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Vector and scalar

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β€’
Nirmal Rizal
Nirmal Rizal

The slide contains very fundamental part of vector with dot product and scalar product. Explained with questionnaire.

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Introduction to scalar and vector Let us begin with the following questions: a. You are at rest position and an object is moving at a speed of 10km/hr. What would be the distance (gap) between you and the object after 30 minutes? It is pretty simple. The distance will be 5 km. b. You are moving to the opposite of an object at a speed of 20 km/hr. The object is moving at a speed of 10km/hr in the North. (i) What would be the distance covered by you in 30 minutes? (ii) What would be the distance (gap) between you and the object in 30 minutes? (iii) What would be the displacement of the object in 30 minutes? (iv) What would be the displacement of the object w.r.t the origin in 15 minutes? (i) It is 10 km. (ii) In 30 minutes, you will cover a distance of 10 km (in the South) and the object will cover a distance of 5 km (in the North). So, a gap of 15 km will be made. (iii) Actually, the question is not precise. It should have been asked in this fashion, 'What would be the displacement of the object w.r.t you after 30 minutes'? The displacement w.r.t you will be 15 km in the case. (iv) It is 2.5 km to the North of the origin. (Note: displacement is a vector & direction must be specified for vectors) Q. How are magnitude and direction imparted to a vector quantity? Magnitude and direction are imparted on the Cartesian plane. To impart the magnitude, a reference point or an initial point is required as the magnitude is the distance between two points. While, to impart the direction, a line is required that passes through the reference point. It will become clear from the following diagram:
The magnitude of π‘Ž βƒ— = |π‘Ž| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = OA. OA can be evaluated if the coordinates of O and A are Known. If O = (0,0) and A = (a,b), OA = βˆšπ‘Ž2 + 𝑏2. It is not mandatory that the direction (ΞΈ) must be always about the horizontal line or x-axis. It can be about any line. But the line must be specified. For example, a ball is moving at a velocity of 5km/hr in the North or in the N45E. Here the line is specified. N45E is 45Β° about the North in clockwise direction. Here the line is specified as the North. So now on, we must be able to decide that the vector π‘Ž βƒ— has the magnitude |π‘Ž| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— and it is directed somewhere. Q. Look at the diagram, there are two vectors equal in magnitude. Are the vectors equal? No. Not only magnitude but also the direction of two vectors must be same in order tobe equal. Representation of vectors & resolution of vectors Vectors can be represented in the coordinate form (p,q) and in the form of unit vectors. If 𝑖 βƒ— and 𝑗 βƒ— are unit vectors along x-axis and y-axis respectively, 𝑝 βƒ— = p𝑖 βƒ— & π‘ž βƒ— = q𝑗 βƒ— ΞΈ O A X Y π‘Ž βƒ— o
This can be written only if 𝑝 βƒ— and 𝑖 βƒ— have same direction and if π‘ž βƒ— and 𝑗 βƒ— have same direction. Here, 𝑝 βƒ— and π‘ž βƒ— are called horizontal and vertical components of the vector π‘Ž βƒ— respectively. The phenomenon of breaking down the vector into 2 components (in 2D) is called vector resolution. In the coordinate form vector π‘Ž βƒ— is represented as π‘Ž βƒ—βƒ—βƒ—βƒ— = (p, q). Note: Direction also must be specified in both forms. Unit vectors have magnitude 1. In the above figure, 𝑝 βƒ— & π‘ž βƒ— are assumed to be greater than their corresponding unit vectors: 𝑝 βƒ— = p𝑖 βƒ— & π‘ž βƒ— = q𝑗 βƒ— Using triangle law of vector addition for above example, Here, the magnitude of π‘Ž βƒ— = |π‘Ž| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = |𝑝 βƒ— + π‘ž βƒ—| = |𝑝𝑖 βƒ— + π‘žπ‘— βƒ—| = distance of OA 𝑖 βƒ— (p,q) ΞΈ O A X Y π‘Ž βƒ— 𝑝 βƒ— π‘ž βƒ— 𝑖 βƒ— 𝑗 βƒ— 𝑝 βƒ— =p𝑖 βƒ— π‘ž βƒ— =q𝑗 βƒ— π‘Ž βƒ— = 𝑝 βƒ— + π‘ž βƒ— = p𝑖 βƒ—+ q𝑗 βƒ— O A
From above, we can infer that a vector (p, q) = π’‘π’Š βƒ— + 𝒒𝒋 βƒ— has: a horizontal component = 𝒑 βƒ—βƒ—βƒ—βƒ— = π’‘π’Š βƒ— and a vertical component = 𝒒 βƒ—βƒ—βƒ—βƒ— = 𝒒𝒋 βƒ— Or, we can say that a vector is equal to the summation of it's components. Q. Can the resultant of two different vector quantities (example force and velocity) be evaluated? No. We cannot decide whether the resultant (𝑅 βƒ—βƒ—) would be force (𝐹 βƒ—) or velocity ( 𝑉 βƒ—βƒ—) or any other quantity. Q. Give a justification for the statement 'Both the horizontal and vertical components of a vector are vectors'. Directions of the components of a vector can be identified w.r.t the vector itself or w.r.t the x- axis or w.r.t the y-axis. Q. How would a vector origin? A vector gets originated due to:  The resultant of 2 vectors  A single vector  Cross/vector product between 2 vectors Let us have a look on the diagram that shows the distribution of forces at 4 different points outside the bounded region. An object of mass 0.5 kg is inside the bounded region and exerting equal force (5N) opposite to the forces as shown below: 𝐹 βƒ— 𝑉 βƒ—βƒ— 𝑅 βƒ—βƒ—
At all the points, the mass of the object will remain unchanged as mass being scalar quantity is not affected by direction. But the resultant force at all the points will remain changed. Resultant of vectors 1. Imagine a ball is being pulled with two forces in opposite directions. Where would the ball go? The ball will move to the right side since the net force or the resultant of two forces is directed along the right side. The net/resultant force = 75-10 = 65N and the direction is along the direction of P. 2N 4N 6N 8N 5N 5N 5N 5N 𝑄 βƒ—βƒ— = 10N 𝑃 βƒ—βƒ— = 75N
2. Imagine a ball is being pulled with two forces in same direction. Where would the ball go? The ball will move to the right side since the net force or the resultant of two forces is directed along the right side. The net/resultant force = 75+10 = 85N and the direction is along the direction of P. In the problem 1: the angle between two forces is 180Β° and the resultant force is directed along the greater force. And, the magnitude of the resultant force = greater force-smaller force. In the problem 2: the angle between two forces is 0Β° and the resultant force is directed along the given forces. And, the magnitude of the resultant force = greater force + smaller force. 3. Imagine a ball is being pulled with two forces in different directions. Where would the ball go? To answer this we need to observe the previous problems. From those problems we can infer that the magnitude and the direction of the resultant force depend on: the angle between two forces and the magnitude of the forces. 75N 10N 10N 75N
4. What would be the direction of the resultant force? Just think of intuitive answer. The direction of the resultant force is always closer to the greater force and farther from the smaller force as the impact of the greater force is greater than that of the smaller force. The ball neither moves along 10N nor along 75N. Hence the ball moves midway between two forces being closer to the 75 N. 5. What would be the magnitude of the resultant force (𝑅 βƒ—βƒ—) if ΞΈ = 60Β°? The resultant force (𝑅 βƒ—βƒ—) will be such that R2 = |𝑃| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +|𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +2|𝑃| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—|𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—CosΞΈ where: 𝑃 βƒ—βƒ— and 𝑄 βƒ—βƒ— are two forces ΞΈ is the angle between two forces |𝑃| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— & |𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— are the magnitudes of 𝑃 βƒ—βƒ— & 𝑄 βƒ—βƒ— respectively. In this example, 𝑃 βƒ—βƒ— = 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—, 𝑄 βƒ—βƒ— = 10𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—, ΞΈ = 60Β° R2 = |75| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +|10| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +2|75| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—|10| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—Cos60 = 100+5625+2.10.75.0.5 = 5725+750= 6475 N2 ΞΈ 10N 75N 𝑅 βƒ—βƒ—
Thus, 𝑅 βƒ—βƒ— = √6475 N = 80.46 N Since, the force is a vector quantity it must have direction too. Let Ο’ be the angle between 𝑅 βƒ—βƒ— & 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—. Then Ο’ will be the direction of 𝑅 βƒ—βƒ— w.r.t. the line of 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—. We can assign the direction for 𝑅 βƒ—βƒ— w.r.t. 10𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— too. The direction of 𝑅 βƒ—βƒ— is such that: tan Ο’ = QSin ΞΈ/(P+QCos ΞΈ) In this example, tan Ο’ = 10.Sin60/(75+10.Cos60) or, tan Ο’ = 8.66/80, or Ο’ = π‘‘π‘Žπ‘›βˆ’1 (8.66/80) = 6Β°. Meaning, 𝑅 βƒ—βƒ— & 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—are at a gap of 6Β° while, 𝑅 βƒ—βƒ— & 10𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—are at a gap of 54Β°. It shows that the impact of 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— is higher on the ball. 6. What would happen if ΞΈ = 80Β° instead of 60Β°in example 5? The angular gap of two vectors is greater for ΞΈ = 80Β°. Whatever the angular gap is, similar phenomenon occurs. Meaning, the greater force will have greater impact on the ball and the smaller force will have smaller impact. ΞΈ = 80Β° will occur if 10 N further rotates towards N in anticlockwise. During rotation of 10 N (to maintain ΞΈ = 80Β°), the resultant will get pulled a bit towards 10 N. This will lead in increasing of the value Ο’ = 6Β°. 7. Let us decrease the value of 𝑃 βƒ—βƒ— progressively until 𝑃 βƒ—βƒ— becomes 0 keeping 𝑄 βƒ—βƒ— and ΞΈ unchanged. What would happen to 𝑅 βƒ—βƒ—? The magnitude and direction of 𝑅 βƒ—βƒ— also changes progressively. Ultimately, 𝑅 βƒ—βƒ— becomes 𝑄 βƒ—βƒ— as there will be no 𝑃 βƒ—βƒ— to impart impact on the ball.
In the last example, the magnitude of 𝑅 βƒ—βƒ— = |𝑅| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = |𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = 10N and the direction of 𝑅 βƒ—βƒ— w.r.t 0N is Ο’ = 60Β° or 0Β° w.r.t. 10N. We can also evaluate the resultant vector by resolving the vectors in horizontal and vertical components as shown below: 10N 75N 𝑅 βƒ—βƒ— 10N 40N 𝑅 βƒ—βƒ— 10N 20N 𝑅 βƒ—βƒ— 10N 0N R = 10N H V
There are two vectors denoted by blue and red line arrows. Both have horizontal and vertical components. Just add the horizontal component of one vector to the horizontal component of another vector. Do similar in case of their vertical components. The two components will be formed. The angle between these newly formed components will be 90Β°. Just put these components in the formula of resultant vector. Note: To study the net impact of 2 vectors on a body, the resultant of two vectors is necessary to identify. The magnitude and direction of the resultant vector are to be evaluated. The calculation of the magnitude is different to that of the scalar quantities. Dot product and vector product (i) Derivation of dot/scalar product between two vectors in 2D Let 𝒂 βƒ— βƒ—βƒ— and 𝒃 βƒ— βƒ—βƒ— be two vectors with the origin 'o' as the common reference or initial point. Let ΞΈ2 and ΞΈ1 be their corresponding directions as show in the figure. Let ΞΈ2 be greater than ΞΈ1 for instance. Hence, the angle between the vectors = ΞΈ2 – ΞΈ1 = ΞΈ. Points P and Q have been projected onto the x-axis normally. Thus, the two right angled triangles, βˆ†OPM and βˆ†OQN get developed. From βˆ†OPM and βˆ†OQN,
CosΞΈ2 = m/OP = m/βˆšπ’ŽπŸ + π’‘πŸ = m/|𝒂 βƒ— βƒ—βƒ—|……………………….……..1 CosΞΈ1 = n/OQ = n/βˆšπ’πŸ + π’’πŸ = n/|𝒃 βƒ— βƒ—βƒ—|……………………..………………..….2 SinΞΈ2 = p/OP = p/βˆšπ’ŽπŸ + π’‘πŸ = p/|𝒂 βƒ— βƒ—βƒ—|………………………………….3 SinΞΈ1 = q/OQ = q/βˆšπ’πŸ + π’’πŸ = q/|𝒃 βƒ— βƒ—βƒ—|………………………….….…4 Taking Cos to both sides of the expression ΞΈ2 – ΞΈ1 = ΞΈ, Cos ΞΈ = Cos (ΞΈ2 – ΞΈ1) = Cos ΞΈ2 Cos ΞΈ1+ Sin ΞΈ2SinΞΈ1………………….…….5 From 1, 2, 3, 4 & 5, mn+pq = |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Cosθ…………………………….……..6 Let π’Š βƒ— and 𝒋 βƒ— be two unit vectors along x-axis and y-axis respectively. Vectors can be represented in the form of unit vectors as: 𝒂 βƒ— βƒ—βƒ— = (m, p) = mπ’Š βƒ—+p𝒋 βƒ— =π’Ž βƒ—βƒ—βƒ—βƒ—+𝒑 βƒ— ⃗⃗………………………………………………………….7 𝒃 βƒ— βƒ—βƒ— = (n, q) = nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ— = 𝒏 βƒ—βƒ—βƒ—+𝒒 βƒ— ⃗⃗…………………………………………..………………8 From 7 & 8, mπ’Š βƒ— = π’Ž βƒ—βƒ—βƒ—βƒ—, nπ’Š βƒ— = 𝒏 βƒ—βƒ—βƒ—, p𝒋 βƒ— = 𝒑 βƒ— βƒ—βƒ— and q𝒋 βƒ— βƒ—βƒ— =𝒒 βƒ— βƒ—βƒ— ……………………………..………………9 Multiplying 7 and 8, 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = (mπ’Š βƒ—+p𝒋 βƒ—) . (nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ—) 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = mn (π’Š βƒ— .π’Š βƒ— ) + mq (π’Š βƒ— .𝒋 βƒ— ) + pn (𝒋 βƒ— .π’Š βƒ— ) + pq (𝒋 βƒ— .𝒋 βƒ— )…………….………..10 For, π’Š βƒ— .π’Š βƒ— = 𝒋 βƒ— .𝒋 βƒ— = 𝟏 and π’Š βƒ— .𝒋 βƒ— = 𝒋 βƒ— .π’Š βƒ— = 0 equation 10 becomes, 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = mn + pq………………………………………………………………… ……11 From 11 & 6, 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = mn+pq = |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Cos ΞΈ iff π’Š βƒ— .π’Š βƒ— = 𝒋 βƒ— .𝒋 βƒ— = 𝟏 and π’Š βƒ— .𝒋 βƒ— = 𝒋 βƒ—.π’Š βƒ— = 0 (ii) Derivation of cross/vector product between two vectors in 2D Taking Sin to both sides of the expression ΞΈ2 – ΞΈ1 = ΞΈ, Sin ΞΈ = Sin (ΞΈ2 – ΞΈ1) = Sin ΞΈ2 Cos ΞΈ1 - CosΞΈ2SinΞΈ1………………….…….12
From 1, 2, 3, 4 & 5, pn-mq = |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Sinθ…………………………….……..13 Let π’Š βƒ— and 𝒋 βƒ— be two unit vectors along x-axis and y-axis respectively. Vectors can be represented in the form of unit vectors as: 𝒂 βƒ— βƒ—βƒ— = (m, p) = mπ’Š βƒ—+p𝒋 βƒ— =π’Ž βƒ—βƒ—βƒ—βƒ—+𝒑 βƒ— ⃗⃗………………………………………………………14 𝒃 βƒ— βƒ—βƒ— = (n, q) = nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ— = 𝒏 βƒ—βƒ—βƒ—+𝒒 βƒ— ⃗⃗…………………………………………..……………15 From 7 & 8, mπ’Š βƒ— = π’Ž βƒ—βƒ—βƒ—βƒ—, nπ’Š βƒ— = 𝒏 βƒ—βƒ—βƒ—, p𝒋 βƒ— = 𝒑 βƒ— βƒ—βƒ— and q𝒋 βƒ— βƒ—βƒ— =𝒒 βƒ— βƒ—βƒ— ……………………………..……………16 Multiplying 7 and 8, 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ— = (mπ’Š βƒ—+p𝒋 βƒ—) * (nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ—) 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ—= mn (π’Š βƒ—*π’Š βƒ—) + mq (π’Š βƒ— *𝒋 βƒ— ) + pn (𝒋 βƒ— * π’Š βƒ—) + pq (𝒋 βƒ—*𝒋 βƒ—)…………….……….17 For, π’Š βƒ— * π’Š βƒ— = 𝒋 βƒ— *𝒋 βƒ— = 𝟎, π’Š βƒ— *𝒋 βƒ— = βˆ’πŸ & 𝒋 βƒ— *π’Š βƒ— = 1 equation 10 becomes, 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ—= pn - mq………………………………………………………………… ……18 From 11 & 6, 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ—= pn - mq= |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Sin ΞΈ 𝒏 Μ‚ iff π’Š βƒ— * π’Š βƒ— = 𝒋 βƒ— *𝒋 βƒ— = 𝟎, π’Š βƒ— *𝒋 βƒ— = βˆ’πŸ & 𝒋 βƒ—*π’Š βƒ— = 1 Since, |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Sin ΞΈ is a formula of area and area is a vector whose direction is normal to the plane, the expression is multiplied by the unit vector 𝒏 Μ‚.

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Vector and scalar

  • 1. Introduction to scalar and vector Let us begin with the following questions: a. You are at rest position and an object is moving at a speed of 10km/hr. What would be the distance (gap) between you and the object after 30 minutes? It is pretty simple. The distance will be 5 km. b. You are moving to the opposite of an object at a speed of 20 km/hr. The object is moving at a speed of 10km/hr in the North. (i) What would be the distance covered by you in 30 minutes? (ii) What would be the distance (gap) between you and the object in 30 minutes? (iii) What would be the displacement of the object in 30 minutes? (iv) What would be the displacement of the object w.r.t the origin in 15 minutes? (i) It is 10 km. (ii) In 30 minutes, you will cover a distance of 10 km (in the South) and the object will cover a distance of 5 km (in the North). So, a gap of 15 km will be made. (iii) Actually, the question is not precise. It should have been asked in this fashion, 'What would be the displacement of the object w.r.t you after 30 minutes'? The displacement w.r.t you will be 15 km in the case. (iv) It is 2.5 km to the North of the origin. (Note: displacement is a vector & direction must be specified for vectors) Q. How are magnitude and direction imparted to a vector quantity? Magnitude and direction are imparted on the Cartesian plane. To impart the magnitude, a reference point or an initial point is required as the magnitude is the distance between two points. While, to impart the direction, a line is required that passes through the reference point. It will become clear from the following diagram:
  • 2. The magnitude of π‘Ž βƒ— = |π‘Ž| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = OA. OA can be evaluated if the coordinates of O and A are Known. If O = (0,0) and A = (a,b), OA = βˆšπ‘Ž2 + 𝑏2. It is not mandatory that the direction (ΞΈ) must be always about the horizontal line or x-axis. It can be about any line. But the line must be specified. For example, a ball is moving at a velocity of 5km/hr in the North or in the N45E. Here the line is specified. N45E is 45Β° about the North in clockwise direction. Here the line is specified as the North. So now on, we must be able to decide that the vector π‘Ž βƒ— has the magnitude |π‘Ž| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— and it is directed somewhere. Q. Look at the diagram, there are two vectors equal in magnitude. Are the vectors equal? No. Not only magnitude but also the direction of two vectors must be same in order tobe equal. Representation of vectors & resolution of vectors Vectors can be represented in the coordinate form (p,q) and in the form of unit vectors. If 𝑖 βƒ— and 𝑗 βƒ— are unit vectors along x-axis and y-axis respectively, 𝑝 βƒ— = p𝑖 βƒ— & π‘ž βƒ— = q𝑗 βƒ— ΞΈ O A X Y π‘Ž βƒ— o
  • 3. This can be written only if 𝑝 βƒ— and 𝑖 βƒ— have same direction and if π‘ž βƒ— and 𝑗 βƒ— have same direction. Here, 𝑝 βƒ— and π‘ž βƒ— are called horizontal and vertical components of the vector π‘Ž βƒ— respectively. The phenomenon of breaking down the vector into 2 components (in 2D) is called vector resolution. In the coordinate form vector π‘Ž βƒ— is represented as π‘Ž βƒ—βƒ—βƒ—βƒ— = (p, q). Note: Direction also must be specified in both forms. Unit vectors have magnitude 1. In the above figure, 𝑝 βƒ— & π‘ž βƒ— are assumed to be greater than their corresponding unit vectors: 𝑝 βƒ— = p𝑖 βƒ— & π‘ž βƒ— = q𝑗 βƒ— Using triangle law of vector addition for above example, Here, the magnitude of π‘Ž βƒ— = |π‘Ž| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = |𝑝 βƒ— + π‘ž βƒ—| = |𝑝𝑖 βƒ— + π‘žπ‘— βƒ—| = distance of OA 𝑖 βƒ— (p,q) ΞΈ O A X Y π‘Ž βƒ— 𝑝 βƒ— π‘ž βƒ— 𝑖 βƒ— 𝑗 βƒ— 𝑝 βƒ— =p𝑖 βƒ— π‘ž βƒ— =q𝑗 βƒ— π‘Ž βƒ— = 𝑝 βƒ— + π‘ž βƒ— = p𝑖 βƒ—+ q𝑗 βƒ— O A
  • 4. From above, we can infer that a vector (p, q) = π’‘π’Š βƒ— + 𝒒𝒋 βƒ— has: a horizontal component = 𝒑 βƒ—βƒ—βƒ—βƒ— = π’‘π’Š βƒ— and a vertical component = 𝒒 βƒ—βƒ—βƒ—βƒ— = 𝒒𝒋 βƒ— Or, we can say that a vector is equal to the summation of it's components. Q. Can the resultant of two different vector quantities (example force and velocity) be evaluated? No. We cannot decide whether the resultant (𝑅 βƒ—βƒ—) would be force (𝐹 βƒ—) or velocity ( 𝑉 βƒ—βƒ—) or any other quantity. Q. Give a justification for the statement 'Both the horizontal and vertical components of a vector are vectors'. Directions of the components of a vector can be identified w.r.t the vector itself or w.r.t the x- axis or w.r.t the y-axis. Q. How would a vector origin? A vector gets originated due to:  The resultant of 2 vectors  A single vector  Cross/vector product between 2 vectors Let us have a look on the diagram that shows the distribution of forces at 4 different points outside the bounded region. An object of mass 0.5 kg is inside the bounded region and exerting equal force (5N) opposite to the forces as shown below: 𝐹 βƒ— 𝑉 βƒ—βƒ— 𝑅 βƒ—βƒ—
  • 5. At all the points, the mass of the object will remain unchanged as mass being scalar quantity is not affected by direction. But the resultant force at all the points will remain changed. Resultant of vectors 1. Imagine a ball is being pulled with two forces in opposite directions. Where would the ball go? The ball will move to the right side since the net force or the resultant of two forces is directed along the right side. The net/resultant force = 75-10 = 65N and the direction is along the direction of P. 2N 4N 6N 8N 5N 5N 5N 5N 𝑄 βƒ—βƒ— = 10N 𝑃 βƒ—βƒ— = 75N
  • 6. 2. Imagine a ball is being pulled with two forces in same direction. Where would the ball go? The ball will move to the right side since the net force or the resultant of two forces is directed along the right side. The net/resultant force = 75+10 = 85N and the direction is along the direction of P. In the problem 1: the angle between two forces is 180Β° and the resultant force is directed along the greater force. And, the magnitude of the resultant force = greater force-smaller force. In the problem 2: the angle between two forces is 0Β° and the resultant force is directed along the given forces. And, the magnitude of the resultant force = greater force + smaller force. 3. Imagine a ball is being pulled with two forces in different directions. Where would the ball go? To answer this we need to observe the previous problems. From those problems we can infer that the magnitude and the direction of the resultant force depend on: the angle between two forces and the magnitude of the forces. 75N 10N 10N 75N
  • 7. 4. What would be the direction of the resultant force? Just think of intuitive answer. The direction of the resultant force is always closer to the greater force and farther from the smaller force as the impact of the greater force is greater than that of the smaller force. The ball neither moves along 10N nor along 75N. Hence the ball moves midway between two forces being closer to the 75 N. 5. What would be the magnitude of the resultant force (𝑅 βƒ—βƒ—) if ΞΈ = 60Β°? The resultant force (𝑅 βƒ—βƒ—) will be such that R2 = |𝑃| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +|𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +2|𝑃| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—|𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—CosΞΈ where: 𝑃 βƒ—βƒ— and 𝑄 βƒ—βƒ— are two forces ΞΈ is the angle between two forces |𝑃| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— & |𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— are the magnitudes of 𝑃 βƒ—βƒ— & 𝑄 βƒ—βƒ— respectively. In this example, 𝑃 βƒ—βƒ— = 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—, 𝑄 βƒ—βƒ— = 10𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—, ΞΈ = 60Β° R2 = |75| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +|10| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—2 +2|75| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—|10| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—Cos60 = 100+5625+2.10.75.0.5 = 5725+750= 6475 N2 ΞΈ 10N 75N 𝑅 βƒ—βƒ—
  • 8. Thus, 𝑅 βƒ—βƒ— = √6475 N = 80.46 N Since, the force is a vector quantity it must have direction too. Let Ο’ be the angle between 𝑅 βƒ—βƒ— & 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—. Then Ο’ will be the direction of 𝑅 βƒ—βƒ— w.r.t. the line of 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—. We can assign the direction for 𝑅 βƒ—βƒ— w.r.t. 10𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— too. The direction of 𝑅 βƒ—βƒ— is such that: tan Ο’ = QSin ΞΈ/(P+QCos ΞΈ) In this example, tan Ο’ = 10.Sin60/(75+10.Cos60) or, tan Ο’ = 8.66/80, or Ο’ = π‘‘π‘Žπ‘›βˆ’1 (8.66/80) = 6Β°. Meaning, 𝑅 βƒ—βƒ— & 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—are at a gap of 6Β° while, 𝑅 βƒ—βƒ— & 10𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—are at a gap of 54Β°. It shows that the impact of 75𝑁 βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— is higher on the ball. 6. What would happen if ΞΈ = 80Β° instead of 60Β°in example 5? The angular gap of two vectors is greater for ΞΈ = 80Β°. Whatever the angular gap is, similar phenomenon occurs. Meaning, the greater force will have greater impact on the ball and the smaller force will have smaller impact. ΞΈ = 80Β° will occur if 10 N further rotates towards N in anticlockwise. During rotation of 10 N (to maintain ΞΈ = 80Β°), the resultant will get pulled a bit towards 10 N. This will lead in increasing of the value Ο’ = 6Β°. 7. Let us decrease the value of 𝑃 βƒ—βƒ— progressively until 𝑃 βƒ—βƒ— becomes 0 keeping 𝑄 βƒ—βƒ— and ΞΈ unchanged. What would happen to 𝑅 βƒ—βƒ—? The magnitude and direction of 𝑅 βƒ—βƒ— also changes progressively. Ultimately, 𝑅 βƒ—βƒ— becomes 𝑄 βƒ—βƒ— as there will be no 𝑃 βƒ—βƒ— to impart impact on the ball.
  • 9. In the last example, the magnitude of 𝑅 βƒ—βƒ— = |𝑅| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = |𝑄| βƒ—βƒ—βƒ—βƒ—βƒ—βƒ— = 10N and the direction of 𝑅 βƒ—βƒ— w.r.t 0N is Ο’ = 60Β° or 0Β° w.r.t. 10N. We can also evaluate the resultant vector by resolving the vectors in horizontal and vertical components as shown below: 10N 75N 𝑅 βƒ—βƒ— 10N 40N 𝑅 βƒ—βƒ— 10N 20N 𝑅 βƒ—βƒ— 10N 0N R = 10N H V
  • 10. There are two vectors denoted by blue and red line arrows. Both have horizontal and vertical components. Just add the horizontal component of one vector to the horizontal component of another vector. Do similar in case of their vertical components. The two components will be formed. The angle between these newly formed components will be 90Β°. Just put these components in the formula of resultant vector. Note: To study the net impact of 2 vectors on a body, the resultant of two vectors is necessary to identify. The magnitude and direction of the resultant vector are to be evaluated. The calculation of the magnitude is different to that of the scalar quantities. Dot product and vector product (i) Derivation of dot/scalar product between two vectors in 2D Let 𝒂 βƒ— βƒ—βƒ— and 𝒃 βƒ— βƒ—βƒ— be two vectors with the origin 'o' as the common reference or initial point. Let ΞΈ2 and ΞΈ1 be their corresponding directions as show in the figure. Let ΞΈ2 be greater than ΞΈ1 for instance. Hence, the angle between the vectors = ΞΈ2 – ΞΈ1 = ΞΈ. Points P and Q have been projected onto the x-axis normally. Thus, the two right angled triangles, βˆ†OPM and βˆ†OQN get developed. From βˆ†OPM and βˆ†OQN,
  • 11. CosΞΈ2 = m/OP = m/βˆšπ’ŽπŸ + π’‘πŸ = m/|𝒂 βƒ— βƒ—βƒ—|……………………….……..1 CosΞΈ1 = n/OQ = n/βˆšπ’πŸ + π’’πŸ = n/|𝒃 βƒ— βƒ—βƒ—|……………………..………………..….2 SinΞΈ2 = p/OP = p/βˆšπ’ŽπŸ + π’‘πŸ = p/|𝒂 βƒ— βƒ—βƒ—|………………………………….3 SinΞΈ1 = q/OQ = q/βˆšπ’πŸ + π’’πŸ = q/|𝒃 βƒ— βƒ—βƒ—|………………………….….…4 Taking Cos to both sides of the expression ΞΈ2 – ΞΈ1 = ΞΈ, Cos ΞΈ = Cos (ΞΈ2 – ΞΈ1) = Cos ΞΈ2 Cos ΞΈ1+ Sin ΞΈ2SinΞΈ1………………….…….5 From 1, 2, 3, 4 & 5, mn+pq = |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Cosθ…………………………….……..6 Let π’Š βƒ— and 𝒋 βƒ— be two unit vectors along x-axis and y-axis respectively. Vectors can be represented in the form of unit vectors as: 𝒂 βƒ— βƒ—βƒ— = (m, p) = mπ’Š βƒ—+p𝒋 βƒ— =π’Ž βƒ—βƒ—βƒ—βƒ—+𝒑 βƒ— ⃗⃗………………………………………………………….7 𝒃 βƒ— βƒ—βƒ— = (n, q) = nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ— = 𝒏 βƒ—βƒ—βƒ—+𝒒 βƒ— ⃗⃗…………………………………………..………………8 From 7 & 8, mπ’Š βƒ— = π’Ž βƒ—βƒ—βƒ—βƒ—, nπ’Š βƒ— = 𝒏 βƒ—βƒ—βƒ—, p𝒋 βƒ— = 𝒑 βƒ— βƒ—βƒ— and q𝒋 βƒ— βƒ—βƒ— =𝒒 βƒ— βƒ—βƒ— ……………………………..………………9 Multiplying 7 and 8, 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = (mπ’Š βƒ—+p𝒋 βƒ—) . (nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ—) 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = mn (π’Š βƒ— .π’Š βƒ— ) + mq (π’Š βƒ— .𝒋 βƒ— ) + pn (𝒋 βƒ— .π’Š βƒ— ) + pq (𝒋 βƒ— .𝒋 βƒ— )…………….………..10 For, π’Š βƒ— .π’Š βƒ— = 𝒋 βƒ— .𝒋 βƒ— = 𝟏 and π’Š βƒ— .𝒋 βƒ— = 𝒋 βƒ— .π’Š βƒ— = 0 equation 10 becomes, 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = mn + pq………………………………………………………………… ……11 From 11 & 6, 𝒂 βƒ— βƒ—βƒ— .𝒃 βƒ— βƒ—βƒ— = mn+pq = |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Cos ΞΈ iff π’Š βƒ— .π’Š βƒ— = 𝒋 βƒ— .𝒋 βƒ— = 𝟏 and π’Š βƒ— .𝒋 βƒ— = 𝒋 βƒ—.π’Š βƒ— = 0 (ii) Derivation of cross/vector product between two vectors in 2D Taking Sin to both sides of the expression ΞΈ2 – ΞΈ1 = ΞΈ, Sin ΞΈ = Sin (ΞΈ2 – ΞΈ1) = Sin ΞΈ2 Cos ΞΈ1 - CosΞΈ2SinΞΈ1………………….…….12
  • 12. From 1, 2, 3, 4 & 5, pn-mq = |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Sinθ…………………………….……..13 Let π’Š βƒ— and 𝒋 βƒ— be two unit vectors along x-axis and y-axis respectively. Vectors can be represented in the form of unit vectors as: 𝒂 βƒ— βƒ—βƒ— = (m, p) = mπ’Š βƒ—+p𝒋 βƒ— =π’Ž βƒ—βƒ—βƒ—βƒ—+𝒑 βƒ— ⃗⃗………………………………………………………14 𝒃 βƒ— βƒ—βƒ— = (n, q) = nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ— = 𝒏 βƒ—βƒ—βƒ—+𝒒 βƒ— ⃗⃗…………………………………………..……………15 From 7 & 8, mπ’Š βƒ— = π’Ž βƒ—βƒ—βƒ—βƒ—, nπ’Š βƒ— = 𝒏 βƒ—βƒ—βƒ—, p𝒋 βƒ— = 𝒑 βƒ— βƒ—βƒ— and q𝒋 βƒ— βƒ—βƒ— =𝒒 βƒ— βƒ—βƒ— ……………………………..……………16 Multiplying 7 and 8, 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ— = (mπ’Š βƒ—+p𝒋 βƒ—) * (nπ’Š βƒ—+q𝒋 βƒ— βƒ—βƒ—) 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ—= mn (π’Š βƒ—*π’Š βƒ—) + mq (π’Š βƒ— *𝒋 βƒ— ) + pn (𝒋 βƒ— * π’Š βƒ—) + pq (𝒋 βƒ—*𝒋 βƒ—)…………….……….17 For, π’Š βƒ— * π’Š βƒ— = 𝒋 βƒ— *𝒋 βƒ— = 𝟎, π’Š βƒ— *𝒋 βƒ— = βˆ’πŸ & 𝒋 βƒ— *π’Š βƒ— = 1 equation 10 becomes, 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ—= pn - mq………………………………………………………………… ……18 From 11 & 6, 𝒂 βƒ— βƒ—βƒ—*𝒃 βƒ— βƒ—βƒ—= pn - mq= |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Sin ΞΈ 𝒏 Μ‚ iff π’Š βƒ— * π’Š βƒ— = 𝒋 βƒ— *𝒋 βƒ— = 𝟎, π’Š βƒ— *𝒋 βƒ— = βˆ’πŸ & 𝒋 βƒ—*π’Š βƒ— = 1 Since, |𝒂 βƒ— βƒ—βƒ—||𝒃 βƒ— βƒ—βƒ—| Sin ΞΈ is a formula of area and area is a vector whose direction is normal to the plane, the expression is multiplied by the unit vector 𝒏 Μ‚.