Tiêu đề 3. VECTOR_SYN _ W.S-5 TO 7_VOL-1_(61 TO 90).pdf ✅

VIII – Physics (Vol – I) Olympiad Class Work Book Cross product of vectors CROSS PRODUCT (OR) VECTOR PRODUCT OF TWO VECTORS If two vectors are multiplied such that their product is again a vector, the product is called vector product or cross product. B A The cross product of two vectors A and B   is denoted by A B   and read as A  cross B  . It is defined as a third vector C  whose magnitude is equal to the product of the magnitudes of the two vectors A and B   are the sine of their included angle  . Thus, if C A B      , thenC ABsin   . CORK SCREW RULE The vector C  is normal to the plane of A and B   are points in the direction in which a right handed screw would advance when rotated about an axis perpen- dicular to the plane of the two vectors in the direction from A to B   through the smaller angle  between them or, alternatively, we might state the rule as: If the fingers of the right hand be curled in the direction in which vector A  must be turned through the smaller included angle  to coincide with the direction of vector B  , the thumb points in the direction of C  as shown in figure. Either of these rules is referred to as the right handed screw rule. Thus, if nˆ be the unit vector in the direction of C  , we have C A B ABsin n    ˆ    where 0     .
Olympiad Class Work Book VIII – Physics (Vol – I) EXAMPLES OF CROSS PRODUCT : 1. A force F  acts at a point 1 r  and the torque acting about a point 2 r is given by   r F    and 1 2 r r r      the torque    is also called as moment of force. 2. Linear velocity of a body under circular motion about an axis is given by V r      3.Angular momentum L  is the cross product of position vector r  and linear momentum p  L r p      4. The couple C  acting on bar magnetic moment(M) placed in a magnetic field of strength B  is given by C M B      . Important points about vector product: (i) A B B A         (ii) The cross product of two parallel vectors is zero, as A B ABsin      and    0 for two parallel vectors. Thus ˆ ˆ ˆ ˆ ˆ ˆ i i j j k k 0       (iii) If two vectors are perpendicular to each other, we have    90 and there- fore, sin 1   . So that A B ABn   ˆ   . The vectors A,B   and A B   thus form a right handed system of mutually perpendicular vectors. If follows at once from the above that in case of the orthogonal triad of unit vectors ˆ ˆ i, j and kˆ (each perpen- dicular to each other) ˆ ˆ ˆ ˆ ˆ i j j i k      ˆ ˆ ˆ ˆ ˆ j k k j i      and k i i k j ˆ ˆ ˆ ˆ ˆ      (iv) A B C A B A C                (v) A vector product can be expressed in terms of rectangular components of the two vectors and put in the determinant form as may be seen from the following: Let 1 1 1 A a i b j c k    ˆ ˆ ˆ  and 2 2 2 B a i b j c k    ˆ ˆ ˆ 
VIII – Physics (Vol – I) Olympiad Class Work Book Then  1 1 1 2 2 2    A B a i b j c k a i b j c k        ˆ ˆ ˆ ˆ ˆ ˆ   1 2 1 2 1 2 1 2 1 2                     a a i i a b i j a c i k b a j i b b j j ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1 2 1 2 1 2 1 2                 b c j k c a k i c b k j c c k k ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Since, ˆ ˆ ˆ ˆ ˆ ˆ i i j j k k 0       and ˆ ˆ ˆ i j k   , etc., we have  1 2 1 2 1 2 1 2 1 2 1 2      A B b c c b i c a a c j a b b a k        ˆ ˆ ˆ   or putting it in determinant form, we have 1 1 1 2 2 2 ˆ ˆ ˆ i j k A B a b c a b c     It may be noted that the scalar components of the first vector A  occupy the middle row of the determinant. CROSS PRODUCT BY COMPONENTS METHOD (i) If the vectors are expressed in terms of their rectangular components along X, Y and Z axes, the cross product yields the product in vector form i.e. in terms of its rectangular components again. (ii) If A  = Ax i + A y j + Az k and B  = Bx i + B y j + Bz k where Ax , Ay and Az are the components of the vector A  along X , Y and Z axis. Similarly, Bx , By and Bz are the components of the vector B  along X, Y and Z axis respectively And i, j and k. are the unit vectors along X , Y and Z axis respectively. then A B    = (Ax i + A y j + Az k) × (Bx i + B y j + Bz k) = Ax i × (Bx i + B y j + Bz k) + Ay j × (Bx i + B y j + Bz k) + Az k × (Bx i + B y j + Bz k) = Ax i × Bx i + Ax i × B y j + Ax i × Bz k + A y j × Bx i + A y j × B y j + A y j × Bz k + Az k × Bx i + Az k × B y j + Az k × Bz k = Ax × Bx (i × i) + Ax × B y (i × j) + Ax × Bz (i × k) + Ay × Bx (j × i) + A y × B y (j × j) + Ay × Bz (j × k) + Az × Bx (k × i) + Az × B y (k × j) + Az × Bz (k × k) = Ax × Bx (0) + Ax × B y (k) + Ax × Bz (– j) + Ay × Bx (– k) + Ay × B y (0) + Ay × Bz (i) + Az × Bx (j) + Az × B y (– i) + Az × Bz (0) = i (Ay Bz – Az B y ) – j (Ax Bz – Az Bx ) + k (Ax B y – A y Bx ) x y z x y z i j k A A A B B B 
Olympiad Class Work Book VIII – Physics (Vol – I) (iii)From the definition of the vector product A B    , we find A B A B sin         A B sin A B        i A B A B j A B A B + k A B A B  y z z y x z z x x y y x      A B        GEOMETRICAL INTERPRETATION OF CROSS PRODUCT Half of magnitude of cross product equals the area of triangle with adjacent sides A and B . Area           1 1 A Bsin ABsin 2 2  Area of triangle   1   A B 2 Area of parallelogram                 perpendicular distance Base A Bsin ABsin between parallel sides Also, Area of parallelogram         1 2 1 A B d d 2 A B B sin A B B sin NOTE(S): So, half of the modulus of cross product equals the area of the triangle with adjacent sides  A and  B and magnitude of cross product equals the area of the parallelogram with adjacent sides   Aand B KAMERLINGH ONNES (1853 - 1926) Leiden University, Leiden, the Netherlands “for his investigations on the properties of matter at low temperatures which led to the production of liquid helium’’
VIII – Physics (Vol – I) Olympiad Class Work Book CUQ 1. The direction of the cross product of two vectors can be known by applying. 1) Triangle law of vector addition. 2) Ampere's swimming rule. 3) Cork screw rule 4) All the above 2. The angle between the vectors (A x B) and (B A )  is 1) 0 0 2) 1800 3) 450 4) 900 3. If a b a b   . then the angle between a and b is 1) 900 2) 600 3) 00 4) 450 4. A vector A points vertically upward and B points towards south. Then the vector product A x B is 1) along west 2) along east 3) along north 4) vertically down ward 5. The magnitude of vector product of two perpendicular vectors P Q are 1) equal to PQ 2) less than PQ 3) equal to zero 4) Greater than PQ 6. If A,B and C are non-zero vectors, and if A B=0  and B×C=0 , value of A C is 1) unity 2) zero 3) 2 B 4) AC cosθ 7. In a clockwise system, which of the following is true. 1) j x k = i 2) i.i = 0 3) j x j=1 4) k.i = 1 8. Given that A and B are greater than one. The magnitude of ( ) A B   can not be 1) equal to AB 2) less than B 3) more than AB 4) equal to A/B 9. If A B B A        , then the angle between A  and B  is ............... 1)  2) 3  3) 2  4) 4  10. a b          + b a          = 1) 2 ab Sin  2) 2ab tan  3) Zero vector 4) 2ab cos 