Good quality notes on Physics - Scalars & Vectors, Study notes of Physics

Download Good quality notes on Physics - Scalars & Vectors and more Study notes Physics in PDF only on Docsity!  Digital Pvt. Ltd. [1] Classification of Physical Quantities Scalars and Vectors Physical quantities Those physical quantities which are used to define law of physics is known as physical quantities. On the basis of directions, there are mainly two types of physical quantities. (1) Scalar (2) Vector Scalar Quantities A physical quantity which can be described completely by its magnitude only and does not require a direction is known as a scalar quantity. It obeys the ordinary rules of algebra. Ex : Distance, mass, time, speed, density, volume, temperature, electric current etc. Vector Quantities A physical quantity which requires magnitude and a particular direction, when it is expressed. (a) has a specified direction. (b) obeys parallelogram law of vector addition, then only it is said to be a vector. If any of the above conditions is not satisfied the physical quantity cannot be a vector. If a physical quantity is a vector it has a direction, but the converse may or may not be true, i.e. if a physical quantity has a direction, it may or may not a be vector. e.g. time, pressure, Electric Current etc. have directions but are not vectors because they do not obey parallelogram law of vector addition. Example of vector quantity : Displacement, velocity, acceleration, force etc. Illustration. Which of the following is a vector quantity? (1) Velocity (2) Force (3) Acceleration (4) All of the above Solution. (4) All of the above are vector quantities Part - 01 TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Representation of Vectors-Symbolic and Geometrical Representation of vector Geometrically, the vector is represented by a line with an arrow indicating the direction of vector as Mathematically, vector is represented by A Important points If a vector is displaced parallel to itself it does not change (see Figure) If a vector is rotated through an angle other than multiple of 2 (or 360°) it changes (see Figure) Illustration. Which of the following expression correctly represent a vector? (1) 𝑉 (2) ?⃗? (3) 2|𝑉| − 𝐴𝐵 (4) |V| Solution. (2) Vector is represented by arrow on letters Part - 02 TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Types of Vectors Types of vector Parallel vectors : Those vectors which have same direction are called parallel vectors. Angle between two parallel vectors is always 0° Equal Vectors Vectors which have equal magnitude and same direction are called equal vectors. A B= Anti–parallel Vectors : Those vectors which have opposite direction are called anti–parallel vector. Angle between two anti–parallel vectors is always 180° Negative (or Opposite) Vectors Vectors which have equal magnitude but opposite direction are called negative vectors of each other. AB and BA are negative vectors AB BA= − Co-initial vector Co-initial vectors are those vectors which have the same initial point. In figure a, b and c are co-initial vectors. Part - 04 TG: @Chalnaayaaar  Digital Pvt. Ltd. [2] Vector Part-04 Collinear Vectors : The vectors lying in the same line are known as collinear vectors. Angle between collinear vectors is either 0° or 180° Example. (i)   ( = 0º) (ii) → → ( = 0º) (ii)  → ( = 180º) (iv) →  ( = 180º) Coplanar Vectors Vectors located in the same plane are called coplanar vectors. Note : Two vectors are always coplanar. Concurrent vectors Those vectors which pass through a common point are called concurrent vectors In figure a,b andc are concurrent vectors Null or Zero Vector A vector having zero magnitude is called null vector. Note : Sum of two vectors is always a vector so, (A) ( A) 0+ − = 0 is a zero vector or null vector. Unit Vector A vector having unit magnitude is called unit vector. It is used to specify direction. A unit vector is represented by  (Read as A cap or A hat or A caret). Unit vector in the direction of A is A  | A | = Vector unit vector Magnitude of the vector   =    ˆ ˆA AA | A | A= = A unit vector is used to specify the direction of a vector. TG: @Chalnaayaaar  Digital Pvt. Ltd. [3] Vector Part-04 Base vectors In an XYZ co-ordinate frame there are three unit vectors ˆ ˆ ˆi, j andk , these are used to indicate X, Y and Z direction respectively. These three unit vectors are mutually perpendicular to each other. Axial Vector These vectors are used in rotational motion to define rotational effects. Direction of these vectors is always along the axis of rotation in accordance with right hand screw rule or right hand thumb rule. Ex. : Infinitesimal angular displacement (d ) , Angular velocity, ( ) , Angular momentum (J) , Angular acceleration ( ) and Torque ( ) Illustration 1. (1) A &C are parallel (2) A &C and B & C are antiparallel (3) C & D are coinitial vectors (4) A,B,C,D are coplanar Solution. All are correct TG: @Chalnaayaaar  Digital Pvt. Ltd. [2] Vector Part-05 Illustration 1. Add vectors by triangle law : (i) (ii) Solution. (i) By using Triangle law (ii) By using Triangle law Illustration 2. Which of the following options is correct (1) P Q R+ = (2) Q R P+ = (3) P R Q+ = (4) None of these Solution. Using Triangle's law P R Q+ = TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Addition of Vectors-Analytical Method Analytical method : In ∆PQR sin θ = PQ PR ⟹ PQ = B sinθ cos θ = PQ PR ⟹ RQ = B cosθ In ∆OPQ OP2 = OQ2 + PQ2 R2 = (A + Bcosθ)2 + (Bsinθ)2 R2 = A2 + B2cos2 θ + 2ABcosθ + B2sin2 θ R2 = A2 + 2ABcosθ + B2(cos2 θ + sin2θ) |R⃗ | = √A2 + B2 + 2ABcosθ |?⃗? + ?⃗? | = √𝐀𝟐 + 𝐁𝟐 + 𝟐𝐀𝐁𝐜𝐨𝐬𝛉 In ΔOPQ tanα = QP OQ = Bsinθ A+Bcosθ Similarly tanβ = Asinθ B+Acosθ Illustration 1. If magnitude of resultant of 2N and 3N is 4N find angle between 2N and 3N: Solution. 2 24 2 3 2(2)(3)cos= + +  16 = 4 + 9 + 12cos 3 = 12cos 1 1 cos 4 −    =     Part - 06 TG: @Chalnaayaaar  Digital Pvt. Ltd. [2] Vector Part-06 Illustration 2. Two vectors of magnitude 4N and 6N are acting at an angle 60º then find : (i) Magnitude of their resultant vectors (ii) Angle between resultant vector and 4N (iii) Angle between resultant vector and 6N Solution. (i) 2 2| R | 4 6 2(4)(6)cos60º= + + 1 | R | 16 36 48 2   = + +     | R | 76= 2 19N (ii) 6sin(60º ) tan 4 6cos(60º )  = + = 3 6 2 1 4 6 2         +     1 3 3 tan 7 −    =      1 3 3 tan 7 − = (iii) 4sin60º tan 6 4cos60º  = + 3 4 2 1 6 4 2       =   +     1 3 tan 4 − = Special cases 𝛉 = 𝟎° 𝛉 = 𝟏𝟖𝟎° |?⃗⃗? | = √𝐀𝟐 + 𝐁𝟐 + 𝟐𝐀𝐁𝐜𝐨𝐬𝛉 𝐑 = √𝐀𝟐 + 𝐁𝟐 + 𝐀𝐁𝐜𝐨𝐬𝟏𝟖𝟎° |?⃗⃗? |𝐦𝐚𝐱 = 𝐀 + 𝐁 |?⃗⃗? |𝐦𝐚𝐱 = 𝐀 − 𝐁 TG: @Chalnaayaaar  Digital Pvt. Ltd. [2] Vector Part-07 Special Case : If =  = 120º then R = 120º 2a cos a 2 = i.e. If  = 120º then | R | | A B| | A | | B | a= + = = = If resultant of two unit vectors is another unit vector then the angle between them () = 120º. OR If the angle between two unit vectors () = 120º, then their resultant is another unit vector. Addition of more than two vectors (law of polygon) If some vectors are represented by sides of a polygon in same order, then their resultant vector is represented by the closing side of polygon in the opposite order. R A B C D= + + + In a polygon if all the vectors taken in same order are such that the head of the last vector coincides with the tail of the first vector then their resultant is a null vector. A B C D E 0+ + + + = 2. If n coplanar vectors of equal magnitude are at equal angular separation ( 360° 𝑛 ), then their resultant is always ZERO If n=3 & |?⃗?| = |?⃗?| = |𝑐| = 𝑎 If n=4 & |?⃗?| = |?⃗?| = |𝑐| = |𝑑| = 𝑎 TG: @Chalnaayaaar  Digital Pvt. Ltd. [3] Vector Part-07 Illustration 1. If 100 coplanar vectors each having magnitude 10 units are equally inclined with each other then find the magnitude of their resultant? (1) 0 (2) 10 (3) 100 (4) 1000 Solution. Because all the coplanar vectors of equal magnitude are at equal angular separation then their resultant is always ZERO Note : The resultant of 3 vectors can be zero if they satisfy following conditions : |A – B|  C  | A + B| Illustration 2. Which of the following groups can give zero resultant(Equilibrium) Vectors ?⃗⃗? ?⃗⃗⃗? ?⃗⃗⃗? (1) 2N 4N 8N (2) 12N 10N 23N (3) 1N 2N 3N (4) 8N 1N 2N Solution. |P − Q| ≤ R ≤ |P + Q| (1) 2 ≤ R ≤ 6 (Not Possible) (2) 2 ≤ R ≤ 22 (Not Possible) (3) 1 ≤ R ≤ 3 (Possible) (4) 7 ≤ R ≤ 9 (Not Possible) Illustration 3. It is regular hexagon and O is centre then find OA OB OC OD OE OF ?+ + + + + = Solution. | OA| | OB| | OC| | OD| | OE | | OF|= = = = = All the vectors are at equal angle so their resultant will be OA OB OC OD OE OF 0+ + + + + = TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Vector Subtraction, Miscellaneous problems on Vector addition and Subtraction Subtraction of two vectors Let A and B are two vectors. Their difference i.e. A B− can be treated as sum of the vector A and vector ( B)− . A B A ( B)− = + − To subtract B from A , reverse the direction of B and add to vector A according to law of triangle. 2 2 2 2| A B| A B 2ABcos( ) A B 2ABcos− = + + − = + −  & Bsin tan A Bcos   = −  Where  is the angle between A and B . Properties of vector subtraction : Vector subtraction dos not follow commutative law i.e. A B B A−  − Vector subtraction does not follow associative law i.e. (A B) C A (B C)− −  − − If two vectors have equal magnitude i.e. | A | | B | a= = and  is the angle between them, then 2 2 2| A B | a a 2a cos 2a sin 2  − = + −  = Special case : If  = 60º then 2a sin 2  =a i.e. | A B| | A | | B | a− = = = at  = 60º If difference of two unit vectors is another unit vector then the angle between them is 60º or If two unit vectors are at angle of 60º, then their difference is also a unit vector. In physics whenever we want to calculate change in a vector quantity, we have to use vector subtraction. For example, change in velocity, 2 1v v v = − Part - 08 TG: @Chalnaayaaar  Digital Pvt. Ltd. [4] Vector Part-08 Illustration 6. If |?̂? + ?̂?| = √𝟑, then find |?̂? − 𝟒?̂?| (1) √13 (2) √3 (3) 3 (4) √21 Solution. |Â| = 1, |B̂| = 1 √A2 + B2 + 2ABcosθ = √3 √1 + 1 + 2cosθ = √3 √2 + 2cosθ = √3 2cosθ = 1 | − 4B̂| = √(Â) 2 + (4B̂) 2 − 2(Â)(4B̂)cos60° = √1 + 16 − 4 | − 4B̂| = √13 Illustration 7. If |?̂? − ?̂?| = 𝟏 = |?̂?| = |?̂?| , then ˆ ˆA B ?+ = Solution. ˆ ˆA B 1− = 2 2A B 2ABcos 1+ −  = 1 1 2(1)(1)cos 1+ −  = 1 cos 2  = 60 =  Now, ˆ ˆA B+ = 2 2A B AB cos+ +  = 1 1 1 2 2 + +   60 =  = 3 TG: @Chalnaayaaar  Digital Pvt. Ltd. [5] Vector Part-08 Illustration 8. If a particle moves with constant speed v on circumference, find magnitude of change in velocity from A to B. (1) 3V (2) V (3) 0 (4) 2V Solution. |V⃗ A − V⃗ B| = 2Vsin θ 2 2 × V × sin ( 60° 2 ) = V TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Resolution of Vectors in 2D-in plane Resolution of Vector 2D To break or split a vector in components is called resolution of vector Any vector can be split in two or more than two components. Note : Maximum number of components of a vector can be infinite. Vector Resolution in A Plane (2D) If the angle between components is 90°, then they are known as perpendicular/orthogonal/rectangular components. Ax : Component of A⃗ along x axis Ay : Component of A⃗ along y axis In 𝚫𝐎𝐐𝐏, cos θ = OQ OP = OQ A ⇒ OQ = Ax = A cos θ sin θ = PQ OP = PQ A ⇒ PQ = Ay = A sin θ (According to law of vector addition, A⃗ = OP⃗⃗⃗⃗ ⃗ = OQ⃗⃗ ⃗⃗ ⃗ + OR⃗⃗⃗⃗ ⃗ A⃗ = Axî + Ayĵ A⃗ = Acosθ î + Asinθj ̂ |A⃗ | = √Ax 2 + Ay 2 𝐭𝐚𝐧𝛉 = 𝐀𝐲 𝐀𝐱 𝐭𝐚𝐧𝛟 = 𝐀𝐱 𝐀𝐲 Here θ is angle between A⃗ and its x component Here ϕ is angle between A⃗ and its y component Part - 09 TG: @Chalnaayaaar  Digital Pvt. Ltd. [4] Vector Part-09 Illustration 5. Resolve the vector in component form Solution. (i) ˆ ˆA 8i 6j= + (ii) ˆ ˆF 10 3i 10j= − + Illustration 6. Find resultant of given vectors Solution. 2 2| R | 1 4 17= + = TG: @Chalnaayaaar  Digital Pvt. Ltd. [5] Vector Part-09 Illustration 7. Solution. | R | 5N= Illustration 8. A vector makes an angle of 𝟑𝟎° with the horizontal. If the horizontal component of the vector is 250 N, find the magnitude of vector and its vertical component. Solution. Acos30º = 250 A = 500 3 Ay = Asin30º = 250 3 TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Resolution of Vectors in 3D-in space Resolution of vectors in 3-D If A⃗ makes angle α with x axis angle β with y axis angle γ with z axis and Component of ?⃗? along 𝐱 axis is 𝐀𝐱 Component of ?⃗? along 𝐲 axis is 𝐀𝐲 Component of ?⃗? along 𝐳 axis is 𝐀𝐳 the 𝐜𝐨𝐬 𝛂 = 𝐀𝐱 𝐀 ⟹ 𝐀𝐱 = 𝐀𝐜𝐨𝐬 𝛂 𝐜𝐨𝐬 𝛃 = 𝐀𝐲 𝐀 ⟹ 𝐀𝐲 = 𝐀𝐜𝐨𝐬 𝛃 𝐜𝐨𝐬 𝛄 = 𝐀𝐳 𝐀 ⟹ 𝐀𝐳 = 𝐀𝐜𝐨𝐬 𝛄 Here, cos α, cos β and cos γ are known as direction cosines of A⃗ As, Ax, Ay and Az are components of A⃗ so, A⃗ = Axî + Ayĵ + Azk̂ A = √Ax 2 + Ay 2 + Az 2 Putting values of Ax, Ay and Az A = √A2 cos2 α + A2 cos2 β + A2 cos2 γ ⇒ A = A√cos2 α + cos2 β + cos2 γ ⇒ cos2 α + cos2 β + cos2 γ = 1 ⇒ (1 − sin2 α) + (1 − sin2 β) + (1 − sin2 γ) = 1 ⇒ sin2 α + sin2 β + sin2 γ = 2 Part - 10 TG: @Chalnaayaaar  Digital Pvt. Ltd. [4] Vector Part-10 Illustration 5. ˆ ˆ ˆ ˆA i 3j; B 2j 3k= − = − then find | A 2B |− Solution. ˆ ˆ2B (4j 6k)= − ˆ ˆ ˆ ˆ| A 2B| i 3j 4j 6k− = − − + = ˆ ˆ ˆi 7j 6k 1 49 36 86− + = + + = Illustration 6. Determine the vector which when added to the resultant of P and Q , gives ZERO resultant (equilibrium). ?⃗? = ?̂? + 𝟐?̂? + ?̂?, ?⃗⃗? = 𝟐?̂? − ?̂? + 𝟐?̂? Solution. Let's new vector = x x (P Q) 0+ + = ˆ ˆ ˆx (3i j 3k) 0+ + + = ˆ ˆ ˆx 3i j 3k= − − − Illustration 7. Determine the vector which when added to the resultant of P⃗ and Q⃗ , gives unit vector along x-axis where ?⃗? = ?̂? + 𝟐?̂? + ?̂? , ?⃗⃗? = 𝟐?̂? − ?̂? + 𝟐?̂?. (1) −2î + ĵ − 3k̂ (2) 2î + ĵ + 3k̂ (3) 2î − ĵ + 3k̂ (4) −2î − ĵ − 3k̂ Solution. Let's new vector = A ˆA (P Q) i+ + = ˆ ˆ ˆ ˆ ˆ ˆ ˆA (i 2j k 2i j 2k) i+ + + + − + = ˆ ˆ ˆ ˆA i 3i j 3k= − − − = ˆ ˆ ˆ2i j 3k− − − Ans. (4) TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Multiplication of Vectors by Scalar and Dot product Multiplication of a Vector by Scalar To multiply a vector by a scalar, simply multiply the similar components, that is the vectors magnitude by the scalars magnitude. This will result in a new vector with the same direction but the product of two magnitudes. Illustration 1. If ?⃗? = 𝟐𝐦/𝐬 then 𝟑?⃗? will be ? Solution. Note : Whenever a vector is multiplied by a positive number then the new vector has same direction but its magnitude changes depending on number. Illustration 2. If ?⃗? = 𝟐𝐦/𝐬 then −𝟑?⃗? will be ? Solution. Note : whenever a vector is multiplied by a negative number then the new vector has opposite direction but its magnitude changes depending on number. How to check parallel vectors ? if A⃗ = Axî + Ayĵ + Azk̂ & B⃗ = Bxî + Byĵ + Bzk̂ Ax Bx = Ay By = Az Bz = Positive constant Part - 11 TG: @Chalnaayaaar  Digital Pvt. Ltd. [2] Vector Part-11 Scalar Product The scalar product of two vectors A and B equals to the product of their magnitudes and the cosine of the angle  between them. A⃗ ⋅ B⃗ = |A⃗ ||B⃗ |cosθ Example : W F S=  Where F = force and S = displacement, Power (P) = F V Where F = Force and V = velocity Angle dependence θ < 90°(Acute) θ > 90°(Obtuse) cos θ → Positive cos θ → Negative A⃗ ∙ B⃗ = |A⃗ ||B⃗ | cos θ A⃗ ∙ B⃗ = positive A⃗ ∙ B⃗ = negative Range of dot product Key notes When two vectors are perpendicular Here, ?⃗? ⊥ ?⃗? TG: @Chalnaayaaar  Digital Pvt. Ltd. [5] Vector Part-11 Illustration 7. Find Work done if Force 𝐅 = 𝟑?̂? + 𝟐?̂? + ?̂? & Displacement 𝐬 = 𝟐?̂? − 𝟓?̂? + 𝟑?̂?. (1) 1 J (2) –1 J (3) 2 J (4) –2 J Solution. W F S=  = ˆ ˆ ˆ ˆ ˆ ˆ(3i 2j k) (2i 5j 3k)+ +  − + = 6 – 10 + 3 = –1J Ans. (2) Illustration 8. Find Power if Force 𝐅 = 𝟐?̂? − 𝟐?̂? + ?̂? & velocity ?⃗? = 𝟑?̂? + 𝟐?̂? − ?̂?. (1) 5W (2) 2W (3) 1W (4) 0W Solution. P F V=  = ˆ ˆ ˆ ˆ ˆ ˆ(2i 2j k) (3i 2j k)− +  + − = 6 – 4 – 1 = 1W Illustration 9. If angle between ?⃗? and 𝐛 is 𝟔𝟎°, then find value of (?̂? − 𝟐?̂?) ∙ (𝟐?̂? + 𝟒?̂?). (1) –6 (2) 6 (3) 0 (4) 12 Solution. ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ(a 2a a 4b 2b 2a 8b b) +  −  −  = ˆ ˆˆ ˆ(2 4(a b) 4(b a) 8) +  −  −   = [–6 + 0] = –6 Illustration 10. If |?̂? + ?̂?| = √𝟑, then find value of (?̂? + 𝟐?̂?) ∙ (𝟐?̂? + ?̂?). (1) 6 (2) 13 (3) 6.5 (4) 7.5 Solution. 2 2ˆ ˆˆ ˆ ˆ| a b | | a | | b | 2 | a || b | cos 3+ = + +  = 1 + 1 + 2 cos = 3, cos = 1 2 ,  = 60º ˆ ˆˆ ˆ(a 2b) (2a b)+  + = ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ2 a a a b 4b a 2b b  +  +  +  = 2 × 1 + 5 × ˆâ b + 2 × 1 = 2 + 5 × 1 × 1 × 1 2 + 2 = 4 + 5 13 2 2 = TG: @Chalnaayaaar  Digital Pvt. Ltd. [6] Vector Part-11 Illustration 11. If ?⃗? + ?⃗? + 𝐂 = ?⃗? and |?⃗? | = |?⃗? | = |𝐂 | = 𝟏, find ?⃗? ∙ ?⃗? + ?⃗? ∙ 𝐂 + 𝐂 ∙ ?⃗? (1) − 3 2 (2) 3 2 (3) 2 (4) 1 Solution. A B C 0+ + = A B C+ = − A B C+ = − 2 2A B 2A B | C |+ +  = − A2 + B2 + 22A B | C| = − (|A| = |B| = |C| = 1) 1 1 1 1 A B 2 2 − −  = = − By same logic 1 B C 2 −  = 1 C A 2 −  = So, A B B C C A +  +  1 1 1 3 2 2 2 2 − − − −      + + =            TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Application of Dot Product Application of DOT product 1. According to definition A B ABcos =  the angle between the vectors  = cos–1 A B AB         Illustration 1. Find angle between ˆ ˆA = 3i + 4j and ˆ ˆB =12i - 5j ? Solution. We have cos  = 2 2 2 2 ˆ ˆ ˆ ˆA B (3i 4j) (12i 5j) AB 3 4 12 5  +  − = + + 36 20 16 cos 5 13 65 −  = =  1 16 cos 65 − = 2. To Check Orthogonal Vectors for  = 90º A⃗ ∙ B⃗ = ABcosθ = 0 Note: If scalar product of two nonzero vectors is zero, then vectors are orthogonal or perpendicular Part - 12 TG: @Chalnaayaaar  Digital Pvt. Ltd. [1] Vector Product Vector product The vector product or cross product of any two vectors A and B , denoted as A B (read A cross B ) is defined as : ˆA B ABsin n =  Here  is the angle between the vectors and the direction of n̂ is given by the right-hand-thumb rule. Right-Hand-Thumb Rule To find the direction of n̂ , draw the two vectors A and B with both the tails coinciding. Now place your stretched right palm perpendicular to the plane of A and B in such a way that the fingers are along the vector A and when the fingers are closed they go towards B . The direction of the thumb gives the direction of n̂ . Examples of vector product (i) Torque : r F =  (ii) Angular momentum : J r p=  (iii) Velocity : v r=  (iv) Acceleration : a r=  Here r is position vector and F,p, and  are force, linear momentum, angular velocity and angular acceleration respectively. Part - 13 TG: @Chalnaayaaar  Digital Pvt. Ltd. [2] Vector Part-13 Properties : • Vector product of two vectors is always a vector perpendicular to the plane containing the two vectors, i.e. orthogonal (perpendicular) to both the vectors A and B • Vector product of two vectors is not commutative i.e. A B B A   But | A B| | B A | ABsin =  =  Note A B B A = −  i.e., in case of vectors A B and B A magnitudes are equal but directions are opposite [See the figure] • The vector product is distributive when the order of the vectors is strictly maintained, i.e. A (B C) A B A C + =  +  • According to definition of vector product of two vectors A B = AB sin n̂   = sin–1 | A B | AB  • The vector product of two vectors will be maximum when sin = max. = 1, i.e.,  = 90º max| A B| ABsin90º AB = = i.e. vector product is maximum if the vectors are orthogonal (perpendicular) • The vector product of two non-zero vectors will zero when |sin| = 0, i.e. when  = 0º or 180º, | A A | = 0 or | A ( A) | 0 − = Therefore if the vector product of two non-zero vectors is zero, then the vectors are collinear. • The self cross product, i.e., product of a vector by itself is a zero vector or a null vector. i.e. ˆA A (AAsin0º )n 0 = = • In case of unit vector ˆ ˆ ˆn :a a = 1 × 1 × sin0º n̂ 0= so that ˆ ˆi i 0 = , ˆ ˆj j 0 = , ˆ ˆk k 0 = • In case of orthogonal unit vectors ˆ ˆi, j and k̂ ; according to right hand thumb rule ˆ ˆ ˆi j k = , ˆ ˆ ˆj k i = , ˆ ˆ ˆk i j = and ˆ ˆ ˆj i k = − , ˆ ˆ ˆk j i = − , ˆ ˆ ˆi k j = − TG: @Chalnaayaaar  Digital Pvt. Ltd. [3] Vector Part-13 • In terms of components, x y z y z z y x z z x x y y x x y z ˆ ˆ ˆi j k ˆ ˆ ˆA B A A A i(A B A B ) j(A B A B ) k(A B A B ) B B B  = = − − − + − , where x y z ˆˆ ˆA A i A j A k= + + and x y z ˆˆ ˆB B i B j B k= + + • If A,B and C are coplanar, then A (B C) 0  = . [ (B C) is perpendicular to A ] • Angle between (A B)+ and (A B) is 90º as A B is perpendicular to plane containing A & B . • A scalar or a vector, cannot be divided by a vector. • Vectors of different types can be multiplied to generate new physical quantities which may be a scalar or a vector. If, in multiplication of two vectors, the generated physical quantity is a scalar, then their product is called scalar or dot product and if it is a vector, then their product is called vector or cross product. Illustration 1. A is East wards and B downwards. Find the direction of A B ? Solution. Applying right hand thumb rule we find that A B is along North. Illustration 2. If A B | A B| =  , find angle between A and B Solution. AB cos = AB sin tan  = 1   = 45º Illustration 3. Find A B if ˆ ˆ ˆA i 2j 4k= − + and ˆ ˆ ˆB 2i j 2k= − + . Solution. ˆ ˆ ˆi j k ˆ ˆ ˆ ˆA B 1 2 4 i( 4 ( 4)) j(2 12) k( 1 ( 6)) 10j 5k 3 1 2  = − = − − − − − + − − − = + − Illustration 4. If | A | 2;| B | 4= = and | A B | 4 = . Then find A B Solution. |A⃗ × B⃗ | = ABsinθ sinθ = 4 8 = 1 2 θ = 30° A⃗ ⋅ B⃗ = ABcos30° = 2 × 4 × √3 2 = 4√3 TG: @Chalnaayaaar