Title 10. VECTORS.pdf ✅

Vectors Types of Vectors: (a)Zero or Null vector: It is that vector whose initial and terminal points are coincident. It is denoted by 0 . Therefore its magnitude is 0 (zero). z Any non-zero vector is called a proper vector. (b) Co-initial vectors: The vectors (two or more) which have a same starting point are called the co-initial vectors. (c)Co-terminus vectors: The vectors (two or more) having the same terminal point are called the co-terminus vectors. (d) Negative of a vector: The vector which has the same magnitude as the r → but opposite direction. It is denoted by – r . Hence if, AB r BA r → → → → = = or − i.e., AB BA PQ QP → → → → =− =− , etc. (e) Unit vector: It is a vector with the unit magnitude. The unit vector in the direction of vector r → is given by r r r = | | such that | |r = 1, so, if r xi y j zk = + + then its unit vector is: r x x y z i y x y z j z x y z k = + + + + + + + + 2 2 2 2 2 2 2 2 2 . [SQP 2020-21] z Unit vector perpendicular to the plane a → and b → is: ± × × a b a b | |. (f) Reciprocal of a vector: It is a vector which has the same direction as the vector r → but magnitude equal to the reciprocal of the magnitude of r → . It is denoted as r → –1. Hence, r r − = 1 1 . (g)Equal vectors: Two vectors are said to be equal if they have the same magnitude as well as direction, regardless of the position of their initial points. Thus a b a b = ⇔  =   | | | | a b and have same direction Also, if a b a i a j a k b i b j b k a b a b a b = ⇒ 1 2 + + 3 1 = + 2 3 + ⇒ 1 1 = = , , 2 2 3 3 = . a b a i a j a k b i b j b k a b a b a b = ⇒ 1 2 + + 3 1 = + 2 3 + ⇒ 1 1 = = , , 2 2 3 3 = . (h)Collinear or Parallel vector: Two vectors a and b are collinear or parallel if there exists a non-zero scalar l such that a = λb . z It is important to note that the respective coefficients of i j k , , in a → and b → are proportional provided they are parallel or collinear to each other. z The d.r’s of parallel vectors are same (or are in proportion). z The vectors a → and b → will have same or opposite direction as l is positive or negative respectively. z The vectors a → and b → are collinear if a b × = 0 . (i) Free vectors: The vectors which can undergo parallel displacement without changing their magnitude and direction are called free vectors. KEY-FACT Vector calculus and its sub objective vector fields was invented by two men J. Willard Gibbs and Oliver Heaviside at the end of 19th century. KEY-TERM Magnitude: It is defined as the maximum extent of size and the direction of an object. Magnitude is used as a common factor in vector and scalar quantities. KEY-FORMULAE The position vector of a point say P dividing a line segment joining the points A and B whose position vectors are a → and b → respectively, in the ratio m : n. (a) Internally, OP m b n a m n → → → = + + [Board- 2016] (b) Externally, OP m b n a m n → → → = − − • Also if point P is the mid-point of line segment AB, then OP → a b → → = + 2 .
MATHEMATICS, Class-XII MNEMONICS Types Of Vectors (A) Interpretation : Types of Vectors- 1. Zero Vector - Initial and terminal points coincide 2. Unit Vector - Magnitude is unity 3. Coinitial Vectors - Same initial points 4. Collinear vectors - Parallel to the same Line 5. Equal Vectors - Same magnitude and direction 6. Negative of a vector - Same magnitude, opp. direction Properties Of Vectors(B) "Neither choose East nor choose North, always choose North-East and save your time". Interpretation : The vector sum of two co-initial vectors is given by the diagonal of the parallelogram whose adjacent sides are given vectors. A B C D AB + AC = AD AB AC AD → → → + = Properties Of Vectors(C) Interpretation: The vector sum of the three sides of a triangle taken in AB BCorder is CA + + = 0 i.e., AB BC CA + + = 0 SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) 1. The position vectors of two points A and B are OA = 2i j k − − and OB = 2 2 i j k − + , respectively. Find the position vectors of a point which divides the line segment joining A and B in the ratio 2 : 1. R&U [Delhi Set-1, 2020] Topper's Answer, 2020 Sol. 2. Find a unit vector in the direction opposite to − 3 4 j . A [SQP 2020-21] Concept Applied To find a vector in the opposite direction to a given vector, you can simply multiply the given vector by –1. Vector in opposite direction. Sol. j [Marking Scheme SQP, 2020-21] 1 Commonly Made Error Some students find any vector instead of unit vector. Some others find the unit vector in the same direction. Read the question carefully and practice more problems involving unit vectors. Answering Tip This Question is for practice and its solution is given at the end of the chapter.
Vectors 3. Give an example of vectors a → and b → such that | | a → = | | b → but a b → → ≠ . R&U [SQP 2017-18] 4. Find the position vector of a point which divides the join of points with position vectors ( 2 a b ) → → - and (2 a b + ) → → externally in the ratio 2 : 1. R&U [Delhi Set 1, 2, 3, 2016] Sol. Required vector = 1 2 2 2 1 2 ( ) a b ( ) a b → → → → − − + − 1⁄2 = ( ) a b ( ) a b → → → → − − + − 2 4 2 1 = 3 4 a b → → + 1⁄2 5. The two vectors j k ∧ ∧ + and 3 + i j 4k ∧ ∧ ∧ - represent the two sides AB and AC, respectively of DABC. Find the length of the median through A. A [Delhi Set 1, 2, 3, 2016] Sol. AB j k → ∧ ∧ = + and AC i j k → ∧ ∧ ∧ = − 3 4 + A B C D E Now ABEC represent a parallelogram with AE as the diagonal. AE → = AB AC → → + 1⁄2 = ( ) j k ( ) i j k i k ∧ ∧ ∧ ∧ ∧ ∧ ∧ + + 3 4 − + = + 3 5 Now, | | AE → = ( ) 3 5( ) 9 25 34 2 2 + = + = units \ | | AD → = 1 2 34 units 1⁄2 Short Answer Type Questions-I (2 marks each) 1. If points A, B and C have position vectors 2 i j ^ ^, and 2k ^ respectively, then show that DABC is an isosceles triangle. [Outside Delhi Set-3, 2023] Sol. A i ( ). ( B j) ( C k) ^ ^ ^ 2 2 and → AB → = ^ ^ j i − 2 | | AB → = 1 2 5 2 2 + =| |unit 1⁄2 BC → = 2 k j ^ ^ − | | BC → = 2 1 5 2 2 + =| |units 1⁄2 AC → = 2 2 k i ^ ^ − | | AC → = 2 2 2 2 2 2 + =| |units 1⁄2 \ AB = BC = 5 unit Hence DABC is an isosceles triangle 1⁄2 Commonly Made Error Few students directly take the magnitudes of positions vectors of points A, B and C to show that triangle ABC is isosceles and loses their marks. Always remember that to show the triangle is isos- celes, show the two sides of triangle are equal, for this evaluates the sides of triangle by using the given position vectors. Answering Tip 2. Shown below are two vectors in their component forms. u i p j k = − 3 5 + v i j qk = −6 1 + + 4 For what values of p and q are the vectors collinear? Show your steps. [Practice Questions 2022] Concept Applied a a i b j c k = + 1 1 + 1 and b a i b j c k = + 2 2 + 2 are parallel, then a a b b c c 1 2 1 2 1 2 = = Parallel vectors: If two vectors Sol. Here, u is parallel to v . We know that if two vectors a a i b j c k = + 1 1 + 1 and b a i b j c j = + + 2 2 2 are parallel, then a a 1 2 = b b 1 2 = c c 1 2 Therefore, we have 3 −6 = −p 14 = 5 q 1 \ 3 −6 = −p 14 Þ p = 7 1⁄2 and 3 −6 = 5 q Þ q = –10 1⁄2 Thus, when p = 7 and q = –10, then vectors are collinear. These Question is for practice and its solution is given at the end of the chapter.