Notes-Part-1-Class-11-Science-Physics-Chapter-2-Mathematical Methods-Maharashtra Board

Examples of different types of vectors :

(i) Null or zero vector : A null or zero vector is defined to be a vector of zero magnitude.

So that for any vector \(\vec{X}\), \(\vec{X}\) + \(\vec{0}\)  = \(\vec{X}\); \(\vec{0}\) is called the additive identity.

• A null vector is not a proper vector because we cannot speak of its direction.
• Examples ; If a particle moves from A to B and then back to A (not necessarily along the same path), its net displacement is a null or zero vector.
• Other examples of a zero vector are (i) the velocity of a stationary particle (ii) the acceleration of a particle moving with uniform velocity.

(ii) Resultant vector: The resultant of two or more vectors is that single vector, which produces the same effect, as produced by all the vectors together is called resultant vector.

• The process of finding the resultant of two or more vectors is called the composition or addition of vectors.

(iii) Negative vector (opposite vector): A negative vector of a given vector is a vector of the same magnitude but opposite in direction to that of the given vector.

Figure shows a vector \(\vec{B}\) which is equal in magnitude to the vector \(\vec{A}\)  but oppositely directed.

Symbolically, we write \(\vec{B}\) = \(−\vec{A}\) . From the procedure for addition of two vectors and the definition of the null vector, it follows that  \(\vec{A} + (−\vec{A})\) = \(\vec{0}\), \(−\vec{A}\) is called the additive inverse of \(\vec{A}\).

We may also write \(\vec{B}\) = (— 1)\(\vec{A}\) , i.e., \(\vec{B}\) is a scalar multiple of \(\vec{A}\) in the special case when the scalar is the negative integer (— 1). By the same token, We may say \(\vec{A}\) = (—1)\(\vec{B}\) , i.e., \(\vec{A}\) is the negative of the vector \(\vec{B}\).

(iv) Equal vectors : Two vectors are said to be equal if they represent the same physical quantity and have the same magnitude and direction. 

• Two vectors of the same type are parallel if they have the same direction; they are equal only when they also have the same magnitude.

(v) Position vector : A vector which gives the position of a particle at a point with respect to the origin of a chosen coordinate system is called the position vector of the particle.

In Fig   \(\vec{r}\) = \(\vec{OP}\)  is the position vector of the particle present at P.

(vi) Unit vector : A unit vector is a dimensionless vector with the magnitude 1 (pure number).

• A unit vector is used only to specify a given direction or the direction of a vector quantity.
• A unit vector is usually denoted by a lower case letter with a caret symbol ( ^{^} ) above it, e.g. \(\hat{u}\)  . For example, given a nonzero vector \(\vec{F}\), a unit vector \(\hat{u}\)   in its direction is given by the ratio of the vector to its magnitude \(\hat{u}=\frac{\vec{F}}{F}\)

• Thus, the vector  \(\vec{F}\) = \(F\hat{u}\) 
• Since a unit vector in the direction of a given vector is given by the ratio of the vector to its magnitude, the dimensions and units of a unit vector is none,

Rectangular unit vectors : The unit vectors specifying the positive directions of the Cartesian coordinate axes, x, y and z, are called the rectangular unit Vectors, they are usually labelled as \(\hat{i}\), \(\hat{j}\)  and \(\hat{k}\)  respectively, as shown in Fig.

Since a vector does not change when translated parallel to itself, it is not necessary that these unit vectors be located at the origin of the coordinate system or along the respective axis. They may be moved around preserving their directions.

Examples of a vector, where a vector of different type is obtained when a vector is multiplied by a scalar :

• The product of time t (a scalar) and velocity \(\vec{v}\)  (a vector) gives displacement \(\vec{s}\)  = t \(\vec{Q}\) , which is a different physical quantity.
•  \(\vec{P}\)  = m\(\vec{v}\), where \(\vec{P}\) is momentum of a particle, m is mass of the particle and \(\vec{v}\) is velocity of the particle.
• \(\vec{F}\) = m\(\vec{a}\) , where \(\vec{F}\) is net force on a particle, m is mass of the particle and \(\vec{a}\) is acceleration of the particle.

Example: (i) Force \(\vec{F_1}\)  and \(\vec{F_2}\) force  can be added to give the resultant force

\(\vec{F}\) = \(\vec{F_1}\) + \(\vec{F_2}\)

But a force vector cannot be added to a velocity vector.

(ii) If vectors \(\vec{AB}\) and \(\vec{BC}\)  having the same or opposite direction but different magnitudes. The magnitude of their resultant is the addition of individual magnitudes, i.e., |\(\vec{AC}\)| = |\(\vec{AB}\)|+|\(\vec{BC}\)|  and direction of the resultant is the same as that of the individual vectors as shown in Fig

(iii) If the individual vectors are anti-parallel (i.e., in the opposite direction), the magnitude of their resultant is the difference of the individual magnitudes, and the direction is that of the larger vector i.e., |\(\vec{AC}\)| = |\(\vec{AB}\)|−|\(\vec{BC}\)|  as shown in Fig.

Explanation : A vector is not changed by translation parallel to itself. Preserving the directions of two given vectors, (\vec{P}\) and (\vec{Q}\), the tail of one vector is joined to the head of the other, regardless of order, as shown in Fig. Then, the arrow from the tail of the first vector, to the head of the second gives the resultant \(\vec{R}\) = \(\vec{P}\) + \(\vec{Q}\)  in magnitude and direction

Since the order in which the vectors are drawn to represent the two sides of the triangle is immaterial, it at once follows that \(\vec{P}\) + \(\vec{Q}\) = \(\vec{Q}\) + \(\vec{P}\) , i.e., vector addition is commutative.

• Dimensional homogeneity of an equation requires that two or more physical quantities to be added — all vectors or all scalars — must have the same dimensions and must also be of the same type.

Q. Explain how triangle law of vector addition can be extended to graphically add more than two vectors.

Ans. Given any three vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) representing the same physical quantity, we may draw two sides OP and OQ of ΔOPQ in sequence such that  \(\vec{OP}\) = \(\vec{A}\) and \(\vec{PQ}\) = \(\vec{B}\)  = , as shown in Fig.

Then, by the triangle law of addition,  \(\vec{OQ}\) = \(\vec{A\) + \(\vec{B}\)

Now, drawing the side QR of ΔOQR, such that \(\vec{QR}\) = \(\vec{C\)

\(\vec{OR}\) = \(\vec{OQ\) + \(\vec{QR}\) = (\(\vec{A\) + \(\vec{B}\) ) + \(\vec{C\)

∴ \(\vec{D\) = \(\vec{OR}\) = \(\vec{A\) + \(\vec{B}\) + \(\vec{C\)  is the resultant of \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\)

Therefore, if several given vectors representing the same physical quantity, are drawn in sequence, regardless of order but preserving their magnitudes and directions, to form the sides of a polygon, then the vector drawn from the tail of the first vector to the head of the last vector completing the polygon represents the resultant of the given vectors.

This is called the polygon law of vector addition.

In Fig.  vector \(\vec{OA}\) = \(\vec{P}\) and vector (\vec{OB}\) = \(\vec{Q}\), represent two vectors originating from point O, inclined to each other at an angle θ.

The parallelogram OACB completed, then according to this law, the diagonal \vec{OC}\) = \(\vec{R}\)  represents the resultant vector.

To find the magnitude of (\vec{R}\) drop a perpendicular from C to reach OA (extended) at D. 

In right angled triangle ODC, by application by Pythagoras theorem,

OC² = OD²+DC² = (OA+AD)² + DC²

OC² = OA²+2OA.AD+AD²+DC²

In the right angled triangle ADC, by application of Pythagoras theorem

AD²+DC²=AC²

∴ OC²=OA²+2OA. AD+ AC² --- (1)

Also,

\(\vec{OA}\) = \(\vec{P}\), \(\vec{AC}\) = \(\vec{OB}\) = \(\vec{Q}\)  and \(\vec{OC}\) = \(\vec{R}\)

In Δ ADC, cos θ = AD/AC

∴ AD = AC cos θ = Q cos θ

Substituting in Eq. (1)

R² = P² + Q² + 2 P Q cos θ.

R =  \(\sqrt{P^2 + Q^2 + 2PQ\,cos\,θ}\)    …….(2)

Equation (2) gives us the magnitude of resultant vector

To find the direction of the resultant vector , we will have to find the angle (α) made by  with

In ΔODC, tan α = DC/OD = \(\frac{DC}{OA+OD}\)  …..(3)

From the figure, sin θ = DC/AC

DC = AC sin θ  = Q sin θ

Also,

AD = AC cos θ = Q cos θ

and OA = \(\vec{P}\)

Substituting in Eq. (3), we get

tan α = \(\frac{Q\,sin\,θ}{P+Q\,cos\,θ}\)

∴ α = \(tan^{-1}(\frac{Q\,sin\,θ}{P+Q\,cos\,θ})\)   …..(4)

Equation (4) gives us the direction of resultant vector

If β is the angle between \(\vec{R}\) and \(\vec{Q}\) , it can be. similarly derived that

β = \(tan^{-1}(\frac{P\,sin\,θ}{Q+P\,cos\,θ})\)