Unit: Functions involves Parameters, Vectors & Matrices
Chapter: Vectors for Motion of Object
Reference: – Basic vectors, Scalar, Vector, Vector representation, Vector addition, Vector subtraction, Position, Displacement & Velocity, Velocity vectors, Acceleration, Projectile motion, Effect of gravity, Tangent vectors, Normal vectors, Arc length, path Integral.
After studying this chapter, you should be able to:
Introduction to Vectors for motion & Their types.
Position, Displacement & velocity.
Effect of Gravity, Tangent vectors & Normal vectors.
Arc length, Path Integral, Applications & Properties.
Introduction to Vectors for motion
The vectors are defined as an object containing both magnitude and direction. Vector describes the movement of an object from one point to another. Vector math can be geometrically picturized by the directed line segment. The length of the segment of the directed line is called the magnitude of a vector and the angle at which the vector is inclined shows the direction of the vector. The beginning point of a vector is called "Tail," and the end side (having arrow) is called "Head."
Definition: A quantity that has magnitude, as well as direction, is called a vector.
Notice that a directed line segment is a vector (above figure), denoted as ¯AB or simply as a ⃗, and read as ‘vector ¯AB’ or ‘vector a ⃗’.
The point A from where the vector ¯AB starts is called its initial point, and the point B where it ends is called its terminal point. The arrow indicates the direction of the vector.
Position Vector
The position vector is used to specify the position of a certain body. The position vector of an object is measured from the origin, in general.
Position vector (r ⃗) = xi ̂ + yj ̂ + zk ̂
where,
i ̂ = unit vector along x direction
j ̂ = unit vector along y direction
k ̂ = unit vector along z direction
Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector ¯OP having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. Using the distance formula, the magnitude of ¯OP (or r ⃗) is given by
|¯OP|= √(x^2 + y^2 + z^2 )
In practice, the position vectors of points A, B, C, etc., with respect to the origin O are denoted by a ⃗,b ⃗,c ⃗ etc., respectively.
Direction Cosines:
Let us consider a point P lying in space, and if its position vector makes positive angles (anticlockwise direction) of α, β and γ with the positive x, y, and z-axis, respectively, then these angles are known as direction angles and on taking the cosine of these angles, we get direction cosines. Taking direction cosines makes it easy to represent the direction of a vector in terms of angles with respect to the reference.
The coordinates of the point P may also be expressed as the product of the magnitude of the given vector and the cosines of direction on the three axes, i.e.
x = l|r ⃗ |
y = m|r ⃗|
z = n|r ⃗|
Where l, m, n represent the direction cosines of the given vector on the axes x, y, z, respectively.
Note:
(i) The coordinates of the point P may also be expressed as (lr, mr, nr). The numbers lr, mr and nr, proportional to the direction cosines, are called as direction ratios of vector r ⃗, and denoted as a, b and c, respectively.
(ii) l2 + m2 + n2 = 1 but a2 + b2 + c2 1.
Types of Vectors:
Zero Vector: A vector whose initial and terminal points coincide is called a zero vector or (null vector). It is denoted by 0.
The zero vector cannot be assigned a definite direction as it has zero magnitudes.
The vectors ¯AA,¯BB represents the zero vector.
Unit Vector: A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector.
The unit vector in the direction of a given vector a ⃗ is denoted by a ̂.
Coinitial Vectors: Two or more vectors having the same initial point are called coinitial vectors.
Collinear Vectors: Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.
For example: Consider 3 vectors as shown in the figure, they all are parallel to each other but their magnitudes are different as well as the directions. But they are said to be collinear vectors because they are parallel to each other.
Equal Vectors: Two vectors a ⃗ and b ⃗ are said to be equal if they have the same magnitude and direction regardless of the positions of their initial points, and written as a ⃗ and b ⃗.
For example: Consider 2 vectors whose magnitudes and their directions are the same irrespective of origin, then they are known as equal vectors.
Negative of a Vector: A vector whose magnitude is the same as that of a given vector but direction is opposite to that of it is called negative of the given vector.
Example:
Represent the following graphically:
i. a displacement of 40 km, 30° east of north
ii. a displacement of 50 km south – east
iii. a displacement of 70 km, 40° north of west.
Solution:
i. a displacement of 40 km, 30° east of north
Step 1: Draw north, south, east and west as shown below:
Step 2: Plot a line (OP) ̅ 30° east of north as shown below:
Step 3: Define scale and mark 40km on line (OP) ̅
Let the scale be 10km = 1cm
∴ (OP) ̅ represents the displacement of 40 km, 30o East of North
ii. a displacement of 50 km south – east
Step 1: Draw north, south, east and west as shown below:
Step 2: As the displacement should be south – east, the angle between the displacement and east (or south) will be 45°. Now, plot a line (OP) ̅ 45° east of south as shown below:
Step 3: Define scale and mark point R such that OR = 50km on line (OP) ̅. Let the scale be 10km = 1cm
∴ (OR) ̅ represents the displacement of 50 km south – east
iii. A displacement of 70 km, 40o north of west.
Step 1: Draw north, south, east and west as shown below:
Step 2: Plot a line (OP) ̅ 40° north of west as shown below:
Step 3: Define scale and mark point R such that OR = 70km on line(OP) ̅ .Let the scale be 10km = 1cm
∴ (OP) ̅ represents the displacement of 70 km, 40o north of west
Example: Represent graphically a displacement of 40 km, 30° west of south.
Solution: The vector ¯OP represents the required displacement.
Example: Classify the following measures as scalars and vectors.
(i) 200 g/cm3 (ii) 20 km/hr
(iii) 10 Newton (iv) 1000 cm3
(v) 500 minute (vi) 50 m/s towards north
Solution
(i) Density-scalar (ii) Speed-scalar
(iii) Force-vector (iv) Volume-scalar
(v) Time-scalar (vi) Velocity-vector
Example: In the given figure, which of the vectors are:
(i) Collinear (ii) Equal (iii) Coinitial
Solution
(i) Collinear vectors: a ⃗,c ⃗ and d ⃗.
(ii) Equal vectors: a ⃗ and c ⃗.
(iii) Coinitial vectors: b ⃗,c ⃗ and d ⃗.
Vector Addition
A variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. Two vectors can be added together to determine the result (or resultant). This process of adding two or more vectors has already been discussed in an earlier unit.
The two vectors a and b can be added, giving the sum to be a + b. This requires joining them head to tail.
Characteristics of Vector Math Addition
Commutative Law: a + b = b + a
Associative law: (a + b) + c = a + (b + c)
Triangle law of vector addition
The Triangle law of vector addition is appropriate to deal with such a situation. If two vectors are represented by two sides of a triangle in sequence, then the third closing side of the triangle, in the opposite direction of the sequence, represents the sum (or resultant) of the two vectors in both magnitude and direction.
A vector ¯AB simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C.
The net displacement made by the girl from point A to the point C, is given by the vector ¯AC and expressed as
¯AC=¯AB+¯BC
This is known as the triangle law of vector addition.
Note:
(i) The associative property of vector addition enables us to write the sum of three vectors a ⃗,b ⃗,c ⃗ as a ⃗+b ⃗+c ⃗ without using brackets.
(ii) a ⃗+0 ⃗=0 ⃗+a ⃗=a ⃗
Here, the zero vector a ⃗ is called the additive identity for the vector addition.
Vector Subtraction
To subtract two vectors, you put their feet (or tails, the non-pointy parts) together, then draw the resultant vector, which is the difference of the two vectors, from the head of the vector you're subtracting to the head of the vector you're subtracting it from.
A reverse vector (-a), which is opposite of (a) has a similar magnitude as (a) but pointed in the opposite direction.
Multiplication of a Vector by a Scalar
Multiplication of a vector by a scalar quantity is called “Scaling.” In this type of multiplication, only the magnitude of a vector is changed not the direction.
S(a + b) = Sa + Sb
(S + T)a = Sa + Ta
a.1 = a
a.0 = 0
a.(-1) = -a
Geometric visualization of multiplication of a vector by a scalar
A scalar, however, cannot be multiplied by a vector. To multiply a vector by a scalar, simply multiply the similar components, that is, the vector's magnitude by the scalar's magnitude. This will result in a new vector with the same direction but the product of the two magnitudes.
When = –1, then a ⃗=-a ⃗, which is a vector having a magnitude equal to the magnitude a ⃗ of and direction opposite to that of the direction of a ⃗. The vector -a ⃗ is called the negative (or additive inverse) of vector a ⃗ and we always have
a ⃗+(-a ⃗ )=(-a ⃗ )+a ⃗=0 ⃗
Also, if =1/(|a ⃗|), provided a ⃗≠0 i.e. a ⃗ is not a null vector, then
|a ⃗| =|||a ⃗| = 1/(|a ⃗|) |a ⃗ |=1
So, a ⃗ represents the unit vector in the direction of a ⃗. We write it as
a ̂=1/(|a ⃗|) a ⃗
Note: For any scalar k,k0 ⃗=0 ⃗.
Components of a vector
When you break a vector into its parts, those parts are called its components. For example, in the vector (4, 1), the x-axis (horizontal) component is 4, and the y-axis (vertical) component is 1.
Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively.
Then, clearly
|¯OA| = 1, |¯OB|= 1 and |¯OC| = 1
The vectors ¯OA,¯OB and ¯OC, each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by i ̂,j ̂ and k ̂, respectively (Fig. 10.13).
Now, consider the position vector ¯OP of a point P(x, y, z). Let P1 be the foot of the perpendicular from P on the plane XOY.
We, thus, see that P1 P is parallel to z-axis. As i ̂,j ̂ and k ̂ are the unit vectors along the x, y and z-axes, respectively, and by the definition of the coordinates of P, we have ¯(P_1 P)=¯QR=zk ̂ similarly, ¯(OP_1 )=¯OS=yj ̂ and ¯OQ=xi ̂.
Therefore, it follows that ¯(OP_1 )=¯OQ+¯(〖QP〗_1 )=xi ̂+yj ̂
and ¯OP=¯(OP_1 )+¯(P_1 P)=xi ̂+yj ̂+zk ̂
Hence, the position vector of P with reference to O is given by
¯OP (or r ⃗ )=xi ̂+yj ̂+zk ̂
This form of any vector is called its component form. Here, x, y and z are called as the scalar components of r ⃗, and xi ̂+yj ̂ and zk ̂ are called the vector components of r ⃗ along the respective axes. Sometimes x, y and z are also termed as rectangular components.
The length of any vector r ⃗=xi ̂+yj ̂+zk ̂, is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP1
|¯(〖OP〗_1 )|=√(〖|¯OQ|〗^2+〖|¯(QP_1 )|〗^2 )=√(x^2+y^2 )
and in the right angle triangle OP1P, we have
Hence, the length of any vector r ⃗=xi ̂+yj ̂+zk ̂ is given by
|r ⃗ |=|xi ̂+yj ̂+zk ̂ |=√(x^2+y^2+z^2 )
If a ⃗ and b ⃗ are any two vectors given in the component form a_1 i ̂+a_2 j ̂+a_3 k ̂ and b_1 i ̂+b_2 j ̂+b_3 k ̂, respectively, then.
(i) the sum (or resultant) of the vectors a ⃗ and b ⃗ is given by
a ⃗+b ⃗=(a_1+b_1 ) i ̂+(a_2+b_2 ) j ̂+(a_3+b_3 ) k ̂
(ii) the difference of the vector a ⃗ and b ⃗ is given by
a ⃗-b ⃗=(a_1-b_1 ) i ̂+(a_2-b_2 ) j ̂+(a_3-b_3 ) k ̂
(iii) the vectors a ⃗ and b ⃗ are equal if and only if
a_1=b_1,a_2=a_2 and a_3=a_3
(iv) the multiplication of vector a ⃗ by any scalar is given by
a ⃗=(a_1 ) i ̂+(〖a〗_2 ) j ̂+(〖a〗_3 ) k ̂
The addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:
Let a ⃗ and b ⃗ be any two vectors, and k and m be any scalars. Then
ka ⃗+ma ⃗=(k+ma)a ⃗
k(ma ⃗ )=(km)a ⃗
k(a ⃗+b ⃗ )=ka ⃗+kb ⃗
Note:
(i) One may observe that whatever be the value of , the vector a ⃗ is always collinear to the vector a ⃗. In fact, two vectors a ⃗ and b ⃗ are collinear if and only if there exists a nonzero scalar such that b ⃗=a ⃗. If the vectors a ⃗ and b ⃗ are given in the component form, i.e. a ⃗=a_1 i ̂+a_2 j ̂+a_3 k ̂ and b ⃗=b_1 i ̂+b_2 j ̂+b_3 k ̂ then the two vectors are collinear if and only if
b_1 i ̂+b_2 j ̂+b_3 k ̂=〖(a〗_1 i ̂+a_2 j ̂+a_3 k ̂)
b_1 i ̂+b_2 j ̂+b_3 k ̂=〖(a〗_1)i ̂+(a_2)j ̂+〖(a〗_3)k ̂
b_1 i ̂=〖a〗_1,+b_2=a_2,b_3=〖a〗_3
b_1/a_1 =b_2/a_2 =b_3/a_3 =
(ii) If a ⃗=a_1 i ̂+a_2 j ̂+a_3 k ̂, then a1, a2, a3 are also called direction ratios
of a ⃗.
(iii) In case if it is given that l, m, n are direction cosines of a vector, then li ̂+mj ̂+nk ̂=(cosα ) i ̂+(cosβ) j ̂+(cosγ)k ̂ is the unit vector in the direction of that vector, where , and are the angles which the vector makes with x, y and z axes respectively.
Example: If a=xi+2j−zk and b=3i−yj+k are two equal vectors, then write the value of x + y + z
Solution:
Given: a=xi+2j−zk and b=3i−yj+k are two equal vectors.
To find: value of x + y + z …(i)
According to question,
Comparing both the sides, we get
x = 3, 2 = -y and -z = 1
⇒ x = 3, y = -2 and z = -1
Putting the value of x, y and z in eq. (i), we get
x + y + z = 3 + (-2) + (-1)
= 3 – 2 – 1
= 0
Hence, the value of x + y + z is 0
Example: Find the values of x, y and z so that the vectors
a ⃗=xi ̂+2j ̂+zk ̂ and b ⃗=2i ̂+yj ̂+k ̂ are equal.
Solution: Given vectors are equal, therefore
a ⃗=b ⃗
xi ̂+2j ̂+zk ̂=2i ̂+yj ̂+k ̂
x = 2, y = 2, z = 1.
Example: Let a ⃗=(5i) ̂+10j ̂ and b ⃗=10i ̂+(5j) ̂. Is |a ⃗|=|b ⃗|?
Are the vectors a ⃗ and b ⃗ equal?
Solution: a ⃗=(5i) ̂+10j ̂ and b ⃗=10i ̂+(5j) ̂
|a ⃗ |=√(5^2+10^2 )=√125 and |b ⃗ |=√(10^2+25^2 )=√125
So, |a ⃗|=|b ⃗|.
But, the two vectors are not equal since their corresponding components are distinct.
Example: Find unit vector in the direction of vector a ⃗=5i ̂+3j ̂+2k ̂.
Solution: a ⃗=5i ̂+3j ̂+2k ̂
The unit vector in the direction of a vector a ⃗ is given by a ⃗=1/(|a ⃗|) a ⃗
|a ⃗ |=√(5^2+3^2+2^2 )=√(25+9+4)=√38
Therefore a ⃗=1/√38 (5i ̂+3j ̂+2k ̂ )=5/√38 i ̂+3/√38 j ̂+2/√38 k ̂
Example: Find a vector in the direction of vector a ⃗=5i ̂-(5j) ̂ that has magnitude 11 units.
Solution: a ⃗=5i ̂-(5j) ̂
The unit vector in the direction of the given vector a ⃗ is
|a ⃗ |=√(5^2+5^2 )=√50=5√2
a ̂=1/(|a ⃗|) a ⃗=1/(5√2) ((5i) ̂-5j ̂ )=1/√2 i ̂-1/√2 j ̂
Therefore,
Required vector = 11a ̂=11(1/√2 i ̂-1/√2 j ̂ )=11/√2 i ̂-11/√2 j ̂
Example: Write the direction ratio’s of the vector a ⃗=(2i) ̂+4j ̂-2k ̂ and hence calculate its direction cosines.
Solution: a ⃗=(2i) ̂+4j ̂-2k ̂
We know, direction ratio’s a, b, c of a vector r ⃗=xi ̂+yj ̂+zk ̂ are just the respective components x, y and z of the vector.
So, a = 2, b = 4 and c = –2
|r ⃗ |=√(4+16+4)=√24
Let l, m and n are the direction cosines of the given vector.
l=a/(|r ⃗|)=2/√24 , m=b/(|r ⃗|)=4/√24 ,n=c/(|r ⃗|)=(-2)/√24
Thus, the direction cosines are (2/√24 ,4/√24 ,-2/√24).
Vector joining two points
If P1(x1, y1, z1) and P2(x2, y2, z2) are any two points, then the vector joining P1 and P2 is the vector ¯(P_1 P_2 ).
Joining the points P1 and P2 with the origin O, and applying triangle law, from the triangle OP1P2, we have
¯(〖OP〗_1 )+¯(P_1 P_2 )=¯(OP_2 )
Using the properties of vector addition, the above equation becomes
¯(P_1 P_2 )=¯(〖OP〗_2 )-¯(OP_1 )
¯(P_1 P_2 )=(x_2 i ̂+y_2 j ̂+z_2 k ̂ )–(x_1 i ̂+y_1 j ̂+z_1 k ̂ )
=(x_2-x_1 ) i ̂+(y_2-y_1 ) j ̂+(z_2-z_1 ) k ̂
The magnitude of vector ¯(P_1 P_2 ) is given by
|¯(P_1 P_2 )|=√(〖(x_2-x_1)〗^2+〖(y_2-y_1)〗^2+〖(z_2-z_1)〗^2 )
Example: Find the position vector of the mid-point of the vector joining the points
and
Solution:
Formula to be used – The midpoint of a line joining points a and b is given by .(a+b)/2
The position vector of the midpoint
Example: Find the vector joining the points A(1, 2, 3) and
B(–2, –3, –4) directed from A to B.
Solution: Since the vector is to be directed from A to B, so
A is the initial point and B is the terminal point.
Therefore,
Required vector = ¯AB=(-2-1) i ̂+(-3-2) j ̂+(-4-3)k ̂
¯AB=-3i ̂-5j ̂-7k ̂
Section formula
If O is taken as reference origin and A is an arbitrary point in space then the vector (OA) ⃗ is called as the position vector of the point. Let us consider two points P and Q denoted by position vectors (OP) ⃗ and (OQ) ⃗ with respect to origin O.
Let us consider that the line segment connecting P and Q is divided by a point R lying on PQ. The point R can divide the line segment PQ in two ways: internally and externally. Let us consider both these cases individually.
Case 1: Line segment PQ is divided by R internally
Let us consider that the point R divides the line segment PQ in the ratio m: n; given that m and n are positive scalar quantities we can say that,
m¯RQ=n¯PR
Consider the triangles, ∆ORQ and ∆OPR.
¯RQ=¯OQ-¯OR=b ⃗-r ⃗
and ¯PR=¯OR-¯OP=r ⃗-a ⃗,
Therefore,
m(b ⃗-r ⃗ )=n(r ⃗-a ⃗)
r ⃗=(mb ⃗+na ⃗)/(m+n)
Therefore the position vector of point R dividing P and Q internally in the ratio m:n is given by:
(OR) ⃗=(mb ⃗+na ⃗)/(m+n)
Case 2: Line segment PQ is divided by R externally
(OR) ⃗=(mb ⃗-na ⃗)/(m-n)
Note: If R is the midpoint of PQ, then m = n.
Therefore, (OR) ⃗=(a ⃗+b ⃗)/2
Example:
Show that the points A, B and C having position vectors.
And
Respectively, form the vertices of a right-angled triangle.
Solution:
Tip – For any 2 perpendicular vectors
a ⃗ & b ⃗, a ⃗ . b ⃗= 0
The triangle is right-angled.
Example: Consider two points P and Q with position vectors
¯OP=5a ⃗-3b ⃗ and ¯OQ=a ⃗+2b ⃗. Find the position vector of a point R which divides the line joining P and Q in the ratio 3:1, (i) internally, and
(ii) externally.
Solution
(i) Required position vector is
¯OR=(2(a ⃗+2b ⃗ )+(5a ⃗-3b ⃗))/(3+1)=(7a ⃗+b ⃗)/4
(ii) Required position vector is
¯OR=(2(a ⃗+2b ⃗ )-(5a ⃗-3b ⃗))/(3+1)=(-3a ⃗+7b ⃗)/4
Example: Show that the points A=(2i ̂-j ̂+k ̂ ),B=(i ̂-3j ̂-5k ̂ ), C=(3i ̂-4j ̂-4k ̂ ), are the vertices of a right angled triangle.
Solution: A=(2i ̂-j ̂+k ̂ ),B=(i ̂-3j ̂-5k ̂ ), C=(3i ̂-4j ̂-4k ̂ )
¯AB=(1-2) i ̂+(-3+1) j ̂+(-5-1) k ̂=-i ̂-2j ̂-6k ̂
¯BC=(3-1) i ̂+(-4+3) j ̂+(-4+5) k ̂=2i ̂-j ̂+k ̂
¯CA=(2-3) i ̂+(-1+4) j ̂+(1+4) k ̂=-i ̂+3j ̂+5k ̂
So, |¯AB|^2=41=6+35=|¯BC|^2+|¯CA|^2
Therefore, A, B and C are the vertices of the right-angled triangle.
Key Points
Vectors represent quantities that have both magnitude and direction, such as displacement, velocity, and acceleration.
Scalars, on the other hand, are quantities that only have magnitude, like distance and speed.
Position vectors describe the location of an object in a coordinate system, often using components like x, y, and z.
Displacement vectors represent the change in position of an object and are calculated by subtracting initial position from final position.
Velocity vectors describe the rate at which an object's position changes with respect to time and are calculated by dividing displacement by time.
Acceleration vectors represent the rate of change of velocity and are calculated by dividing the change in velocity by time.
Average velocity is the total displacement divided by the total time taken.
Instantaneous velocity is the velocity of an object at a specific instant in time and is found by taking the derivative of the position vector.
Average acceleration is the change in velocity divided by the change in time.
Instantaneous acceleration is the acceleration of an object at a specific instant in time and is found by taking the derivative of the velocity vector.
Constant acceleration equations can be used to analyze motion under uniform acceleration, such as free fall or motion along an inclined plane.
Projectile motion involves objects moving in two dimensions under the influence of gravity, with horizontal and vertical components of velocity.
Parametric equations and vector functions are used to describe the motion of objects in space.
The derivative of a vector function gives the velocity vector, while the second derivative gives the acceleration vector.
Integrals of vector functions can be used to calculate arc length, work done, or other quantities related to motion.