Tiêu đề 4. VECTORS _WS-1 TO WS-7-KEY AND SOL_(91 TO 141).pdf ✅

VIII – Physics (Vol – I) 91 Narayana Group of Schools Olympiad Class Work Book VECTORS WORKSHEET-1 KEY CUQ 1. 2 2. 3 3. 2 4. 4 5. 3 6. 4 7. 4 8. 2 9. 1 10. 4 11 .4 12. 1 JEE MAIN & ADVANCED 1. 4 2. 2 3. 1 4. 4 5. 2 6. 3 7. 2 8. 2 9. 4 10.2 11. 4 12. 3 13. 1 14. 3 15. 2 16. 3 17. 3 18. 4 19. 1 20. 2 21. 1 22. 3, 4 23.     A A A 24. a-q; b-p,q; c-s; d-r,s 25. 3 26. 4 27. 1 28. 1,2,4 29. a-r,s; b-p,q;c-r,s;d-r,s 30. 4 31. 1 32. 3 SOLUTIONS 2. Coordinate of head of the vector     1 1 1 1 r x ,y ,z 2,1,0  coordinate of tail of the vector     2 2 2 2 r x ,y ,z 4,2, 3    Then magnitude of vector       2 2 2 2 1 2 1 2 1 r x x y y z z             2 2 2 r 4 2 2 1 3 0         ;       2 2 2 r 6 1 3      r 36 1 9    ; r 46  3. A 5i j 2k    ˆ ˆ ˆ   2 2 2 A 5 1 2     A 25 1 4    A 30  4. Three vectors with magnitude b units along the coordinate axes represent the adjacent sides of cube. Let r be the vector along diagonal from origin
Olympiad Class Work Book VIII – Physics (Vol – I) ̄r b b O b ˆ ˆ ˆ r bi bj bk      ˆ ˆ ˆ r b i j k    r 3b   unit vector along the diagonal   b i j k ˆ ˆ ˆ r rˆ r 3 b     ˆ ˆ ˆ i j k rˆ 3     5. Position of particle in coordinate system A 3,2,5   then position vector from origin is OA 3i 2j 5k    ˆ ˆ ˆ 6. A 3i 3j 9k B i aj 3k       ˆ ˆ ˆ ˆ ˆ ˆ if 1 2 3 1 2 3 a i a j a k and b i b j b k ˆ ˆ ˆ ˆ ˆ ˆ     are parallel vectors 1 2 3 1 2 3 a a a 3 3 9 b b b 1 a 3      3 9 9a 9 a 3      a 1 8. Magnitude of required vector B 10  required vector is opposite to A 6i 8j  ˆ ˆ  required vector B B A    ˆ A B 10 A           
VIII – Physics (Vol – I) Olympiad Class Work Book     2 2 6i 8j ˆ ˆ B 10 6 8         6i 8j ˆ ˆ B 10 10         B 6i 8j ˆ ˆ 9. Magnitude of position vector r=4 Slope 1 0 tan 30 3       The position vector ˆ ˆ r rcos i rsin j     0 0 ˆ ˆ r 4cos30 i 4sin30 j   3 1 ˆ ˆ r 4 i 4 j 2 2     ˆ ˆ r 2 3i 2j   ˆ ˆ    r 2 3i j     10. A 0.4i 0.3j ck ˆ ˆ ˆ    is a unit vector i.e. A 1 2 0.16 0.09 c 1    2 c 1 0.25   2 c 0.75 c 0.75 0.866     11.     x ,y ,z 1,2,3 1 1 1      x ,y ,z 5,4,2 2 2 2   displacement vector       2 1 2 1 2 1 ˆ ˆ ˆ r x x i y y j z z k             ˆ ˆ ˆ        r 5 1 i 4 2 j 2 3 k   ˆ ˆ ˆ r 4i 2j k m    12. Direction of car is along ˆ ˆ r 4i 3j    speed v = 10 m/s velocity of car r V v. r     2 2 4i 3j ˆ ˆ v 10 4 3     
Olympiad Class Work Book VIII – Physics (Vol – I)     4i 3j ˆ ˆ ˆ ˆ v 10 8i 6j m/s 3       13. Displacement from original position A is zero 40m 5 3 0 0m m 14. If ˆ ˆ ˆ r 0.2i 0.3j Zk    is a unit vector then its magnitude r 1     2 2 2 r 0.2 0.3 Z 1     2 0.04 0.09 Z 1    2 z 1 0.13 0.87    so the answer is none 15. P i 3j 7k Q 5i 2j 4k       ˆ ˆ ˆ ˆ ˆ ˆ       PQ Q P 5 1 i 2 3 j 4 7 k          ˆ ˆ ˆ PQ 4i 5j 11k    ˆ ˆ ˆ   2 2 2      PQ 4 5 11 PQ 16 25 121    PQ 162  16. 2i yj 3k ˆ ˆ ˆ   is parallel to 2i 3j 12k ˆ ˆ ˆ   then y 3 3 y 3 12 4      17. A 3i 4j  ˆ ˆ parallel vector B should be of the form B A    should be positive value   B 3i 4j   ˆ ˆ it 1 5  