PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1IPd5nQpQ-DLlnWpZa_XplRl43V-3ztPQ/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true two dimensional and three dimensional space, scalar and vector products, scalar and vector triple product. Application of vectors to plane geometry. Vector algebra deals with addition / subtraction / product of vector quantities. Application of vectors to many geometrical problems cuts short the procedure. Hence geometrical significance of vectors should be well understood. VECTORS AND SCALARS The physical quantities (we deal with) are generally of two types: Scalar Quantity: A quantity which has magnitude but no sense of direction is called Scalar quantity (or scalar), e.g., mass, volume, density, speed etc. Vector Quantity: A quantity which has magnitude as well as a sense of direction in space is called a vector quantity, e.g., velocity, force, displacement etc. NOTATION AND REPRESENTATION OF VECTORS CHAPTER INCLUDES : Vectors and Scalars Representation of vectors Magnitude of a vectors Types of vectors Angle between vectors Vectors are represented by a , b , c and their magnitude (modulus) is represented Addition of vectors by a, b, c, or | a |, |b |,|c | , The vectors are represented by directed line segments. P O For example, line segment OP represents a vector with magnitude OP (length of line segment), arrow denotes its direction. O is initial point and P is terminal point. SOME SPECIAL VECTORS Null vectors: A vector with zero magnitude and indeterminate direction, denoted by O . Unit vector: A vector with unit magnitude (one unit), denoted by aˆ where | aˆ | = 1 unit. Equal vectors: Two vectors a and b are said to be equal if they have same sense of direction and | a |= |b | , denoted by a = b . Subtraction of vectors Section formulae Resolution of vectors Linearly dependent & independent system of vectors Multiplication of a vectors by a scalar Scalar (dot) product of two vectors Vectors (cross) product of two vectors Scalar triple product Vector triple product Application of vectors to geometry Solved examples TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Like and unlike vectors: Vectors having same sense of directions are called Like vector and opposite sense of directions are called Unlike vector. Negative of a vector: Negative of a vector a , denoted by – a , is a vector whose magnitude is | a | and direction is opposite of a . Collinear vectors: Vectors having same line of action. Parallel vectors: Vectors having same line of action or are parallel to a fixed straight lines. Coplanar vectors: The vectors which lie in the same plane. At least three coplanar unequal vectors are required to make the sum zero and at least four if non-coplanar. Free vectors: A vector not restricted to pass through a fixed point. Localized vectors: A vector restricted to pass through a fixed point. Co-initial vectors: Vectors having same initial point. Position vectors: Let O be fixed point in space, then vector OP (P is any point in space) is called position vector of P w.r.t. O. If A and B are any two point in space then AB = p.v. of B – p.v. of A = OB − OA . ANGLE BETWEEN TWO VECTORS The angle between two vectors a and b represented by OA and OB is ∠AOB = θ B 0 ≤ θ ≤ π If θ = π , then vectors are called orthogonal or perpendicular vectors b 2 θ if θ = 0 or π then vectors are called parallel or coincident vectors. θ a A ADDITION OF VECTORS Let a and b be two vectors. Draw OA representing vector a . Taking terminal point of a as initial point of vector b , draw AB B representing vector b . Then vector OB is called sum of vectors b a and b , denoted by a + b (triangle law of addition) O A Triangle Law of Addition a Three vectors are in equilibrium if represented by, three sides of a closed triangle taken in order, (in magnitude and direction). Converse of triangle law is also true. Law of polygons : If several vectors, when added, form a closed polygon, their resultant is zero. Parallelogram law of addition – If two vectors are represented in magnitude and direction, by the two adjacent sides of a parallelogram, their resultant is represented in magnitude and direction, by the co-initial diagonal of that parallelogram. R = and tanφ = Q sinθ P + Q cos θ If θ = 0º then R = P + Q (maximum) θ = π then R = P ~ Q (minimum) If θ = π 2 then R = and tanφ = Q P Properties of Vector Addition P Vector addition is commutative. a + b = b + a Vector addition is associative (a + b ) + c = a + (b + c ) O or null vector is additive identity. A + O = A = O + A Additive inverse of A is (−A ) , Since A + (−A ) = O At what angle (θ) shall two vectors P & Q each. What shall be the range of n? (| P |=| Q |) act, so as to have a resultant n times in magnitude of Solution : Let | P |=| Q |= F then magnitude of resultant = nF. By law of parallelogram of addition (nF )2 = F 2 + F 2 + 2F 2cosθ or n2 = 2 + 2 cosθ 2 − cos θ = θ = cos 2 −1⎛ n 2 − 2 ⎞ ⎜ ⎟ ⎝ ⎠ − n2 − 2 Since –1 ≤ cosθ ≤ 1, So 1 ≤ ≤ 1 2 ⇒ –2 ≤ n2 – 2 ≤ 2 ⇒ 0 ≤ n2 ≤ 4 or 0 ≤ n ≤ 2 SUBTRACTION OF TWO VECTORS The subtraction of two vectors a and b is defined as a − b = a + (−b ) i.e. the vector to be subtracted is reversed and added to other vector. a − b ≠ b − a a − b = −(b − a ) not commutative i.e. directions are opposite but magnitudes are same For non - zero vectors a & b a + b ≠ a − b | a + b |=| a − b | mean a ⊥ b – a SECTION FORMULAE Let OA = a ,OB = b then p.v. of point P which divides AB internally in the ratio m : n given by ρ OP = r = mb + n ρ A (m ≠ – n) m m + n If P divides AB externally in the ratio m : n then OP = na − mb n − m a r P n O b B VECTOR RESOLUTION OF A VECTOR (COMPONENTS OF A VECTOR) A vector ρ r = ai + bj + ck has its x, y and z components as a, b & c respectively. The vector resolution can be orthogonal or non-orthogonal. Orthogonal resolution means the components of the vector are mutually perpendicular otherwise non-orthogonal. Orthogonal components of vector R are as shown. P = Rcosθ Q = Rsinθ R = P + Q P Non-orthogonal components of R are as shown P = R sinβ sin(α + β) R sinα Q = sin(α + β) LINEAR COMBINATION OF VECTORS If a and b are two non-collinear vectors, then any vector r in the plane of a and b is uniquely expressed as r = xa + yb , where x and y are scalars. (Similar linear combination (unique) exists for three non-coplanar vectors, r = xa + yb + zc ). LINEARLY DEPENDENT AND INDEPENDENT VECTORS Vectors a1 a2, , an are said to be linearly dependent if for scalars x1, x2,....xn (not all zero), x1 a1 + x2 a2 + .... + xn an = 0 independent. . If x1 a1 + .... + xn an = 0 ⇒ x1 = x2 = xn = 0 then vectors are called linearly If the vectors xiˆ + ˆj + kˆ , iˆ + yˆj + kˆ and iˆ + ˆj + zkˆ are coplanar, when x ≠ 1, y ≠ 1 and z ≠ 1, then show that 1 + 1− x Solution : 1 1− y + 1 = 1 1− z For the given vectors to be coplanar. x 1 1 1 y 1 = 0 1 1 z x − 1 1 − y 0 ⇒ 0 1 y − 1 1 1 − z z = 0 (Operature R1 − R2 & R2 − R3 ) 1 1 0 ⇒ (x − 1)(1− y )(1− z) 0 1 1 = 0 1 1 z x − 1 1− y 1− z ⎡⎛ − z ⇒ (x −1)(1− y)(1− z)⎢⎜1− z − 1  1  ⎟ + = 0 1− y ⎠ x −1⎦  1 1 x + 1 1− y = −z 1− z (Θ x ≠ 1, y ≠ 1,z ≠ 1)  1 1 x + 1 1− y + 1 1− z  1 1 z z 1− z = 1− z = 1 1− z Multiplication of a Vector by a Scalar If m is a scalar and a is a vector, then ma (scalar multiple) is a vector whose magnitude is | m | | a| and direction is same as of a (if m +ve) and opposite that of a if m is –ve. Properties m (a + b ) = ma + mb (mn )a = m(na ) = n(ma ) (m + n)a = ma + na Scalar (dot) Product of Two Vectors The scalar product of vectors a and b , denoted by a . b , is defined as a .b =| a || b | cosθ, θ is angle between two vectors. Properties : 1. a . b = b .a 2. a .(b + c ) = a .b + a .c a 3. (ma ). b = m(a .b ) = a .(m b ) If θ = 0 ⇒ a .b =| a ||b | (like vectors) If θ = π ⇒ a .b = − | a ||b | (unlike vectors) If aˆ and bˆ are unit vectors then aˆ.bˆ = cosθ (where θ is angle between them). 7. a .a =| a |2 ⇒| a |= If a ⊥ b ⇒ a . b = 0 but a .b = 0 ⇒ a = 0 or b = 0 or a ⊥ b . If iˆ, ˆj and kˆ are unit vectors along the rectangular coordinate are OX, OY and OZ then iˆ.iˆ = ˆj. = kˆ.kˆ = 1 , iˆ. ˆj = ˆj.kˆ = kˆ.iˆ = 0 . If a = a1iˆ + a2 ˆj + a3kˆ and b = b1iˆ + b2 ˆj + b3kˆ then a .b = a1b1 + a2b2 + a3b3 and cos θ = a1b1 + a2b2 + a3b3 ⎛ r .a ⎞  ⎛ r .a ⎞  Components of a vector r in the direction of vector a is ⎜ ⎟ a ⎜ | a |2 ⎟ and ⊥ to a is r − ⎜ ⎟ a . ⎜ | a |2 ⎟ ⎝ ⎠ ⎝ ⎠ Work done by a force F in displacing a particle from A to B (AB = d ) W = F . AB = F .d Determine the values of c such that for all x (real) the vectors cxiˆ − 6 ˆj + 3kˆ and xiˆ + 2 jˆ + 2cx kˆ make an obtuse angle with each other. Solution : If θ is the angle between the given vectors then cos θ = cx 2 − 12 + 6 cx If θ is obtuse then cosθ < 0 ⇒ cx2 + 6cx – 12 < 0 Which is possible if c < 0 and 36c2 + 48c < 0 ⇒ c < 0 and 12c (3c + 4) < 0 ⇒ c < 0 and c > – 4/3 c 2 x 2 + 45 ∀ x ∈ |R x 2 + 4c 2 x 2 + 4 ⇒ − 4 < c < 0 3 (but for c = 0, cx2 + 6cx – 12 < 0 ∀x) Hence − 4 < c ≤ 0 . 3 VECTOR (CROSS) PRODUCT OF TWO VECTORS Vector product of two vectors a and b is defined as a × b =| a | | b | sin θ nˆ where θ is the angle between a and whose direction is that of unit vector nˆ which is ⊥ to both a and b in such away that handed triad (right handed screw system). Properties a , b , nˆ form a right In general, a × b ≠ b × a . In fact a × b = − b × a . For scalar m, ma × b = m(a × b ) = a × mb . a ×(b + c ) = a × b + a × c A If a ||b then θ = 0 or π ⇒   a × b = 0 (but a × b = 0 ⇒ a = 0 or b = 0 a || b ). In particular a × a = 0 . If a ⊥ b then a × b = | a | |b |nˆ (or | a × b | =| a | | b | ) ^j 6. iˆ × iˆ = ˆj × ˆj = kˆ × kˆ = 0 and iˆ × jˆ = kˆ, kˆ × iˆ = jˆ (use cyclic system) ˆj × kˆ = iˆ and k Unit vector perpendicular to a and b is given by ± a × b | a × b | If θ is angle between a and b then sinθ = | a × b | | a || b | If a = a1iˆ + a2 ˆj + a3kˆ and b = b1iˆ + b2 ˆj + b3kˆ then iˆ ˆj kˆ a × b = a1 a2 a3 b1 b2 b3 (i) If a and b are adjacent sides of a parallelogram, then Area of parallelogram = | a × b| (ii) If diagonals of parallelogram are a and b , then area of parallelogram = 1 | a × b| Vector moment of a force about a point: The vector moment or torque M of a force F acting at A about the point O is given by M = r × F A M F SCALAR TRIPLE PRODUCT Scalar triple product: If a , b denoted [a b c ] . Properties and c are three vectors then a .(b × c ) is called scalar triple product of a , b and If a , b and c are adjacent sides of a parallelopiped then volume = [a b c ] [a b c ] =[b c a ] = [c a b ] [a b c ] = − [a c b ] etc. (cyclic permutations of a , b and c makes no change in value) but i.e., Dot and cross can be interchanged, keeping the same cyclic order i.e. a .( b × c ) =(a × b ).c For scalar m, [ma b c ] =[a m b c ] = [a b mc ] = m[a b c ] The value of scalar triple product is zero if any two vectors are equal or parallel If a = a1iˆ + a2 jˆ + a3 kˆ , b = b1iˆ + b2 ˆj + b3kˆ and c = c1iˆ + c2 ˆj + c3kˆ a1 a2 a3 then [a b c ] = b1 b2 b3 c1 c2 c3 Condition of coplanarity: Three non-colinear, non-zero vectors a , b Vector Triple Product and c are called coplanar iff [a b c ] = 0 Vector triple product: If a , b and c are three vectors their vector triple product is defined by a × ( b × c ) = (a .c ) b − ( a .b )c and (a × b ) × c = (a .c ) b − ( b .c )a APPLICATION OF VECTORS TO GEOMETRY Vector Equation of Straight Line Case I : Vector equation of a straight line passing through a point a and parallel to vector b . Let O be the origin and OA = a . Let P be any point on the line with OP = r . b Since AP || b , for some scalar t, AP = t b A t b P ⇒ r − a = t b ⇒ a r (required equation) O Case II : Vector equation of a straight line passing through two points a and b . Vector Equation of a Plane Case I : Vector equation of a plane passing through the point a and parallel to two given vectors b and c . Let O be the origin, OA = a . Let AB and AC be the lines on the plane | | to vectors b and c . Let P(r ) be any point on the plane, draw PL | | AC, (AL | | AB). Now AL = sb and LP = t c (s, t are scalars) Now AP and OP = AL + LP = OA + AP = s b + t c = a + s b + t c Required equation Another form : O Case II : Vector equation of a plane passing through the point a , b , c is given by (where s, t are scalars) Another form : Case III: Vector equation of plane passing through a point a and perpendicular to vector n : Perpendicular distance of a point from a line Let straight line passes through A ( a ) and is | | to b . Then perpendicular distance of point b ( r ) from the line = Another form : Perpendicular distance of a point from a plane A( a ) N M P( r ) Case I : When plane passes through a point a and is | | to b and c . Distance of point P (r ) from the plane Case II : When plane passes through the points a , b and c . Shortest distance between two non-intersecting lines : Let there be two non-intersecting lines passing through a and b and are | | to c and d . If PQ be the shortest distance between them, then PQ = projection of AB on the vector n (where n Then = c × d ) Vector equation of the bisector of the angle between two straight lines : Let the two lines AB and AC meet at A whose p.v. is a . Let b and c be vectors | | to AB and AC respectively. Also let P be a point on internal bisector of ∠BAC where OP = r . From P draw line | | AB which meets AC at M. Now ∠PAM = ∠PAB = ∠APM ⇒ AM = PM = t (say) AM is collinear with c and MP is collinear with b . Then AM = t c | c | ,MP = t b | b | ⎛ c b ⎞ Now in ΔAPM, AP = AM + MP = t⎜ + ⎜ | c | ⎟, | b | ⎟ C In ΔOAP, OP = OA + AP ⎛ b c ⎞ ⇒ r = a + t⎜ + ⎜ | b | ⎟, | c | ⎟ required equation of O internal bisector ⎛ b c ⎞ Equation of external bisector is given by r = a + t⎜ − ⎜ | b | ⎟ | c | ⎟  Show that the perpendicular distance of the point A whose position vector is a from the line r {(a − b ).c }c = b + tc is Solution : We have b − a + | c |2 BC = projection of vector BA along the given line (a − b ). c A( a ) = c | c |2 Now in ΔABC we have AC = AB + BC = (b − a ) + (a − b ).c c | c |2 B( b ) r = b + t c C Therefore perpendicular distance =| AC |= − a + (a − b ). c | c |2 SOLVED EXAMPLES If a ,b ,c are unit vectors such that a is perpendicular to the plane of b and c then find | a + b + c | when the angle between b and c is π/3. Solution : We have | a |=| b |=| c |= 1,a .b = a .c = 0 and b .c = cos π = 1 3 2 Now | a + b + c |2 =(a + b + c ).(a + b + c ) ⇒ | a + b + c |= 2 =| a |2 + | b |2 + | c |2 +2b.c = 1+ 1+ 1+ 2. 1 = 4 2 (Θ a.c = a .b = 0) ( r − a ) × b b Find the distance of the point B(iˆ + 2 jˆ + 3kˆ) from the line which passes through A(4iˆ + 2 ˆj + 2kˆ) parallel to the vector C = 2iˆ + 3 ˆj + 6kˆ. and which is Solution : Given, p.v. of A = a = 4iˆ + 2 ˆj + 2kˆ p.v. of B = b = iˆ+ 2 ˆj + 3kˆ A C N M and p.v. of C = c = 2iˆ + 3 ˆj + 6kˆ Now AB = b − a = −3iˆ + kˆ. Then AB × c = −3iˆ + 20 ˆj − 9kˆ. Now | AB × c |=| AB ||c | sinθ= | AB || c | BN | AB | =|c |BN ⇒ BN = | 2iˆ+ 3 ˆj + 6kˆ | = = ❑ ❑ ❑