇒ \( \vec{A} = A_x\hat{i} + A_y\hat{j} \)
The quantities A_x and A_y are called the x- and y-components of the vector \( \vec{A} \). A_x is itself not a vector, but \( A_x\hat{i} \) is a vector, and so is \( A_y\hat{j} \).
A_x = A cos θ and A_y = A sin θ
It is clear from the above equations that a component of a vector can be positive, negative or zero depending on the value of θ. A vector \( \vec{A} \) can be specified in a plane in two ways:
(a) its magnitude A and the direction θ it makes with the x-axis; or
(b) its components A_x and A_y: \( A = \sqrt{A_x^2 + A_y^2},\ \ \theta = \tan^{-1}\frac{A_y}{A_x} \)

Note: If A = A_x ⇒ A_y = 0 and if A = A_y ⇒ A_x = 0, i.e. the component of a vector perpendicular to itself is always zero. The rectangular components of each vector and those of the sum \( \vec{C} = \vec{A} + \vec{B} \) are shown in figure (a). We see that \( \vec{C} = \vec{A} + \vec{B} \) is equivalent to both
C_x = A_x + B_x and C_y = A_y + B_y
In figure (b), vector \( \vec{R} \) has been resolved along two axes x and y that are not perpendicular to each other. Applying the sine law in the triangle shown, we have
\( \frac{R}{\sin[180^\circ - (\alpha + \beta)]} = \frac{R_x}{\sin\beta} = \frac{R_y}{\sin\alpha} \)
or \( R_x = \frac{R\sin\beta}{\sin(\alpha + \beta)} \) and \( R_y = \frac{R\sin\alpha}{\sin(\alpha + \beta)} \)
If α + β = 90°, R_x = R sin β and R_y = R sin α.

8. Procedure to Solve a Vector Equation

\( \vec{A} = \vec{B} + \vec{C} \)    …(1)
(a) There are 6 variables in this equation, which are the following:
(1) Magnitude of \( \vec{A} \) and its direction
(2) Magnitude of \( \vec{B} \) and its direction
(3) Magnitude of \( \vec{C} \) and its direction
(b) We can solve this equation if we know the values of 4 variables [Note: two of them must be directions].
(c) If we know the directions of any two vectors, then we put them on the same side and the other on the other side.
For example: If we know the directions of \( \vec{A} \) and \( \vec{B} \) and the direction of \( \vec{C} \) is unknown, then we write the equation as \( \vec{C} = \vec{A} - \vec{B} \).
(d) Then we make the vector diagram according to the equation and resolve the vectors to find the unknown values.

9. Short Method

If there are two vectors \( \vec{A} \) and \( \vec{B} \) and their resultant makes an angle α with \( \vec{A} \) and β with \( \vec{B} \), then
A sin α = B sin β
This means the component of \( \vec{A} \) perpendicular to the resultant is equal in magnitude to the component of \( \vec{B} \) perpendicular to the resultant.

10. Addition and Subtraction in Component Form

Suppose there are two vectors in component form:
\( \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} \)
\( \vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k} \)
Then \( \vec{A} \pm \vec{B} = (A_x \pm B_x)\hat{i} + (A_y \pm B_y)\hat{j} + (A_z \pm B_z)\hat{k} \)
If a third vector \( \vec{C} = C_x\hat{i} + C_y\hat{j} + C_z\hat{k} \) is also added or subtracted, then
\( \vec{A} \pm \vec{B} \pm \vec{C} = (A_x \pm B_x \pm C_x)\hat{i} + (A_y \pm B_y \pm C_y)\hat{j} + (A_z \pm B_z \pm C_z)\hat{k} \)
Note: The modulus of vector \( \vec{A} \) is given by \( |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \)

11. Multiplication of Vectors (Scalar and Vector Products)

11.1 Scalar product: The scalar product or dot product of any two vectors \( \vec{A} \) and \( \vec{B} \), denoted as \( \vec{A} \cdot \vec{B} \) (read \( \vec{A} \) dot \( \vec{B} \)), is defined as the product of their magnitudes with the cosine of the angle between them. Thus,
\( \vec{A} \cdot \vec{B} = AB\cos\theta \) (here θ is the angle between the vectors)

Properties:
(1) It is always a scalar, which is positive if the angle between the vectors is acute (i.e. < 90°) and negative if the angle between them is obtuse (i.e. 90° < θ ≤ 180°).
(2) It is commutative, i.e. \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \)
(3) It is distributive, i.e. \( \vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} \)
(4) As by definition \( \vec{A} \cdot \vec{B} = AB\cos\theta \), the angle between the vectors is \( \theta = \cos^{-1}\left[\frac{\vec{A} \cdot \vec{B}}{AB}\right] \)
(5) \( \vec{A} \cdot \vec{B} = A(B\cos\theta) = B(A\cos\theta) \). Geometrically, B cos θ is the projection of \( \vec{B} \) onto \( \vec{A} \) and vice versa.
Component of \( \vec{B} \) along \( \vec{A} \) \( = B\cos\theta = \frac{\vec{A} \cdot \vec{B}}{A} = \hat{A} \cdot \vec{B} \) (projection of \( \vec{B} \) on \( \vec{A} \))
Component of \( \vec{A} \) along \( \vec{B} \) \( = A\cos\theta = \frac{\vec{A} \cdot \vec{B}}{B} = \vec{A} \cdot \hat{B} \) (projection of \( \vec{A} \) on \( \vec{B} \))

(6) The scalar product of two vectors will be maximum when cos θ = max = 1, i.e. θ = 0°, i.e. the vectors are parallel ⇒ \( (\vec{A} \cdot \vec{B})_{max} = AB \)
(7) If the scalar product of two non-zero vectors vanishes, then the vectors are perpendicular.
(8) The scalar product of a vector by itself is termed the self dot product and is given by \( (\vec{A})^2 = \vec{A} \cdot \vec{A} = AA\cos 0^\circ = A^2 \) ⇒ \( A = \sqrt{\vec{A} \cdot \vec{A}} \)
(9) In case of a unit vector \( \hat{n} \): \( \hat{n} \cdot \hat{n} = 1 \times 1 \times \cos 0^\circ = 1 \) ⇒ \( \hat{n} \cdot \hat{n} = \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \)
(10) In case of orthogonal unit vectors \( \hat{i}, \hat{j} \) and \( \hat{k} \): \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0 \)
(11) \( \vec{A} \cdot \vec{B} = (\hat{i}A_x + \hat{j}A_y + \hat{k}A_z) \cdot (\hat{i}B_x + \hat{j}B_y + \hat{k}B_z) = A_xB_x + A_yB_y + A_zB_z \)

Components of b along and perpendicular to a: Let \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) represent two (non-zero) given vectors a and b respectively. Draw BM perpendicular to \( \overrightarrow{OA} \).
From ΔOMB, \( \overrightarrow{OB} = \overrightarrow{OM} + \overrightarrow{MB} \) ⇒ \( \vec{b} = \overrightarrow{OM} + \overrightarrow{MB} \)
Thus \( \overrightarrow{OM} \) and \( \overrightarrow{MB} \) are the components of b along a and perpendicular to a.
Now \( \overrightarrow{OM} = (OM)\hat{a} = (OB\cos\theta)\hat{a} = |\vec{b}|\cos\theta\,\hat{a} = |\vec{b}| \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \frac{\vec{a}}{|\vec{a}|} = \frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\vec{a} \)
\( \overrightarrow{MB} = \vec{b} - \overrightarrow{OM} = \vec{b} - \frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\vec{a} \)
Hence, the components of b along and perpendicular to a are \( \frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\vec{a} \) and \( \vec{b} - \frac{(\vec{a} \cdot \vec{b})}{|\vec{a}|^2}\vec{a} \) respectively.

11.2 Vector product: The vector product or cross product of any two vectors \( \vec{A} \) and \( \vec{B} \), denoted as \( \vec{A} \times \vec{B} \) (read \( \vec{A} \) cross \( \vec{B} \)), is defined as
\( \vec{A} \times \vec{B} = AB\sin\theta\,\hat{n} \)
Here θ is the angle between the vectors and the direction \( \hat{n} \) is given by the right-hand-thumb rule. Hence \( \hat{n} = \frac{\vec{A} \times \vec{B}}{AB\sin\theta} \), where \( \hat{n} \) is perpendicular to both \( \vec{A} \) and \( \vec{B} \).

Right-hand-thumb rule: To find the direction of \( \hat{n} \), draw the two vectors \( \vec{A} \) and \( \vec{B} \) with both the tails coinciding. Now place your stretched right palm perpendicular to the plane of \( \vec{A} \) and \( \vec{B} \) in such a way that the fingers are along the vector \( \vec{A} \) and when the fingers are closed they go towards \( \vec{B} \). The direction of the thumb gives the direction of \( \hat{n} \).

Properties:
(1) The vector product of two vectors is always a vector perpendicular to the plane containing the two vectors, i.e. orthogonal to both the vectors \( \vec{A} \) and \( \vec{B} \), though the vectors \( \vec{A} \) and \( \vec{B} \) may or may not be orthogonal.
(2) The vector product of two vectors is not commutative, i.e. \( \vec{A} \times \vec{B} \neq \vec{B} \times \vec{A} \). But \( |\vec{A} \times \vec{B}| = |\vec{B} \times \vec{A}| = AB\sin\theta \)
(3) The vector product is distributive when the order of the vectors is strictly maintained, i.e. \( \vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C} \)
(4) The magnitude of the vector product of two vectors will be maximum when sin θ = max = 1, i.e. θ = 90°: \( |\vec{A} \times \vec{B}|_{max} = AB \)
(5) The magnitude of the vector product of two non-zero vectors will be minimum when |sin θ| = minimum = 0, i.e. θ = 0° or 180°, and \( |\vec{A} \times \vec{B}|_{min} = 0 \), i.e. if the vector product of two non-zero vectors vanishes, the vectors are collinear.
(6) The self cross product, i.e. the product of a vector by itself, vanishes, i.e. is a null vector: \( \vec{A} \times \vec{A} = AA\sin 0^\circ\,\hat{n} = \vec{0} \)
(7) In case of a unit vector \( \hat{n} \): \( \hat{n} \times \hat{n} = \vec{0} \) ⇒ \( \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0} \)
(8) In case of orthogonal unit vectors \( \hat{i}, \hat{j} \) and \( \hat{k} \), in accordance with the right-hand-thumb rule:
\( \hat{i} \times \hat{j} = \hat{k},\ \ \hat{j} \times \hat{k} = \hat{i},\ \ \hat{k} \times \hat{i} = \hat{j} \)

(9) In terms of components,
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = \hat{i}(A_yB_z - A_zB_y) + \hat{j}(A_zB_x - A_xB_z) + \hat{k}(A_xB_y - A_yB_x) \)

12. Position Vector

The position vector of a point is the vector whose tail is at the origin and whose head is at the given point itself. The position vector of a point defines the position of the point with respect to the origin.
\( \overrightarrow{OP} = \vec{r} \), \( \vec{r} = x\hat{i} + y\hat{j} \)

13. Displacement Vector

The change in the position vector of a particle is known as the displacement vector.
\( \overrightarrow{OP} = \vec{r}_1 = x_1\hat{i} + y_1\hat{j} \)
\( \overrightarrow{OQ} = \vec{r}_2 = x_2\hat{i} + y_2\hat{j} \)
\( \overrightarrow{PQ} = \vec{r}_2 - \vec{r}_1 = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} \)
Thus we can represent a vector in space starting from (x₁, y₁) and ending at (x₂, y₂) as \( (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} \).

Solved Examples

Example 1. Three vectors \( \vec{A}, \vec{B}, \vec{C} \) are shown in the figure. Find the angle between (i) \( \vec{A} \) and \( \vec{B} \), (ii) \( \vec{B} \) and \( \vec{C} \), (iii) \( \vec{A} \) and \( \vec{C} \).

Answer:
To find the angle between two vectors we connect the tails of the two vectors. We can shift \( \vec{B} \) and \( \vec{C} \) such that the tails of \( \vec{A}, \vec{B} \) and \( \vec{C} \) are connected as shown in the figure.

Now we can easily observe that the angle between \( \vec{A} \) and \( \vec{B} \) is 60°, between \( \vec{B} \) and \( \vec{C} \) is 15° and between \( \vec{A} \) and \( \vec{C} \) is 75°.

Example 2. A physical quantity (m = 3 kg) is multiplied by a vector \( \vec{a} \) such that \( \vec{F} = m\vec{a} \). Find the magnitude and direction of \( \vec{F} \) if
(i) \( \vec{a} \) = 3 m/s² eastwards
(ii) \( \vec{a} \) = −4 m/s² northwards
Answer:
(i) \( \vec{F} = m\vec{a} \) = 3 × 3 m s⁻² eastwards = 9 N eastwards
(ii) \( \vec{F} = m\vec{a} \) = 3 × (−4) N northwards = −12 N northwards = 12 N southwards

Example 3. A boy moves 4 m south and then 5 m in the direction 37° E of N. Find the resultant displacement.
Answer:
Take east as +x and north as +y. First displacement \( \vec{s}_1 = -4\hat{j} \) m.
Second displacement: 5 m at 37° east of north, so \( \vec{s}_2 = 5\sin 37^\circ\,\hat{i} + 5\cos 37^\circ\,\hat{j} = 3\hat{i} + 4\hat{j} \) m
Resultant \( \vec{s} = \vec{s}_1 + \vec{s}_2 = 3\hat{i} + 0\hat{j} \) m
Hence the resultant displacement is 3 m towards east.

Example 4. Two vectors of 10 units and 5 units make an angle of 120° with each other. Find the magnitude and the angle of the resultant with the vector of 10 unit magnitude.
Answer:
\( |\vec{a} + \vec{b}| = \sqrt{a^2 + b^2 + 2ab\cos\theta} = \sqrt{100 + 25 + 2 \times 10 \times 5 \times \left(-\frac{1}{2}\right)} = \sqrt{75} = 5\sqrt{3} \)
\( \tan\alpha = \frac{5\sin 120^\circ}{10 + 5\cos 120^\circ} = \frac{5\sqrt{3}/2}{10 - 5/2} = \frac{5\sqrt{3}}{15} = \frac{1}{\sqrt{3}} \) ⇒ α = 30°
[Here the angle between the two vectors is 120° and not 60°.]

Example 5. Two vectors of equal magnitude 2 are at an angle of 60° to each other. Find the magnitude of their sum and difference.
Answer:
\( |\vec{a} + \vec{b}| = \sqrt{2^2 + 2^2 + 2 \times 2 \times 2\cos 60^\circ} = \sqrt{4 + 4 + 4} = 2\sqrt{3} \)
\( |\vec{a} - \vec{b}| = \sqrt{2^2 + 2^2 + 2 \times 2 \times 2\cos 120^\circ} = \sqrt{4 + 4 - 4} = 2 \)

Example 6. Find \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \) in the diagram shown in the figure. Given A = 4 units and B = 3 units.

Answer:
Addition: \( R = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \sqrt{16 + 9 + 2 \times 4 \times 3\cos 60^\circ} = \sqrt{37} \) units
\( \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} = \frac{3\sin 60^\circ}{4 + 3\cos 60^\circ} = 0.472 \) ∴ α = tan⁻¹(0.472) = 25.3°
Thus, the resultant of \( \vec{A} \) and \( \vec{B} \) is \( \sqrt{37} \) units at angle 25.3° from \( \vec{A} \).
Subtraction: \( S = \sqrt{A^2 + B^2 - 2AB\cos\theta} = \sqrt{16 + 9 - 2 \times 4 \times 3\cos 60^\circ} = \sqrt{13} \) units
and \( \tan\alpha = \frac{B\sin\theta}{A - B\cos\theta} = \frac{3\sin 60^\circ}{4 - 3\cos 60^\circ} = 1.04 \) ∴ α = tan⁻¹(1.04) = 46.1°

Thus, \( \vec{A} - \vec{B} \) is \( \sqrt{13} \) units at 46.1° from \( \vec{A} \) in the direction shown in the figure.

Example 7. A unit vector along east is defined as \( \hat{i} \). A force of 10⁵ dynes acts westwards. Represent the force in terms of \( \hat{i} \).
Answer:
\( \vec{F} = -10^5\,\hat{i} \) dynes

Example 8. Resolve the vector \( \vec{A} = A_x\hat{i} + A_y\hat{j} \) along and perpendicular to the line which makes an angle 60° with the x-axis.
Answer:

So the component along the line = |A_y cos 30° + A_x cos 60°|
and the component perpendicular to the line = |A_x sin 60° − A_y sin 30°|

Example 9. Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined at 30° to the horizontal.
Answer:

Component perpendicular to the plane: \( W_\perp = W\cos 30^\circ = (10)\frac{\sqrt{3}}{2} = 5\sqrt{3} \) N
and component parallel to the plane: \( W_\parallel = W\sin 30^\circ = (10)\left(\frac{1}{2}\right) = 5 \) N

Example 10. Resolve horizontally and vertically a force F = 8 N which makes an angle of 45° with the horizontal.
Answer:

Horizontal component of \( \vec{F} \): \( F_H = F\cos 45^\circ = (8)\left(\frac{1}{\sqrt{2}}\right) = 4\sqrt{2} \) N
and vertical component of \( \vec{F} \): \( F_V = F\sin 45^\circ = (8)\left(\frac{1}{\sqrt{2}}\right) = 4\sqrt{2} \) N

Example 11. Find the net displacement of a particle from its starting point if it undergoes two successive displacements given by \( \vec{S}_1 \) = 20 m, 37° north of west and \( \vec{S}_2 \) = 50 m, 53° north of east.
Answer:

Take east as the x-axis and north as the y-axis. \( \vec{S} = \vec{S}_1 + \vec{S}_2 \)
S_x = S_1x + S_2x = −20 cos 37° + 50 cos 53° = −16 + 30 = 14
S_y = S_1y + S_2y = 20 sin 37° + 50 sin 53° = 12 + 40 = 52
\( S = \sqrt{S_x^2 + S_y^2} = \sqrt{(14)^2 + (52)^2} = 53.85 \) m
Angle from the west–east axis (x-axis): \( \tan\theta = \frac{S_y}{S_x} = \frac{52}{14} = \frac{26}{7} \) ⇒ \( \theta = \tan^{-1}\left(\frac{26}{7}\right) \) north of east

Example 12. Find the magnitude of \( \vec{B} \) and the direction of \( \vec{A} \), if \( \vec{B} \) makes an angle 37° and \( \vec{C} \) makes 53° with the x-axis, \( \vec{A} \) has magnitude equal to 10 and \( \vec{C} \) has magnitude 5. (Given \( \vec{A} + \vec{B} + \vec{C} = 0 \))

Answer:
\( -\vec{A} = \vec{B} + \vec{C} \), with \( \vec{A} = A_x\hat{i} + A_y\hat{j} \)
⇒ A_x = −(B cos 37° + C cos 53°) and A_y = −(B sin 37° + C sin 53°)
\( |\vec{A}|^2 = A_x^2 + A_y^2 \) ⇒ \( 10^2 = \left(\frac{4B}{5} + 5 \times \frac{3}{5}\right)^2 + \left(\frac{3B}{5} + 5 \times \frac{4}{5}\right)^2 = \left(\frac{4B}{5} + 3\right)^2 + \left(\frac{3B}{5} + 4\right)^2 \)
⇒ \( 100 = \frac{16}{25}B^2 + \frac{9}{25}B^2 + 25 + 2\left(\frac{4 \times 3}{5} + \frac{3 \times 4}{5}\right)B \) ⇒ \( B^2 + \frac{48}{5}B - 75 = 0 \)
⇒ \( B = -4.8 + \sqrt{(4.8)^2 + 75} = -4.8 + \sqrt{98.04} \approx 5.1 \) (magnitude cannot be negative)
Direction of \( \vec{A} \): A_x = −(0.8 × 5.1 + 3) ≈ −7.08, A_y = −(0.6 × 5.1 + 4) ≈ −7.06
\( \tan\theta = \frac{A_y}{A_x} = \frac{-7.06}{-7.08} \approx 1 \); both components are negative, so θ ≈ 180° + 45° = 225° from the +x axis (i.e. \( \vec{A} \) points about 45° below the negative x-axis).

Example 13. Find the magnitudes of F₁ and F₂, if F₁ and F₂ make angles 30° and 45° with F₃ and the magnitude of F₃ is 10 N. (Given \( \vec{F}_1 + \vec{F}_2 = \vec{F}_3 \))

Answer:
Along \( \vec{F}_3 \): \( |\vec{F}_3| = F_1\cos 30^\circ + F_2\cos 45^\circ \); perpendicular to \( \vec{F}_3 \): F₂ sin 45° = F₁ sin 30°
⇒ \( 10 = \frac{\sqrt{3}F_1}{2} + \frac{F_2}{\sqrt{2}} \) and \( \frac{F_2}{\sqrt{2}} = \frac{F_1}{2} \)
⇒ \( 10 = \frac{\sqrt{3}F_1}{2} + \frac{F_1}{2} \) ⇒ \( F_1 = \frac{20}{\sqrt{3} + 1} \) N and \( F_2 = \frac{F_1}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{3} + 1} \) N

Example 14. If two vectors \( \vec{A} \) and \( \vec{B} \) make angles 30° and 60° respectively with their resultant and \( \vec{B} \) has magnitude equal to 10, then find the magnitude of \( \vec{A} \).
Answer:

By the short method, the components perpendicular to the resultant are equal: B sin 60° = A sin 30°
⇒ 10 sin 60° = A sin 30° ⇒ \( A = 10 \times \frac{\sqrt{3}}{2} \times 2 = 10\sqrt{3} \)

Example 15. If \( \vec{A} \) and \( \vec{B} \) have an angle between them equal to 60° and their resultant makes an angle 45° with \( \vec{A} \), and \( \vec{A} \) has magnitude equal to 10, then find the magnitude of \( \vec{B} \).
Answer:

Here α = 45° and β = 60° − 45° = 15°
So A sin α = B sin β ⇒ 10 sin 45° = B sin 15°
So \( B = \frac{10\sin 45^\circ}{\sin 15^\circ} = \frac{10/\sqrt{2}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} = \frac{20}{\sqrt{3} - 1} = 10(\sqrt{3} + 1) \)

Example 16. Obtain the magnitude of \( 2\vec{A} - 3\vec{B} \) if \( \vec{A} = \hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{B} = 2\hat{i} - \hat{j} + \hat{k} \).
Answer:
\( 2\vec{A} - 3\vec{B} = 2(\hat{i} + \hat{j} - 2\hat{k}) - 3(2\hat{i} - \hat{j} + \hat{k}) = -4\hat{i} + 5\hat{j} - 7\hat{k} \)
∴ Magnitude of \( 2\vec{A} - 3\vec{B} = \sqrt{(-4)^2 + (5)^2 + (-7)^2} = \sqrt{16 + 25 + 49} = \sqrt{90} \)

Example 17. Find \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \) if \( \vec{A} \) makes an angle 37° with the positive x-axis and \( \vec{B} \) makes an angle 53° with the negative x-axis as shown, and the magnitude of \( \vec{A} \) is 5 and of \( \vec{B} \) is 10.

Answer:
Components: A_x = A cos 37° = 4, A_y = A sin 37° = 3; B_x = −B cos 53° = −6, B_y = B sin 53° = 8
For \( \vec{A} + \vec{B} \): x-component = 4 − 6 = −2, y-component = 3 + 8 = 11
So the magnitude of the resultant = \( \sqrt{11^2 + (-2)^2} = \sqrt{125} = 5\sqrt{5} \), at angle \( \theta = \tan^{-1}\left(\frac{11}{2}\right) \) from the negative x-axis towards up.
For \( \vec{A} - \vec{B} \): x-component = 4 − (−6) = 10, y-component = 3 − 8 = −5
So the magnitude of the resultant = \( \sqrt{10^2 + (-5)^2} = 5\sqrt{5} \), at angle \( \theta = \tan^{-1}\left(\frac{5}{10}\right) \) from the positive x-axis towards down.

Example 18. If the vectors \( \vec{P} = a\hat{i} + a\hat{j} + 3\hat{k} \) and \( \vec{Q} = a\hat{i} - 2\hat{j} - \hat{k} \) are perpendicular to each other, find the value of a.
Answer:
If vectors \( \vec{P} \) and \( \vec{Q} \) are perpendicular, \( \vec{P} \cdot \vec{Q} = 0 \) ⇒ \( (a\hat{i} + a\hat{j} + 3\hat{k}) \cdot (a\hat{i} - 2\hat{j} - \hat{k}) = 0 \)
⇒ a² − 2a − 3 = 0 ⇒ a² − 3a + a − 3 = 0 ⇒ a(a − 3) + 1(a − 3) = 0 ⇒ a = −1, 3

Example 19. Find the component of \( 3\hat{i} + 4\hat{j} \) along \( \hat{i} + \hat{j} \).
Answer:
The component of \( \vec{A} \) along \( \vec{B} \) is given by \( \frac{\vec{A} \cdot \vec{B}}{B} \); hence the required component \( = \frac{(3\hat{i} + 4\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{7}{\sqrt{2}} \)

Example 20. Find the angle between \( \vec{A} = 3\hat{i} + 4\hat{j} \) and \( \vec{B} = 12\hat{i} + 5\hat{j} \).
Answer:
We have \( \cos\theta = \frac{\vec{A} \cdot \vec{B}}{AB} = \frac{(3\hat{i} + 4\hat{j}) \cdot (12\hat{i} + 5\hat{j})}{\sqrt{3^2 + 4^2}\sqrt{12^2 + 5^2}} = \frac{36 + 20}{5 \times 13} = \frac{56}{65} \)
\( \theta = \cos^{-1}\left(\frac{56}{65}\right) \)

Example 21. (i) For what value of m is the vector \( \vec{A} = 2\hat{i} + 3\hat{j} - 6\hat{k} \) perpendicular to \( \vec{B} = 3\hat{i} - m\hat{j} + 6\hat{k} \)?
(ii) Find the component of vector \( \vec{A} = 2\hat{i} + 3\hat{j} \) along the direction of \( \hat{i} + \hat{j} \).
Answer:
(i) \( \vec{A} \cdot \vec{B} = 0 \) ⇒ 6 − 3m − 36 = 0 ⇒ m = −10
(ii) Component \( = \frac{(2\hat{i} + 3\hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{5}{\sqrt{2}} \)

Example 22. The velocity of a particle is given by \( \vec{v} = 3\hat{i} + 2\hat{j} + 3\hat{k} \). Find the vector component of its velocity parallel to the line \( \vec{l} = \hat{i} - \hat{j} + \hat{k} \).
Answer:
Component of \( \vec{v} \) along \( \vec{l} \) \( = v\cos\theta\,\hat{l} = v\frac{\vec{v} \cdot \vec{l}}{vl}\hat{l} = \frac{\vec{v} \cdot \vec{l}}{l^2}\vec{l} \)
\( = \frac{(3\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k})}{|\hat{i} - \hat{j} + \hat{k}|^2}(\hat{i} - \hat{j} + \hat{k}) = \frac{4}{3}(\hat{i} - \hat{j} + \hat{k}) \)

Example 23. \( \vec{A} \) is eastwards and \( \vec{B} \) is downwards. Find the direction of \( \vec{A} \times \vec{B} \).
Answer:
Applying the right-hand-thumb rule we find that \( \vec{A} \times \vec{B} \) is along north.

Example 24. If \( \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \), find the angle between \( \vec{A} \) and \( \vec{B} \).
Answer:
\( \vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}| \) ⇒ AB cos θ = AB sin θ ⇒ tan θ = 1 ⇒ θ = 45°

Example 25. Find \( \vec{A} \times \vec{B} \) if \( \vec{A} = \hat{i} - 2\hat{j} + 4\hat{k} \) and \( \vec{B} = 3\hat{i} - \hat{j} + 2\hat{k} \).
Answer:
\( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 4 \\ 3 & -1 & 2 \end{vmatrix} = \hat{i}(-4 - (-4)) - \hat{j}(2 - 12) + \hat{k}(-1 - (-6)) = 10\hat{j} + 5\hat{k} \)

Example 26. (i) \( \vec{A} \) is north-east and \( \vec{B} \) is downwards; find the direction of \( \vec{A} \times \vec{B} \).
(ii) Find \( \vec{B} \times \vec{A} \) if \( \vec{A} = 3\hat{i} - 2\hat{j} + 6\hat{k} \) and \( \vec{B} = \hat{i} - \hat{j} + \hat{k} \).
Answer:
(i) By the right-hand-thumb rule, \( \vec{A} \times \vec{B} \) is towards north-west.
(ii) \( \vec{B} \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & -2 & 6 \end{vmatrix} = \hat{i}(-6 + 2) - \hat{j}(6 - 3) + \hat{k}(-2 + 3) = -4\hat{i} - 3\hat{j} + \hat{k} \)