Vectors 29th May

Slide 1 : VECTORS Vectors www.vidyadrishti.com

Slide 2 : This chapter is the abc of Physics, so study this chapter carefully. If you don’t have a proper concept of this chapter you are going to have problem in Physics. I have represented all concepts exhaustively, which you can’t find in any book or any course material provided by coaching centers.

Slide 3 : Vector Concepts Type of Vectors Components (Highly important) Vector addition & subtraction Dot Product Cross Product Exercises Vectors Topics

Slide 4 : The actual length of path along which you go to your school is known as distance and the shortest path which could have been made is called displacement. Please note, when we say the word distance, we are only interested in the length of the path but the word displacement has two parts: first length and another its direction. For this example displacement direction is towards right. HOME SCHOOL DISTANCE DISPLACEMENT

Slide 5 : The length of the path is also known as magnitude of the path. SCALARS: Those physical quantities which have magnitude only but no direction are called scalar quantities. e.g., distance, speed, work, energy etc. VECTORS: Those physical quantities which have both magnitude as well as direction are called vector quantities. E.g., displacement, velocity, acceleration, force etc. HOME SCHOOL DISTANCE DISPLACEMENT

Slide 6 : In general, we can show any vector by a straight ray in a proper direction. The length of the vector (straight ray) gives the magnitude of the vector. The starting point of the vector is known as tail of the vector and end point is known as head or tip of the vector. HOME SCHOOL DISTANCE DISPLACEMENT TAIL HEAD A B

Slide 7 : We represent a vector by either bold (thick) letters or by arrow over letters. REMEMBER: represents magnitude as well as direction

Slide 8 : From last slide, we have seen that a vector in both magnitude & direction from can be written as AB or or P or How to represent the magnitude of a vector ? We know that vector magnitude is just the length of vector. We represent magnitude of vector AB by AB (not bold letter or arrow) or or |AB| (modulus of vector AB) . Thus, magnitude (length) of vector (or )

Slide 9 : We represent a vector by either bold (thick) letters or by arrow over letters. This form represents both the magnitude as well as direction of vector AB (or P). (magnitude and direction of a vector) Magnitude is nothing but the length of the vector.

Slide 10 : We can shift a vector to any place provided vector’s magnitude and direction remain same. Thus, shifting is nothing but to shift a vector parallely to another position. Shifting of vectors is a very powerful tool during vector addition, multiplication etc. Generally, we need to shift a vector to a position of our convenience. Sometimes we have to shift a vector to origin.

Slide 11 : It seems that it is a easy topic but you should always beware of it. To get angle between two vectors, the tails of two vectors must be at same point . Therefore , for the above picture angle between two vectors=?. NOTE: Angle between two vectors cannot be greater than 180o 0o ? ? ? 180o

Slide 12 : Example: What is the angle between vectors & ? SOLUTION: To get angle between two vectors, the tails of two vectors must be at same point . Clearly, a is not the angle between two vectors as tails of two vectors are not at same point. Then, what to do? Yes! We have to shift vector such that its tail can touch vector tail. Clearly, (180o-a) is the angle between two vectors.

Slide 13 : We shall skip those vectors which are not important. Important types of vectors are: Position vector Equal vectors Negative vectors Zero vector (Null vector) Unit Vector (Very Important) Co-initial vectors

Slide 14 : A vector whose initial point (tail) is at origin is called a position vector. Here, vectors & are position vectors but is not a position vector.

Slide 15 : Two vectors are said to be equal vectors when they have same magnitude and same direction. Clearly, equal vectors must be parallel to each other.

Slide 16 : A vector is said to be negative of another vector if has same magnitude as that of but it is in the opposite direction of . Clearly, negative of a vector is obtained by reversing its direction. NOTE: If the magnitude of is 5m don’t think magnitude of will be -5m. Magnitude of a vector is always taken as positive.

Slide 17 : As the name suggest, zero vector has zero magnitude. Since, it is a vector, it must have some direction. This direction is arbitrary. Thus, a vector having magnitude zero and arbitrary direction is called a zero vector. Zero vector is denoted by O or . Zero vector is a pure concept to carry out vector algebra.

Slide 18 : Loosely speaking, co means same & initial means start. Therefore, co-initial vectors are those vectors which have same starting point, meaning their tails lie at same point. In the above picture are co-initial vectors but

Slide 19 : Unit means one. Therefore, a vector whose magnitude is unity (1 unit) is called a unit vector.

Slide 20 : This coordinate-system is known as right handed coordinate system. x Direction of z axis is obtained by right hand thumb rule. o y z

Slide 21 : Put the fingers of your right-hand along the x-axis. Orient your hand so you can curl your fingers from x-axis towards y-axis. Your extended thumb gives the direction of z-axis . x y z

Slide 22 : This coordinate-system is known as right handed coordinate system. x y z o

Slide 23 : In general we represent 3D coordinate system as x y z 0

Slide 24 : A vector whose magnitude is ‘1’ is called unit vector. Unit vector along X-axis = Unit vector along Y- axis = Unit vector along Z- axis =

Slide 25 :

Slide 26 : Component of a vector A along a direction making an angle ? with it is A cos ? . NOTE: Component plays a very powerful role in Physics.

Slide 27 : Let a vector such that it makes an angle of ? with positive direction of x- axis. Then, x-component of A = A cos? y-component of A = A sin? and we can write in the unit vector notation as A = A cos? i + A sin? j NOTE : Bold letters for vectors

Slide 28 : UNIT VECTOR NOTATION TRICKS: A vector of magnitude a (a> 0) and (parallel to) x-axis = a . A vector of magnitude a (a> 0) and anti-parallel (means opposite direction) to x-axis = -a . A vector of magnitude a (a> 0) and parallel to y-axis = a . A vector of magnitude a (a> 0) and anti-parallel to y-axis = -a . SOME EXAMPLES: A = 4m [parallel to x-axis] B = 2m [parallel to y-axis] C = -3m [anti-parallel to x-axis] D = -5m [anti parallel to y-axis] NOTE: These tricks should always be kept in your mind.

Slide 29 : 1. A vector having magnitude of 5m makes an angle of 450 with x-axis. Find its x and y component. Represent this vector in unit vector notation. x- Component = (5m) cos45 = 3.53m y- Component = (5m) sin45 = 3.53m Unit vector notation : 3.53m + 3.53m .

Slide 30 : Qs . Write in unit vector notation. CAUTION: See coordinate system carefully (i.e., where are x-axis and y-axis). Clearly our vector lies in 4th quadrant. x-component of y-component of Therefore, x y 30o o A 20N

Slide 31 : Qs . Write the component of along incline and perpendicular to incline. SOLUTION: From geometry we can easily find out the angle made by with incline. Clearly, it is 53o. Therefore, component of along incline = (Ans) And, component along perpendicular to incline = (Ans) 5N 37o 53o O A 3N 4N

Slide 32 : Let [in unit vector notation] Then, y O x Angle with positive direction of x-axis: Step 1: Find principle angle as a = tan-1 |y/x| Step 2: If it lies in a) 1st quadrant then, ? = a b) 2nd quadrant then, ? = 180o - a c) 3rd quadrant then, ? = (180o + a) or (a – 180o) d) 4th quadrant then, ? = (360o – a) or ( -a) y x A(x , y) ?

Slide 33 : Qs. Find magnitude and direction with positive x-axis of a vector m . Solution: Clearly, vector lies in 3rd quadrant. [ Because x component = -3 and y component =-3, means ( x, y) = ( -, -), so third quadrant]. First we draw it on x-y plane. Let OA = m. MAGNITUDE: . (Ans) DIRECTION: Principal angle = tan-1 |y/x| = tan-1 |-3/-3|= 45o Therefore angle with positive x axis = 180o + 45o = 225o (Ans) x y o A (-3,0) (0,-3) 3 v2 m 45o 225o

Slide 34 : For the given co-ordinate system find the displacement of the particle in unit vector notation. Also, find the magnitude of its displacement. Radius of the circular path is 7cm [centre is @ C]. SOLUTION: Clearly displacement is . It is in 4th quadrant. x-component of OP = displacement along x-axis = radius = 7cm. y-component of OP = displacement along y-axis = -radius = -7cm. Therefore, displacement cm [Ans]. Magnitude of displacement ? OP = v[72 + (-7)2] = 7v2 cm [Ans]. x y o P C

Slide 35 : A vector whose starting point is origin is called a position vector. Here position vector :

Slide 36 : Qs. A mosquito flies from one corner O of a room and sits at the diagonally opposite corner P of the room. Find the x component, y component and z component of its displacement. Also find displacement vector. Also write displacement magnitude. SOLUTION: x-component of displacement = displacement along x-axis = length of room = 6m. [Ans.]

Slide 37 : y-component of displacement = displacement along y-axis = height of room = 4m. [Ans.] z-component of displacement = displacement along z-axis = breadth of room = 3m. [Ans.] Displacement vector ? m [Ans] Magnitude of displacement = ? ? = 7.81m. (Ans)

Slide 38 : Method 1: Graphically [shifting of vectors is used here] + =

Slide 39 : A + B + C B C A

Slide 40 : + +

Slide 41 : = resultant [vector sum of all vectors]

Slide 42 : In practice , if more than two vectors are given, we use analytical method. PROCEDURE is as follows: STEP 1: Shift all vectors to origin. STEP 2: Write each vectors in unit vector notation. STEP 3: Add/Subtract them analytically. STEP 4: Your answer is in unit vector notation. You can convert it to magnitude and direction form easily . Let us do one problem to illustrate this.

Slide 43 : Find the sum of three vectors shown. SOLUTION: STEP 1: Shift all vectors to origin. STEP 2: Write each vectors in unit vector notation. As vector A lies in 2nd quadrant: [because parallel to x-axis] [because anti-parallel to x-axis] 37o 5N 4N 3N o x y

Slide 44 : STEP 3: Add them analytically. Therefore, their resultant

Slide 45 : See example 7. Find . Write in magnitude and direction form. SOLUTION: First we have to find etc. From example Therefore, Therefore, MAGNITUDE : R = v(x2 + y2) = v(42 + 92) = 9.85 N. (Ans). DIRECTION with positive x-axis: Since, R is in first quadrant, therefore, tan? = y/x = 9/4 = 2.25 ? ? = 66o (Ans)

Slide 46 : If two vectors are represented (both in magnitude and direction) by the two sides of a triangle taken in same order, then the third side taken in opposite order represents their resultant (both in magnitude and direction).

Slide 47 : AB + BC = AC (vector form) (magnitude form) [Direction] CAUTION: Be careful of ? and a. Here ? is angle between two vectors and But, a is the angle made by Resultant with ? AB + BC a

Slide 48 : From C we drop CE perpendicular on AB extended. Now, BE = BC cos? And, CE = BC sin? Now, in right angled triangle ABE, from Pythagoras Theorem, AC2 = AE2 + CE2 ?AC2 = (AB + BE)2 + CE2 ? AC2 = (AB + BC cos? )2 + (BC sin?)2 ? AC2 = AB2 + 2.AB.BC.cos? + BC2 ( sin2? + cos2?) ? AC2 = AB2 + BC2+ 2.AB.BC.cos? [because, from trigonometry sin2? + cos2? =1] ? AC = v (AB2 + BC2+ 2.AB.BC.cos?) [1st part proved] Again, in right angled triangle ABE, tan = p/b = CE/AE ? tan a = CE/(AB+BE) ? tan a = BC cos? / (AB + BC sin?) [2nd part proved] A B C E BC cos? BC sin? a ?

Slide 49 : If two adjacent sides of a parallelogram represents two co-initial vectors (in magnitude and direction), then the diagonal from that co-initial point represents their resultant completely (i.e., in magnitude and direction)

Slide 50 : ? a

Slide 51 : |A + B| =v(A2 + B2 + 2ABcos?) tana = [B sin ? / ( A + Bcos?)] REMEMBER above formulae. Be careful of ? and a. Proof of parallelogram is same as that of triangle law. Actually triangle law and parallelogram law are equivalent. In general, we should use parallelogram /triangle law for adding two vectors only. For addition of more than two vectors we should use analytical method.

Slide 52 : A and B are along same direction. Here ? = 0o = v(A2 + B2 + 2ABcos ?) = v(A2 + B2 + 2AB) [because cos0o =1] = v(A + B)2 = A + B tana = [B sin ? / ( A + Bcos ?)] = 0 ? a = 0o A B A B A + B

Slide 53 : A and B are along opposite direction. Here ? = 180o = v(A2 + B2 + 2ABcos ?) = v(A2 + B2 - 2AB) [because cos180o =-1] = v(A - B)2 = A - B tana = [B sin ? / ( A + Bcos?)] = 0 ? a = 0o A B A B A + B

Slide 54 : A and B are perpendicular to each other. Here ? = 90o = v(A2 + B2 + 2ABcos ?) ? = v(A2 + B2 ) [because cos90o =0] tana = [B sin ? / ( A + Bcos ?)] ? tana = (B/A) A B

Slide 55 : |A|=|B|=a (say). = v(a2 + a2 + 2.a.a.cos ?) ? = av(2(1+cos?)) = av(2x2cos2(?/2)) [because, 1+cos? =2cos2(?/2) & sin?=2sin(?/2) cos(?/2)] = 2acos(?/2) ** tana = [a sin ? / ( a + a cos ?)] = [2sin(?/2) cos(?/2)] /[2cos2(?/2)] ? tana = tan (?/2) ? a=(?/2) ? A+B is along the angle bisector of angle between A & B. A B ?

Slide 56 : Add given vectors. Angle between them= 60o . SOLUTION: We have =P=3m , =Q=4m & ? = 60o. We use parallelogram to add these vectors. Therefore, = v(P2 + Q2 + 2PQcos ?) ? =v(32+42+2.3.4.cos 60o)=v37m=6.08m (Ans) And angle that makes with tana = [Q sin?/( P+Q cos?)] = [ 4sin 60o/(3+4cos60o)] = 0.6928 ? a = 34.71o . (Ans) P 3m Q 4m P Q 60o a

Slide 57 : = (4.5N) cos25 i + (4.5N) sin25 j = (4.078N) i + (1.9018N) j = (6.0N) cos(-50) i + (6.0N) sin(-50) j = (3.8567N) i - (4.5963N) j. Vector sum : In unit vector notation = (7.9N) i - (2.7N) j. Magnitude = (7.9)2 + (2.7)2 = 8.3N. Direction with +ve x-axis Angle = - (tan-1| -2.7/7.9 |) [-ve because lies in 4th quadrant] = -18.860 Direction with +ve y-axis = 90-(-18.86) = 108.860 57

Slide 58 : = 4.5 N = 6 N ? = 50o + 25o = 75o =v(A2 + B2 + 2ABcos?) = v [(4.5)2 + 62 + 2 x 4.5 x 6 x cos 75o) = 8.3 N tana = [B sin ? / ( A + Bcos ?)] = [6 sin 75o / ( 4.5 + 6cos 75o)] ? a = 43.75o . ( It is the angle which A+B makes with A; refer parallelogram law). * Therefore angle of resultant with x axis= -43.75o + 25o = -18.75o 4.5N 6.0N 250 x y 4.5N 250 500 A B A + B

Slide 59 : i.e. vector subtraction is obtained taking –ve of another vector and adding it to first vector. For example to subtract B from A graphically. We take –ve of vector B by reversing its direction. Then add A and –B By parallelogram. [we can also use triangle law] A B -B A -B A-B

Slide 60 : Qs. Find . Magnitudes are: A=4N, B=3N. SOLUTION: We take –ve of vector by reversing its direction. Clearly, angle between and ? = 180o-60o =120o Also, =B=3N Now, we add and using parallelogram law, (Ans) And, Direction of with tana = [B sin ? / ( A + B cos?)] =3sin 120o /(4+3cos 120o) ? tana = 1.0392 ? a =46.1o . (Ans) A B -B 60o 120o A - B a

Slide 61 : H C VERMA: Page # 29, Qs # 1 to 12. Page # 24-25, Examples # 1 to 7.

Slide 62 : 1.Scalar product or Dot product : Dot product gives a scalar quantity. Using definition of dot product we can find the angle between two vectors.

Slide 63 : where is a ‘unit vector’ perpendicular to both and . Direction of is along n. Direction of is obtained by ‘Right hand thumb’ rule. ?

Slide 64 : Put the fingers of your right-hand along 1st vector (here A). Orient your hand so you can curl your fingers from 1st vector (A) towards the 2nd vector (B). Your extended thumb gives the direction of cross product . .

Slide 65 : We read as dot not as into etc. We read as cross . Dot product i.e., scalar product gives a scalar quantity. i.e., as result of dot product we get magnitude only. Cross product i.e., vector product gives a vector quantity. i.e., as result of cross product we get magnitude and direction both. Direction of cross product is obtained by right hand thumb rule.

Slide 66 : Let and Remember above formulae.

Slide 67 : [cross product is not commutative] . or is perpendicular to . What does imply ? Think…….

Slide 68 : because angle between two same unit vectors=0. Therefore, from definition of dot product Similarly, for others. Each of is a unit vector along x, y and z-axis respectively. So, angle between any of these three unit vectors=90o, as they are mutually perpendicular. Therefore, Similarly for others.

Slide 69 : From definition of vector (cross) product [i.e. , ], we have NOTE: Here 0 is null vector. [Proof on next slide]

Slide 70 : From definition of cross product we know that is perpendicular to both A and B and its direction is obtained by right hand thumb rule. So, if we have to , then must be perpendicular to both and . Since i lies along +ve x-axis and lies along +ve y-axis, then from right hand thumb rule must lies along +ve z axis. [Refer slide no. 12, 13 and 53]. And, is a unit vector along +ve z-axis. Therefore, [because is here] Also, Similarly, others can be proved.

Slide 71 : Let (m is a scalar) Or A vector perpendicular to both Unit vector perpendicular to both

Slide 72 : SOLUTION DOT Product: A.B = 2X(-3) + 3X2 + (-4)X2 = -8. CROSS Product :

Slide 73 : Find such that is parallel to and SOLUTION: We know that two vectors and are parallel when ? 2/4 = 3/a = -4/b ? a=6 and b = -8. Therefore,

Slide 74 : Find (a) Angle between A and B and (b) unit vector perpendicular to both A and B: where SOLUTION: (a) We know that Now, v(x2 + y2 + z2) = v[22 + 32 +(-4)2] = v29. v(x2 + y2 + z2) = v[(-3)2+22+22] = v17. Therefore, cos? = -8/(v29 v17) = -8/v493 = -0.3603 ? ? = 111.12o [Ans]

Slide 75 : (b) Unit vector perpendicular to both Unit vector perpendicular to both

Slide 76 : H C VERMA Page # 29, Qs # 13 to 18. Page # 25-26, Examples 8 to 10.