Ch-3 Help-Session. CH-3-082 T082 Q7. Vector A has a magnitude of 40.0 cm and is directed 60.0 degrees above the negative x-axis. Vector B has magnitude.

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Presentation transcript:

Ch-3 Help-Session

CH T082 Q7. Vector A has a magnitude of 40.0 cm and is directed 60.0 degrees above the negative x-axis. Vector B has magnitude of 20.0 cm and is directed along the positive x- axis. Find the resultant vector (i and j are unit vectors along positive x and y axes, respectively). (Ans: 34.6 j cm) Q8. Consider two vectors A = (3 i + 4 j) cm and B = (- 4 i + 3 j) cm. Find the angle between these two vectors. (Ans: 90 degrees) C=A+B; A=  A  cos 120 i +  A  sin 120 j ; A= i j B= 20 i; C= A+B = (-20+20) i j = 34.6 j A.B=  A  B  cos  =A x B x +A y B y  = cos -1 [(A x B x +A y B y )/  A  B  ]  A  =  B  = 5  = cos -1 [(-12+12)/ 25]= cos-1(0)  = 90 degrees

CH-3-082/081 Q9. If vector A is added to vector B, the result is (6 i + 1j) m. If A is subtracted from B, the result is (- 4 i + 7 j) m. Find the magnitude of B. (Ans. 4 m.) T081 Q7. A boat travels a distance of 27.0 km 20.0 o North of East. It then travels 53.0 km in a direction 30.0 o East of North. What is the total distance traveled by the boat in the East direction? ( Ans: 51.9 km) B+A= 6 i + 1j ; B-A= - 4 i + 7 j ; Adding both equations together 2B= 2 i + 8j B=i+4j  B  =  ( ) = 4  A  = 27 km,  A = 20  B  =53 km;  B =60 C x =A x +B x =  A  cos 20 +  B  cos60 C x = = 51.9

CH Q8. Consider two vectors A and B with magnitudes 5 cm and 8 cm, respectively. Vector A is along the positive x-axis and vector B is along the positive y-axis. Find A· (A+B). (Ans: 25 cm2) Q9. Consider the following three vectors A= 5i+ 2j-5k, B = -2i+4j+3k, and C=i-2j-3k. Find the magnitude of the vector D= A – B + 2C. (Ans: 17.7) A=5i; B=8j ; A+B= 5i+8j A. (A+B) = 5x5 = 25 D= A-2B+3C =5i+2j-5k+2i-4j-3k+2i-4j-6k =9i-6j-14k |D|=  ( )=  313= 17.7

CH T072 Q5.Vectors a, b, and c are related through equations and A+B=C; A-B=5C. If C= 3i+4j, what is the magnitude of vector a ? (Ans: 15) Q6. Three vectors F,v and B are related through F = 5(vxB). If vector v=3i -5j and B=-2k, then vector F is: (Ans: 50 i+30 j ) A+B=C; A-B=5C ; Adding both equations together 2A=6C; A=3C A=3C=9i+12j  A  =  ( ) = 15 F = 5(vxB)=5[(3i -5j)x(-2k)] = 5[ -6i xk+10jxk)]=5[6j+10i] = 30 j+50 i

CH Q7. A vector A of magnitude 20 is added to a vector B of magnitude 25. The magnitude of the vector A+B can be: (Ans: 12) Q8. Vectors F and G are defined as F=3i+4j, and G=-i+j. Find the component (projection) of vector G along the direction of vector F.( Ans: 0.20)  A+B  max =25+20=45  A+B  min =25-20 =5 Value of  A+B  =5-45 F.G=  F  G  cos  =F x G x +F y G y  G  cos  = (F x G x +F y G y )/  F  = (-3+4)/  ( )=1/5 =0.2

CH T071 : : In Fig. 3, the unknown vector C is given by: (Ans:B-A) Q8. Two vectors are given by: P=-1.5i+2j,and Q=1.0j. The angle that the vector makes with the positive x-axis is: (A: 135°) A-C = B ; -C=B-A; C=A-B  =cos -1 [(P x Q x +P y Q y )/  P   Q  ] =cos -1 [(2x1)/(  ) x1)]

CH Q5. Q9.: A man walks 5.0 km due North, then 13 km 22.6° South of East, and then 12 km due West. The man is finally at: (Ans: where he started) Add components A=5j; B=13 km 22.6 degrees south of east C= -12i Resultant must be Zero