Tiêu đề VECTOR ALGEBRA (Q).pdf ✅

\\ Chapter – 1 Physical quantities, like displacements, velocity, acceleration etc. which have not only a value to signify how much (quantity wise) they are; – this aspect is called the magnitude, – but in addition have a directional element in them, are known as vectors. Any segment AB of a straight line which has distinct end points A and B is called a directed line segment. A and B are called respectively the initial point and end point of the line segment. A vector may be defined in terms of such a line segment and we write a  AB . a  A O Vectors are generally printed in bold faced type (in printing). When writing in the manuscript form the notation a, AB may be used. A vector, thus, is geometrically represented by a directed line segment. It is said that a  is equivalently OA and they are vectorially indistinguishable. The laws, governing the algebra of vectors, are so designed as to be particularly applicable in cases of physical quantities like velocity, acceleration, etc. Thus two vectors a and b are added according to the parallelogram law of addition; namely: If two vectors a and b are represented in magnitude and direction by two, line segments OA and OB , their sum c  a  b, is represented by the diagonal OC of the completed parallelogram OACB. Sometimes this is also referred to as the triangle law of addition. The parallelogram law of addition is OA  OB  OC ( a  b  c ) The triangle law is OA  AC  OC B C O a A  b  c  ( a  b  c ) This addition operation, cumulatively, may be had for more than two vectors; and we have ( a  b )  c  a  ( b  c )  a  b  c VECTORS AND SCALARS 1 THEORY CONTENT OF ALGEBRA: VECTOR LAW OF ADDITION OF VECTORS 2
For addition of more than two vectors we have a polygon laws of vectors addition which is just an extension of triangle law. OA  AB  BC  CD  DE  EF  OF As a result if terminus of last vector coincides with the initial point of the first vector, then the sum of vectors is a null vector (a vector with zero magnitude). A E C B D F O Illustration 1 Question: If the vectors   a and b represent two adjacent sides of a regular hexagon, express the other sides as vectors in terms of   a and b . Solution: ABCDEF is a regular hexagon. Let       FA a and AB b .          FB FA AB a b    FC 2 b (  FC is parallel to  AB and lengthwise doubled)             BC FC FB 2 b a b =   b a         CD a ; DE b ;      EF a b . a  b  A B C D E F Illustration 2 Question: Prove that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero. Solution: ABC is the triangle and AD is the median through A. If AD be produced to a length DE = AD, then ACEB is a parallelogram. Hence by the parallelogram law of addition of two vectors,     AB AC AE  2AD Similarly,    BA  BC  2BE and    CB  CA  2CF Adding, we have                          AB AC BA BC CB CA 2 AD BE CF But the L.H.S is such that     0     AB BA AB AB . Similarly, the other two pairs also become zero. Hence     AD  BE  CF  0 . Illustration 3 Question: Five forces represented by      AB, AC, AD, AE and AF act at the vertex A of a regular hexagon ABCDEF. Prove that their resultant is a force represented by 6  AO , where O is the centre of the hexagon. B C A D E F E
Solution: AB AC AD AE AF          =      ED  AC  AD  AE  CD =      AC  CD  AE  ED  AD =     AD  AD  AD  3AD = 6 AO  This is the resultant required. The magnitude or modulus of a vector a refers to its absolute value and is denoted by | a | . A vector whose modulus is one unit is called a unit vector, and a vector whose modulus is zero is called a zero vector or a null vector. Such a vector has its length zero, and is therefore geometrically represented by a pair of coincident points. For two vectors a and b , || a |  | b ||  | a + b |  | a | + | b |. Illustration 4 Question: If G be the centroid of a triangle ABC, show that     GA  GB  GC  0 ; and conversely, if     GA  GB  GC  0 , then G is the centroid of the triangle ABC. Solution: Necessary part Take G as the centroid. Let the parallelogram GCFB be completed.    GB  GC  GF (parallelogram law of addition of vectors) =  2GD =   GA (  GA is oppositely directed to  GD and lengthwise doubled) A B C D F H E G     GA  GB  GC = 0. Conversely: Assume that    GA  GB  GC = 0. Let G be joined to the midpoint D of BC and produced to  F     GC  GB  GF  2GD  2   0   GD GA ... (i) This means that   GD and GA have the same directions. Already GD is the join of G to the midpoint of BC. Hence, AGD is a continuous line. So AD is the median. From (i), it is also seen that 1 2  GD AG . G is the point of trisection of the median. Hence G is the centroid. A B C E D F O MAGNITUDE AND MODULUS OF A VECTOR 3 MULTIPLICATION OF A VECTOR BY A SCALAR 4
When a vector is multiplied by a scalar number, its magnitude gets multiplied but direction wise there is no change. Thus k a is a vector in the same direction of a but magnitude made k times. Thus if in the direction of a , a unit vector is usually represented as a ˆ then a = | a | a ˆ. Thus any vector = (its magnitude) unit vector in that direction. It may be also said that a and a ˆ which are direction wise same, are collinear. The position vector r  of any point P with respect to the origin of reference O is a vector OP . For any two points P and Q in the space, the vector PQ can be expressed in terms of their position vectors (p.v.) as O P Q PQ  OQ OP Illustration 5 Question: If a  and b  are position vectors of A and B respectively, find the position vector of a point C in AB produced such that AC  3AB . Solution: AC  3AB  c a    = 3 (b a)    where c  is the position vector of point C  c  = a b a c b a        3  3   3  2 . It is defined as the smaller angle formed when the initial points or the terminal points of two vectors are brought together. Angle between two vectors lies in the interval [0, ]. Q a  b  c Q  d  Illustration 6 Question: If angle between equal vectors a  and b  is , then find the angle between a  and a b    . Solution: The angle between a  and a b    is /2 as the diagonal of rhombus bisect the angle between two sides. Two vectors are parallel if they have same direction. They are also known as like vectors. Non-parallel vectors are known as unlike vectors. Illustration 7 Question: What is the unit vector parallel to a i j k ˆ 2 ˆ 4 ˆ 3    ? What vector should be added to  a so that the resultant is the unit vector i ˆ ? POSITION VECTOR OF A POINT 5 ANGLE BETWEEN TWO VECTORS 6 PARALLEL VECTORS 7