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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If `vecu=veca-vecb,vecv=veca+vecb and |veca|=|vecb|=2,` then `|vecuxxvecv|` is equal toA. `2sqrt(16-(a.b)^(2))`B. `sqrt(16-(a.b)^(2))`C. `2sqrt(4-(a.b)^(2))`D. `2sqrt(4+(a.b)^(2))` |
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Answer» Correct Answer - A Now, `|uxxv|=(a-b)xx(a+b)=2|=2||axxb|` `[because axxa =bxxb=0]` and `|axxb|^(2)+(a*b)^(2)=(ab sin theta)^(2)+(ab cos theta)^(2)=a^(2)b^(2)` `|axxb|=sqrt(a^(2)b^(2)-(a*b)^(2))` So`|uxxv|=2|axxb|=2sqrt(a^(2)b^(2)-(a*b)^(2))` `=2sqrt(2^(2)2^(2)-(a*b)^(2))` `=2sqrt(16-(a*b)^(2))[because |a|=|b|=2]` |
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| 52. |
A ship is sailing towardsthe north at a speed of 1.25 m/s. The current is taking it towards the eastat the rate of 1 m/s and a sailor is climbing a vertical pole on the ship atthe rate of 0.5 m/s. Find the velocity of the sailor in space. |
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Answer» Correct Answer - `1.677` m`//`sec. We take the unit vectors `hati, hatj and hatk` parallel to the east, north and vertically upwards in the direction of pole, respectively. Then the velocity vectors of the current, ship and the sailor are, respectively, `hati, 1.25hatj and 0.5hatk`. Velocity `vecv` of the sailor in space is the resultant of these vectors. Hence, `vecv= hati+ 1.25hatj+ 0.5 hatk` Then `|vecv|= sqrt(1+ (1.25)^(2) + (0.5)^(2))` `" "=sqrt(1+ 1.5625 + .25)` `" " = sqrt(2.8125)` = 1.677 m/s |
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| 53. |
Let `veca,vecb,vecc` be three unit vectors such that `3veca+4vecb+5vecc=vec0`. Then which of the following statements is true? (A) `veca` is parrallel to vecb` (B) `veca` is perpendicular to vecb` (C) `veca` is neither parralel nor perpendicular to `vecb` (D) `veca,vecb,vecc` are copalanarA. `veca` is parallel to `vecb`B. `veca` is perpendicular to `vecb`C. `veca` is neither parallel nor perpendicular to `vecb`D. none of these |
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Answer» Correct Answer - D `3veca + 4vecb + 5vecc =0` Hence `veca, vecb and vecc` are collinear. No other conclusion can be derived from it. |
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| 54. |
If a, b, c are linearly independent vectors and `delta =|(a,b,c),(a.a,a.b,a.c),(a.c,b.c,c.c)|` then (a)`delta = 0` (b) `delta = 1` (C) A=any non-zero volue (d)None of theseA. 1B. 0C. `-1`D. None of these |
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Answer» Correct Answer - B Since a, b and c are coplanar, there must exist three scalars x,y and z not all zero such that `ax+yb+zc=0" "…(i)` ON multiplyaing both sids of Eq (i) by a and b respectively we get `xaa+ya.b+za.c =0:" "…(ii)` `ab.a+yb.b+zb.c=0" "...(iii)` On eliminating x,y and z from Eqs, (i), (ii) and (iii), we get `|{:(a,b,c),(a.a., a.b.,b.c),(b.a, a.b, b.c):}|=0` |
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| 55. |
Four non zero vectors willalways bea. linearly dependent b. linearly independentc. either a or b d. none of theseA. linearly dependentB. linearly independentC. either a or bD. none of these |
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Answer» Correct Answer - A Four or more than four non-zero vectors are always linearly dependent. |
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| 56. |
`ABCDEF` is a regular hexagon, Fig. 2 (c ) .65. What is the value of ` (vec (AB) + vec (AC) + vec (AD) + vec (AE) + vec (AF) ?` .A. `vec(AO)`B. `2vec(AO)`C. `4vec(AO)`D. `6vec(AO)` |
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Answer» Correct Answer - D |
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| 57. |
In going from one city to another, a car travles 75`km` north, 60`km` noth-west and 20`km` east,The magnitude of displacement between the two cities is `(take1//sqrt(2)=0.7)`A. 170`km`B. 137`km`C. 119`km`D. 140`km` |
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Answer» Correct Answer - C `vec(S)=75hat(j)+[60cos 45^(@)hat(j)-60sin 45^(@)hat(i)]+20hat(i)` `(20-60xx0.7)hat(i)+(60xx0.7+75)hat(j)=-22hat(i)+117hat(j)` `S=sqrt(22^(2)+117^(2))=sqrt(484+13686)=sqrt(14173)=119km` |
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| 58. |
A car travles 6km towards north at an angle of `45^(@)` to the east and then travles distance of 4km towards north at an angle of `135^(@)` to east (figure). How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east? |
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Answer» Net movement along x-direction, `S_(x)=(6-4) cos 45^(@) hat(i)=2xx1/sqrt(2)=sqrt(2)km` Net movement along y-direction, `S_(y)=(6+4) sin 45^(@) hat(j)=10xx1/sqrt(2)=5sqrt(2)km` Net movement from starting point, `|vec(s)|=sqrt(s_(a)^(2)+s_(y)^(2))=sqrt((sqrt(2))+(5sqrt(2))^(2))=sqrt(52) km` Angle which makes with the east direction, `tan theta=(Y-compon ent)/(X-compon ent)=(5sqrt(2))/sqrt(2) :. theta=tan^(-1)(5)` |
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| 59. |
A car travles 6km towards north at an angle of `45^(@)` to the east and then travles distance of 4km towards north at an angle of `135^(@)` to east (figure). How far is the point from the starting point? What angle does the straight line joining its initial and final position makes with the east? A. `sqrt(50)km` and `tan^(-1)(5)`B. `10 km` and `tan^(-1)(5)`C. `sqrt(52)km` and `tan^(-1)(5)`D. `sqrt(52)km` and `tan^(-1)(sqrt(5))` |
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Answer» Correct Answer - A |
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| 60. |
If ` vec a , vec b , vec c`are mutually perpenedicularvectors of equal magnitudes, show thatthe vector ` vec a+ vec b+ vec c`is equally inclined to ` vec a , vec b ,a n d vec cdot` |
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Answer» Let `|vec(a)|=|vec(b)| = |vec(c )|=a` Given that, `vec(a).vec(b) = vec(b).vec(c) = vec(c).vec(a) =0` Now `(vec(a) + vec(b) + vec(c ))^(2)` `=|vec(a)|^(2) + |vec(b)|^(2) + |vec(c )|^(2) + 2(vec(a).vec(b) + vec(b).vec(c ) + vec( c) .vec(a))` `=a^(2) + a^(2) +a ^(2) +0=3a^(2)` `implies |vec(a)+vec(b)+vec(c)|=sqrt(3)a` Let `theta` be the angle between `vec(a)` and `(vec(a) + vec(b) + vec(c ))` `therefore vec(a). (vec(a) + vec(b) + vec(c )) = |vec(a)||vec(a)+vec(b)+vec(c )|costheta_(1)` `implies vec(a) .vec(a)+vec(a).vec(b) + vec(a).vec(c ) = a(asqrt(3))costheta_(1)` `implies a^(2) + 0 + 0 = a^(2) sqrt(3) cos theta_(1)` `implies cos theta_(1) = (a^(2))/(a^(2)sqrt(3)) = (1)/(sqrt(3))` `implies theta_(1) = cos^(-1)((1)/(sqrt(3)))` Similarly, angle between `vec(b)` and `vec(a) + vec(b) + vec(c) = cos^(-1) .(1)/(sqrt(3))` and angle between `vec(c ) and vec(a) + vec(b) + vec(c ) = cos^(-1).(1)/(sqrt(3))`. Therefore, `vec(a) + vec(b) + vec(c )` makes equal angles with the vectors `vec(a),vec(b)` and `vec(c )`. Hence Proved. |
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| 61. |
If vectors `vec(A),vec(B)` and `vec(C)` have magnitudes 8,15 and 17 units and `vec(A) +vec(B) = vec(C)`, find the angle between `vec(A)` and `vec(B)`. |
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Answer» Correct Answer - `90^(@)` |
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| 62. |
Find magnitude of A-2 B 3 C, where, `A = 2hati+3hatj, B = hati + hatj and c =hatk.` |
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Answer» Correct Answer - A `A -2B + 3C = (2hati+ 3hatj) - 2(hati+hatj) + 3hatk` `= (hatj+3hatk)` `:. |A - 2B + 3C| = sqrt((1)^2+(3)^2)` `=sqrt(10) units.` |
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| 63. |
`|vec(A)xxvec(B)|^(2)+|vec(A).vec(B)|^(2)=`A. ZeroB. `A^(2)B^(2)`C. `AB`D. `sqrt(AB)` |
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Answer» Correct Answer - B Let `theta` be angle between Vectors `vec(A)` and `vec(B)` `:. |vec(A)xxvec(B)|=AB sin theta` and `|vec(A).vec(B)|= AB cos theta` `|vec(A)xxvec(B)|^(2)= |vec(A).vec(B)|^(2)=(AB sin theta)^(2)+(AB cos theta)^(2)` `=A^(2)B^(2)sin^(2)theta+A^(2)B^(2)cos^(2)theta= A^(2)B^(2)` |
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| 64. |
If `vec(A)xxvec(B)= vec(C )+vec(D)`, then select the correct alternative.A. `vec(B)` is parallel to `vec(C )+vec(D)`B. `vec(A)` is perpendicualr to `vec(C )`.C. Components of `vec(C )` along `vec(A)`= component of `vec(D)` along `vec(A)`D. Component of `vec(C )` along `vec(A)`= -component of `vec(D)` along `vec(A)` |
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Answer» Correct Answer - D According to defination of cross- product `vec(C )+vec(D)` is perpendicualr to both `vec(A)` and `vec(B)`. i.e., `vec(A).(vec(C )+vec(D))=0` or `vec(A).vec(C )+vec(A).vec(D)=0` or A(component of `vec(C )` along `vec(A)`)+A(component of `vec(D)` along `vec(A))=0` or component of `vec(C )` along `vec(A)`= -component of `vec(D)` along `vec(A)` |
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| 65. |
Find the (i) Scalar component and (ii)vector component of `A=3hati+4hatj+5hatk on B=hati+hatj+hatk.` |
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Answer» (i) Scalar component of a A along B is `Acostheta=A.hatB=(3hati+4hatj+5hatk).((hati+hatj+hatk))/(sqrt(3))` ` =(3+4+5)/(sqrt(3))=(12)/(sqrt(3))=4sqrt(3)` (ii) Vector component of A along B is `(Acostheta)hatB=(A.hatB)hatB` `(4sqrt(3))((hati+hatj+hatk))/(sqrt(3))=4hati+ 4hatj+4hatk` |
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| 66. |
If ` vec a= hat i+ hat j+ hat k a n d`` vec b= hat j- hat k ,`find avector ` vec c`such that ` vec a "x" vec c= vec b`and ` vec adot vec c=3` |
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Answer» Here`vec(a)=hat(i) + hat(j) + hat(k) and vec(b) = hat(j)-hat(k)`. `Let vec(c)= c_(1) hat(i) + c_(2) hat(j) + c_(3)hat(k).` Then, `vec(a) * vec(c) = 3 and vec(a) xx vec(c) = vec(b)` ` rArr (hat(i) + hat(j) +hat(k))* (c_(1) hat(i) + c_(2) hat(j) + c_(3)hat(k))=3` `and (hat(i) + hat(j) + hat(k))xx ( c_(1) hat(i) + c_(2) hat(j) + c_(3) hat(k)) = ( hat(j) - hat(k))` `rArr c_(1) + c_(2) + C_(3) " "...(i) and |(hat(i),hat(j),hat(k)),(1,1,1),(c_(1), c_(2) , c_(3))|=(hat(j)-hat(k))`...(ii) Now, (ii) gives : `(c_(3) - c_(2) hat(i)- ( c_(3)-c_(1)) hat(j) + ( c_(2) - c_(1)) hat(k)= (hat(j) - hat(k))` `rArr c_(3)- c_(2) = 0, c_(1)-c_(3) = 1 and c_(2) - c_(1) =-1` Putting `C_(3) = c_(2) " in (i), we get " c_(1) + 2 c_(2) = 3.` On solving `c_(1)+ 2 c_(2) = 3 and c_(1) - c_(2) = 1," we get " c_(2) = 2/3 and c_(1)=5/3`. `:. c_(1)= 5/3, c_(2) = 2/3 and c_(3) = 2/3`. Hence, `vec(c) = 5/3 hat(i) + 2/3hat(j) + 2/3 hat(k) rArr vec(c) = 1/3 (5 hat(i) + 2 hat(j) +2hat(k)).` |
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| 67. |
If `|vec(a)| = 2, |vec(b)|= 7 and (vec(a) xx vec(b))=(3 hat(i) + 2 hat(j) + 6 hat(k)), " find the angle between " vec(a) and vec(b).` |
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Answer» Correct Answer - `pi/6` `vec(a) xx vec(b)= (3 hat(i) + 2 hat(j)+ 6 hat(k))` `rArr|vec(a) xxvec(b)|= sqrt(3^(2)+2^(2)+6^(2))=7 rArr|vec(a)||vec(b)| sin theta = 7` `rArr sin theta= 7/(|vec(a)||vec(b)|)=7/((2 xx7))=1/2 rArr theta = 30^(@).` |
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| 68. |
If `vec(a) = (hat(i) - 2 hat(j)+ 3 hat(k)) and vec(b) = (2 hat(i) + 3 hat(j)- 5 hat(k))` then find `(vec(a) xx vec(b)) and " verify that " (vec(a) xx vec(b))` is perpendicular to each one of` vec(a) and vec(b)`. |
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Answer» We have `(vec(a) xx vec(b))= |(hat(i), hat(j), hat(k)),(1,-2,3),(2,3,-5)|` `=(10-9) hat(i)-(-5-6)hat(j) + (3+4) hat(k) = ( hat(i) + 11 hat(j) + 7 hat(k))`. Now, `(vec(a) xx vec(b) ) * vec(a)= (hat(i) + 11 hat(j) + 7 hat(j))*(hat(i) - 2hat(j) + 3 hat(k)).` `:. (vec(a) xx vec(b))*vec(a) = (1-22+21)=0` `:. (vec(a) xxvec(b)) bot vec(a).` And,`(vec(a) xx vec(b))* vec(b)=(hat(i) +11vec(j)+7vec(k))*(2hat(i)+3 hat(j)-5hat(k))` `=(2+33-35)=0`. `:. (vec(a)xxvec(b))bot vec(b)`. |
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| 69. |
If `vec(a) = 4 hat(i) + 3 hat(j) + 2 hat(k) and vec(b)= 3 hat(i)+ 2 hat(k), "find" |vec(b) xx 2 vec(a)|.` |
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Answer» We have `vec(b) = ( 3 hat (i) + 2 hat(k) ) and 2 vec(a) = ( 8 hat(i) + 6 hat(j) + 4 hat(k))` `:. ( vec(b) xx2 vec(a))=|(hat(i),hat(j), hat(k)),(3,0,2),(8,6,4)|` `= (0-12) hat(i) + (16 - 12) hat(j) + (18-0) hat(k)` `= (-12hat(i) + 4 hat(j) + 18 hat(k)).` `:. |vec(b) xx 2 vec(a)| = |-12 hat(i) + 4 hat(j) + 18 hat(k)|` ` = sqrt((-12)^(2) + 4^(2) + (18)^(2))=sqrt(484 )= 22.` Hence,`| vec(b) xx 2 vec(a)| = 22.` |
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| 70. |
Find a unit vector perpendicular to each of thevectors `( -> a+ -> b)`and `( -> a- -> b)`, where ` -> a= hat i+ hat j+ hat k , -> b= hat i+2 hat j+3 hat k`. |
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Answer» We have `(vec(a)+vec(b)) = (hat(i) + hat(j) + hat(k)) + (hat(i) + 2 hat(j) + 3 hat(k))=(2hat(i) + 3 hat(j) + 4 hat(k)) and` `(vec(a) - vec(b)) = (hat(i) + hat(j) + hat(k))- (hat(i) + 2 hat(j) + 3 hat(k)) = (- hat(j) - 2 hat(k)).` `:. (vec(a) + vec(b)) xx (vec(a) - vec(b))= |(hat(i),hat(j),hat(k)),(2,3,4),(0,-1,-2)|` `=(-6+4)hat(i)-(-4-0)hat(j) + (-2-0)hat(k)` `=(-2hat(i) + 4 hat(j)-2 hat(k)).` `:. |(vec(a) + vec(b)) xx(vec(a)-vec(b))|= sqrt((-2)^(2)+4^(2)+(-2)^(2))=sqrt((24)) = 2sqrt(6).` So, the vectors of magnitude 5 units and perpendicular to each the vectors `(vec(a)+vec(b)) and (vec(a)-vec(b))` are: `pm(5{(vec(a) + vec(b)) xx(vec(a)- vec(b))})/(|(vec(a) + vec(b)) xx (vec(a) -vec(b))|)=pm (5(-2hat(i) +4hat(j) -2 hat(k)))/(2sqrt(6))=pm (5(-hat(i) + 2 hat(j) - hat(k)))/sqrt(6).` |
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| 71. |
Find a vector of magnitude 15, which is perpendicular to both the vectors `(4hat(i) -hat(j)+8hat(k)) and (-hat(j)+hat(k)).` |
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Answer» Let `vec(a) = (4 hat(i)-hat(j) + 8 hat(k) ) and vec(b)= (-hat(j) + hat(k)).` A unit vector perpendicular to both `vec(a) and vec(b) = ((vec(a) xxvec(b)))/(|vec(a) xx vec(b)|`. Now, `vec(a) xx vec(b) = |{:(hat(i),hat(j),hat(k)),(4,-1,8),(0,-1,1):}|` `=(-1+8) hat(i) - ( 4-0) hat(j) + (-4-0) hat(k)` `=(7 hat(i) - 4 hat(j) - 4 hat(k)).` `:. |vec(a) xxvec(b)| = sqrt(7^(2) + (-4)^(2)+(-4)^(2)) =sqrt(81)=9.` So, a unit vector perpendicular to both `vec(a) and vec(b)`. `((vec(a) xxvec(b)))/|(vec(a) xx vec(b))| = (7 hat(i) -4 hat(j)- 4 hat(k))/9.` The required vector `=(15(7hat(i)-4hat(j)-4hat(k)))/9 =5/3(7 hat(i) - 4 hat(j) - 4 hat(k)).` |
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| 72. |
Find a unit vector perpendicular to the plane ABC, where thecoordinates of `A , B , a n d C`are `A(3, -1, 2), B(1, -1, -3)a n d C(4, -3, 1)dot` |
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Answer» Correct Answer - `(-hat(i) - 2 hat(j) + 4 hat(k))/sqrt(21)` Required vector `=((vec(AB) xxvec(AC)))/(|vec(AB) xx vec(AC)|).` |
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| 73. |
Find the area of the trinagle whose two adjacent sides are determined by the vectors (i) `vec(a)= -2hat(i)-5 hat(k)`and `vec(b)=hat(i)- 2 hat(j)- hat(k)` (ii) `vec(a)= 3 hat(i)+ 4 hat(j)` and `vec(b)=-5hat(i)+7 hat(j).` |
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Answer» Correct Answer - (i) `1/2 sqrt(165)` sq units (ii) `41/2 `sq units Area of a triangle `=1/2|vec(a) xx vec(b)| " where "vec(a) and vec(b)` are adjacent sides. |
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| 74. |
Find the sine of the angle between the vectors `vec(a) =(2 hat(i) - hat(j) + 3 hat(k)) and vec(b) = (hat(i) + 3 hat(j) + 2 hat(k)).` |
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Answer» We have `(vec(a) xxvec(b))= |(hat(i),hat(j),hat(k)),(2,-1,3),(1,3,2)|` `= (-2 -9) hat(i) - (4-3)hat(j) + ( 6+1)hat(k)` `=(-11hat(i) - hat(j) + 7 hat(k)).` `|vec(a) xx vec(b)|=sqrt((-11)^(2)+ (-1)^(2) + 7^(2))=sqrt(171) = 3 sqrt(19)` `|vec(a)|= sqrt(2^(2)+ (-1)^(2)+ 3^(2))=sqrt(14) ,` `|vec(b)|=sqrt(1^(2)+3^(2) + 2^(2)) =sqrt(14).` Let`theta" be the angle between "vec(a) and vec(b).` Then, `sin theta =(|vec(a) xx vec(b)|)/(|vec(a)| |vec(b)|) =(3sqrt(19))/((sqrt(14)) (sqrt14))=3/14 sqrt(19).` |
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| 75. |
What is the value of linear velocity, if `vecomega=3hati-4hatj+hatk` and `vecr=5hati-6hatj+6hatk`? |
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Answer» Correct Answer - D (d ) Velocity `v=omegaxxr|underset(5" "-6" "6)overset(hati" "hatj" "hatk)(3""-4" "1)|=-18hati -13hatj+2hatk` |
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| 76. |
AT what angle two forces (P +Q) and (P-Q) act so that resultant is (i) `sqrt(3 P^2 + Q^2)` (ii) `sqrt(2 (P^2 + Q^2))` (iii) ` sqrt(P^2 + Q^2)` . |
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Answer» Correct Answer - `60^(@)` |
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| 77. |
The following sets of three vectors act on a body. Whose resultant cannot be zero ?A. `10,10,10`B. `10,10,20`C. `10,20,23`D. `10,20,40` |
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Answer» Correct Answer - D |
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| 78. |
At what angle should the two forces vectors 2F and `sqrt(2)` F act so that the resultant force is `sqrt(10)F` |
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Answer» `A=2F,B=sqrt(2) F,R=sqrt(10)F` `R^(2)=A^(2)+B^(2)=2AB cos theta` `rArr 10F^(2)=4F^(2)+2(2F)sqrt(2)F cos theta` `rArr cos theta=1//sqrt(2)rArr theta =45^(@)` |
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| 79. |
The minimum number of vectors having different planes which can be added to give zero resultant isA. `2`B. `3`C. `4`D. `5` |
| Answer» Correct Answer - C | |
| 80. |
The following sets of three vectors act on a body, whose resultant can be zero. These are :A. `10 , 10 , 10`B. `10 , 10 , 20`C. `10 , 20 , 23`D. `10 , 20 , 40` |
| Answer» Correct Answer - D | |
| 81. |
The resultant of `vecA and vecB` makes an angle `alpha` with `vecA` and `beta and vecB`,A. `alpha lt beta if A gt B`B. `alpha lt beta if A lt B`C. `alpha = beta if A lt B`D. `alpha lt beta if A = B` |
| Answer» Correct Answer - A | |
| 82. |
A vector coplanar with vectores ` hati + hatj and hatj + hatk ` and parallel to the vector ` 2 hati - 2 hatj - 4 hatk `isA. `hati - hatk`B. `hati - hatj -2hatk `C. `hati +hatj-hatk `D. `3hati+3hatj-6hatk` |
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Answer» Correct Answer - B Any vector coplanar with vectors ` hati + hatj and hatj + hatk ` is `a=x (hati +hatj )+y (hatj +hatk)` `implies a=xhati +(x+y)hatj +yhatk` It is given that a is paraallel to ` 2 hati - 2 hatj - 4 hatk ` ` therefore a= lamda ( 2 hati - 2hatj - 4 hatk )` for some scalar ` lamda `. `implies [ x hati +(x+y) hatj +yhatk ]= lamda (2 hati - 2hatj - 4 hatk)` `implies x= 2 lamda , x+y=- 2 lamda and y=- 4 lamda ` `implies x=2lamda and y=-4 lamda ` `therefore a= 2 lamda ( hati - hatj - 2hatk )` , where `lamda in R.` |
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| 83. |
For the figure shown .A. `vecA+vecB=vecC`B. `vecB+vecC=vecA`C. `vecC+vecA=vecB`D. `vecA+vecB+vecC=0` |
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Answer» Correct Answer - C |
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| 84. |
Let `vecC=vecA+vecB`A. `|vecC|` is always greater then `|vecA|`B. It is possible to have `|vecC|lt|vecA|` and `|vecC|lt|vecB|`C. `C` is always equal to `A+B`D. `C` is never equal to `A+B` |
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Answer» Correct Answer - B |
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| 85. |
Three vector `vec(A)`,`vec(B)`, `vec(C )` satisfy the relation `vec(A)*vec(B)=0`and `vec(A).vec(C )=0`. The vector `vec(A)` is parallel toA. `vec(B)`B. `vec(C )`C. `vec(B)xxvec(C )`D. `vec(B)xxvec(C )` |
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Answer» Correct Answer - C If `vec(A).vec(B)=0,vec(A)` is perpendicular to `vec(B)`. If `vec(A).vec(C )=0,vec(A)` is perpendicular to `vec(C )`. So `vec(A)` is perpendicular to both `vec(B)` and `vec(C )`. Also `vec(B)xxvec(C )` is perpendicular to both `vec(B)` and `vec(C )`. Hence `vec(A)` is parallel to `vec(B)xxvec(C )` |
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| 86. |
Three vector `vec(A)`,`vec(B)`, `vec(C )` satisfy the relation `vec(A)*vec(B)=0`and `vec(A).vec(C )=0`. The vector `vec(A)` is parallel toA. `vec(B)`B. `vec(C )`C. `vec(B) . vec(C )`D. `vec(B) xx vec(C )` |
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Answer» Correct Answer - D `vec(A) . vec(B) = 0 , vec(A) . vec(C ) = 0` `vec(A) is _|_^(ar) to vec(B)` as well as `vec(C )`. |
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| 87. |
Three vector `vec(A)`,`vec(B)`, `vec(C )` satisfy the relation `vec(A)*vec(B)=0`and `vec(A).vec(C )=0`. The vector `vec(A)` is parallel toA. `vec(B)`B. `vec(C )`C. `vec(B).vec(C )`D. `vec(B)xxvec(C )` |
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Answer» Correct Answer - D `vec(A).vec(B)=0`(given) `rArr vec(A)_|_vec(B)` `vec(A).vec(C )=0` (given) `rArr vec(A)_|_vec(C )` `vec(A)` is perpendicular to both `vec(B)` and `vec(C )`. We know from the defination of cross product that `vec(B)xxvec(C )` is perpendicular to both `vec(B)` and `vec(C )`. So `vec(A)` is parallel to `vec(B)xxvec(C )`. |
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| 88. |
Three vector `vec(A)`,`vec(B)`, `vec(C )` satisfy the relation `vec(A)*vec(B)=0`and `vec(A).vec(C )=0`. The vector `vec(A)` is parallel toA. `vec(b)`B. `vec(c)`C. `vec(b).vec(c )`D. `vec(b)xxvec(c )` |
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Answer» Correct Answer - D `vec(a).vec(b)=0` i.e., `vec(a)` and `vec(b)` will be perpendicular to each other `vec(a).vec(c )=0 i.e., vec(a)` and `vec(c )` will be perpendicular to each other. `vec(b)xxvec(c )` will be a vector perpendicular to both `vec(b)` and `vec(c )` So `vec(a)` is parallel to `vec(b)xxvec(c )` |
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| 89. |
Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0. |
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Answer» The Cartesian equation of the plane passing through (x1, y1, z1), the direction ratios of whose normal are a, b, c, is a(x – x1) + b(y – y1) + c(z – z1) = 0 ∴ the cartesian equation of the required plane is o(x + 1) + 2(y + 2) + 5(z – 3) = 0 i.e. 0 + 2y – 4 + 10z – 15 = 0 i.e. y + 2 = 0. |
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| 90. |
Find the Cartesian equation of the plane passing through A( -1, 2, 3), the direction ratios of whose normal are 0, 2, 5. |
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Answer» The cartesian equation of the plane passing ; through (x1, y1, z1), the direction ratios of whose normal are a, b, c, is a(x – x1) + b(y – y1) + c(z – z1) = 0 ∴ the cartesian equation of the required plane is 0(x +1) + 2(y – 2) + 5(z – 3) = 0 i.e. 0 + 2y – 4 + 5z – 15 = 0 i.e. 2y + 5z = 19. |
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| 91. |
The magnitude of the vectors product of two vectors `|vecA| and |vecB|` may beA. greater than ABB. equal to ABC. less than ABD. equal to zero |
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Answer» Correct Answer - B::C::D `|vecA xx vecB| = AB sin theta` |
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| 92. |
The angle made by the vector `vecA=2hati+3hatj` with Y-axis isA. `tan^(-1)3//2`B. `tan^(-1)2//3`C. `sin^(-1)2//3`D. `cos^(-1)2//3` |
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Answer» Correct Answer - B |
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| 93. |
A particle whose speed is `50ms^(-1)` moves along the line from `A(2,1)` to `B(9,25)`. Find its velocity vector in the from of `ahat(i)+bhat(j)`. |
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Answer» Correct Answer - 25 `vec(V)=50.vec(AB) vec(AB)=(9-2)hat(i)+(25-1)hat(j)` `50/25(7hat(i)+24hat(j)) |vec(AB)|=sqrt((7)^(2)+(24)^(2))=14hat(i)+48hat(j)=25` |
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| 94. |
A particle travels with speed `50ms^(-1)` from the point `(3,-7)` in a direction `7hat(i)-24(j)`. Find its position vector after `3s`. |
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Answer» Correct Answer - `45hat(i)-151hat(j)` Speed= `50ms^(-1)` `S= X_(0)+V_(0)t` `=(3hat(i)-7hat(j))+3xx50(7hat(i)-24hat(j))/sqrt((24)^(2)+(7)^(2))` `=3 hat(i)-7hat(j)+(150)/25(7hat(i)-24hat(j))` `=3hat(i)-7hat(j)+42hat(i)-144hat(j)=45hat(i)-151hat(j)` |
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| 95. |
A particle whose speed is `50ms^(-1)` moves along the line from `A(2,1)` to `B(9,25)`. Find its velocity vector in the from of `ahat(i)+bhat(j)`.A. `(7hat(i)+24hat(j)) m//s`B. `2(7hat(i)+24hat(j))m//s`C. `4(7hat(i)+24hat(j))m//s`D. `5(7hat(i)+24hat(j))m//s` |
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Answer» `vec(V)= 50(hat(A)B)` `vec(AB)= (9-2)hat(i)+(25-1)hat(j)=(7hat(i)+24hat(j))` `|vec(AB)|= sqrt((7^(2))+(24)^(2))` `hat(AB)= (7hat(i)+24hat(j))/sqrt((7^(2))+(24^(2)))=((7hat(i)+24hat(j)))/(25)` `vec(V)= 50.((7hat(i)+24hat(j)))/(25)= 2(7hat(i)+24hat(j))m//s` |
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| 96. |
The projection of the vector `vec(A)= hat(i)-2hat(j)+hat(k)` on the vector `vec(B)= 4hat(i)-4hat(j)+7hat(k)` isA. `(19)/9`B. `(38)/9`C. `8/9`D. `4/9` |
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Answer» Correct Answer - A Here, `vec(A)= hat(i)-2hat(j)+hat(k)` `vec(B)= 4hat(i)-4hat(j)+7hat(k)` The projection of `vec(A)` on `vec(B)= A cos theta` `A cos theta= (AB cos theta)/B= (vec(A).vec(B))/B` `=((hat(i)-2hat(j)+hat(k)).(4hat(i)-4hat(j)+7hat(k)))/(sqrt((4)^(2)+(-4)^(2)+(7)^(2)))` `=(4+8+7)/(sqrt(18))=(19)/9` |
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| 97. |
Calculate the are of the triangle determined by the two vectors `vec(A)=3hat(i)+4hat(j)` and `vec(B)=-3hat(i)+7hat(j).` |
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Answer» We know thet the half of magnitude of the cross product of two vectors gives the area of the triangle. `vec(A)xxvec(B)=|(hat(i), hat(j), hat(k)) ,(3,4,0), (-3 ,7 ,0)|` `=hat(i)(0-0)-hat(j)(0-0)+hat(k)(21+12)=33hat(k)` Taking magnitude `|vec(A)xxvec(B)|=sqrt(33^(2))=33`. So area of triangle `=1/2|vec(A)xxvec(B)|=33/2sq.`unit. |
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| 98. |
If `A=3hat(i)+4hat(j)` and `B=7hat(i)+24hat(j)`,find the vector having the same magnitude as B and parallel to A. |
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Answer» `|B|=sqrt(7^(2)+(24)^(2))=sqrt(625)=25` Unit vector in the direction of A will be `hat(A)=(3hat(i)+4hat(j))/5` So required vector `=25((3hat(i)+4hat(j))/(5))=15hat(i)+20hat(j)` |
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| 99. |
A particle travels with speed `50ms^(-1)` from the point `(3,-7)` in a direction `7hat(i)-24(j)`. Find its position vector after `3s`.A. `(45hat(i)-125hat(j))m`B. `(45hat(i)-151hat(j))m`C. `(45hat(i)-125hat(j))m`D. `(35hat(i)-115hat(j))m` |
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Answer» Correct Answer - B Given speed `v= 50 m//s` in the direction `vec(a)= 7hat(i)-24hat(j)` `vec(v)= vhat(a)` is the unit vector in the direction `7hat(i)-24hat(j)` `hat(a)= ((7hat(i)-24hat(j)))/(sqrt((24)^(2)+(7)^(2)))=((7hat(i)-24hat(j)))/(25)` `vec(v)= vhat(a)=50.((7hat(i)-24hat(j)))/(25)= 2(7hat(i)-24hat(j))m//s` Initially the particle is at `vec(r)_(0)= (3hat(i)-7hat(j))m` Position of the particle after 3 sec, `vec(r )= vec(r )_(0)+vec(v)t` `implies vec(r )= (3hat(i)-7hat(j))+3xx2(7hat(i)-24hat(j))= (45hat(i)-151hat(j))m` |
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| 100. |
If `A=3hat(i)+4hat(j)` and `B=7hat(i)+24hat(j)`,find the vector having the same magnitude as B and parallel to A.A. `5hat(i)+20hat(j)`B. `15hat(i)+10hat(j)`C. `20hat(i)+15hat(j)`D. `15hat(i)+20hat(j)` |
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Answer» Correct Answer - D `|vec(B)|= sqrt(7^(2)+(24)^(2))= sqrt(625)= 25` Unit vector in the direction of A will be `hat(A)= (3hat(i)+4hat(j))/5` So required vector `= 25((3hat(i)+4hat(j))/5) = 15hat(j)+20hat(j)` |
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