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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. |
Assertion: If I is the incentre of `/_ABC, then`|vec(BC)|vec(IA)+|vec(CA)|vec(IB)+|vec(AB)|vec(IC)=0` Reason: If O is the origin, then the position vector of centroid of `/_ABC` is (vecOA)+vec(OB)+vec(OC))/3` (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.A. Statement-II and statement II ar correct and Statement III is the correct explanation of statement IB. Both statement I and statement II are correct but statement II is not the correct explanation of statement IC. Statement I is correct but statement II is incorrectD. Statement II is correct but statement I is incorrect |
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Answer» Correct Answer - B We know that, `OI=(|CB|OA+|CA|OB+|AB|OC)/(|BC|+|CA|+|AB|)` and `OG=(OA+OB+OC)/(3)` |
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| 52. |
Statement 1: If `| vec a+ vec b|=| vec a- vec b|,t h e n vec aa n d vec b`are perpendicularto each other.Statement 2: If thediagonal of a parallelogram are equal magnitude, then the parallelogram is arectangle.A. Statement-II and statement II ar correct and Statement III is the correct explanation of statement IB. Both statement I and statement II are correct but statement II is not the correct explanation of statement IC. Statement I is correct but statement II is incorrectD. Statement II is correct but statement I is incorrect |
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Answer» Correct Answer - A a+b=a-b are diagonals of a parallelogram whose sides are a and b. `|a+b|=|a-b|` Thus, diagonals of the parallelogram have the same length. So, the parallelogram is rectangle, i.e., `a bot b`. |
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| 53. |
If points `hati+hatj, hati -hatj and p hati +qhatj+rhatk` are collinear, thenA. p=1B. r=0C. `q in R`D. `q ne 1` |
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Answer» Correct Answer - A::B::D Points `A(hati+hatj),B(hati-hatj) and C(phati+qhatj+rhatk)` are collinear Now, `AB=-2hatj` and `BC=(p-1)hati+(q+1)hatj+rhatk` Vectors AB and BC must be collinear `implies p=1,r=0 and q ne -1` |
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| 54. |
If a+b is along the angle bisector of a and b, where `|a|=lamda|b|`, then the number of digits in value of `lamda` is |
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Answer» Correct Answer - 1 Since, angle bisector of a and b `impliesh(hata+hatb)=h((a)/(|a|)+(b)/(|b|))` . . . (i) Given, a+b is alongg angle bisector `implies mu((a)/(|a|)+(b)/(|b|))=a+b` only, when `|a|=|b|=mu` `therefore|a|=|b| implies lamda=1`. |
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| 55. |
The vector ` vec c `, directed along the internal bisector of the angle between the vectors `vec c = 7 hati - 4 hatj - 4hatk and vecb = -2hati - hatj + 2 hatk " with " |vec c| = 5 sqrt(6),` isA. `(5)/(3)(hati-7hatj+2hatk)`B. `(5)/(3)(5hati+5hatj+2hatk)`C. `(5)/(3)(hati+7hatj+2hatk)`D. `(5)/(3)(-5hati+5hatj+2hatk)` |
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Answer» Correct Answer - A Let `a=7hati-4hatj-4hatk` and `b=-2hati-hatj+2hatk` Now, required vector `c=lamda((a)/(|a|)+(b)/(|b|))` `=lamda((7hati-4hatj-4hatk)/(9)+(-2hati-hatj+2hatk)/(3))` `=(lamda)/(9)(hati-7hatj+2hatk)` `|c|^(2)=(lamda^(2))/(81)xx54=150` `implies lamda=+-15` `impliesc=+-(5)/(3)(hati-7hatj+2hatk)`. |
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| 56. |
If `A(-4,0,3)a n dB(14 ,2,-5),`then which one of the following points lieon the bisector of the angle between ` vec O Aa n d vec O B(O`is the origin ofreference )?a. `(2,2,4)`b. `(2, 11 ,5)`c. `(-3,-3,-6)`d. `(1,1,2)`A. (2,2,4)B. (2,11,5)C. (-3,-3,-6)D. (1,1,2) |
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Answer» Correct Answer - A::C::D `OA=-4hati+3hatk,OB=14hati+2hatj-5hatk` `a=(-4hati+3hatk)/(5),b=(14hati+2hatj-5hatk)/(15)` `r=(lamda)/(15)[12hati+9hatj+14 hati+2hatj-5hatk]` `=(lamda)/(15)[2hati+2hatj+4hatk]=(2lamda)/(15)[hati+hatj+2hatk]`. |
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| 57. |
The vertices of a triangle are A(1,1,2), B (4,3,1) and C (2,3,5). The vector representing internal bisector of the angle A is |
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Answer» Let AD is the bisector of `/_A`.Then, `(BD)/(DC) = (AB)/(AC)`eq(1) Given the vertices of triangle, `AB = sqrt(3^2+2^2+1^2) = sqrt(14)` `AC = sqrt(1^2+2^2+3^2) = sqrt(14)` As AB = BC, So, BD = DC(from eq(1)) It means D is middle point of BC. So, vertices of D will be (3,3,3). So, vector AD will be `2hati+2hatj+hatk`. |
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| 58. |
The triple product of ` (vec d + vec a) .[ vec a xx ( vec b xx ( vec c xx vec d ))]` is equal to: |
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Answer» `(vecd+veca).[vecaxx(vecbxx(veccxxvecd))]` `=(vecd+veca).[vecaxx((vecb.vecd)vecc-(vecb.vecc)vecd)]` `=(vecd+veca).[(vecb.vecd)(vecaxxvecc)-(vecb.vecc)(vecaxxvecd)]` `=(vecb.vecd)([vecd.veca.vecc + veca.veca.vecc]) - (vecb.vecc)([veca.vecd.vecd+veca.veca.vecd])` We know, if in a scalar triple product, two same values are there, value of products is `0`. So, our expression becomes, `=(vecb.vecd)([vecd.veca.vecc+0])-(vecb.vecc)([0-0])` `=(vecb.vecd)[vecd.veca.vecc]` |
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| 59. |
Let `vec a=hat i+hat j+hat k and vec b=i and vec c=c_1 hat i + hat c_2 j +c_3 hat k` Then(a) if `c_1 = 1 and c_2=2` , find `c_3` which makes ` vec a , vec b, vec c ` coplanar(b) if `c_2 = -1 and c_3= 1` , show that no value of `c_3` can makes ` vec a , vec b, vec c ` coplanar. |
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Answer» Here, `veca = hati+hatj+hatk` `vecb = hati` `vecc = c_1hati+c_2hatj+c_3hatk` (i) As these trhree vectors are coplanar, ,their scalar product should be `0`. `:. |[1,1,1],[1,0,0],[c_1,c_2,c_3]| = 0` `=>-c_3+c_2 = 0` `:.c_3 = c_2 = 2` (ii)From first part, `[vecavecbvecc] = -c_3+c_2` `= -1+1 = -2` So, scalar product of these three vectors is always `-2`. So, no value of `c_1` can make these vectors coplanar. |
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| 60. |
Given three vectors `a=6hati-3hatj,b=2hati-6hatj and c=-2hati+21hatj` such that `alpha=a+b+c`. Then, the resolution of the vector `alpha` into components with respect to a and b is given byA. 3a-2bB. 3b-2aC. 2a-3bD. a-2b |
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Answer» Correct Answer - C `alpha=a+b+c=6hati+12hatj` Let `alpha=xa+ybimplies 6x+2y=6` and `-3x-6y=12` `therefore x=2,,y=-3` `therefore alpha=2a-3b`. |
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| 61. |
The vertices of a triangle ABC are A (1,-2, 2), B (1, 4, 0) and C (-4, 1, 1) respectively. If M be the foot of perpendicular drawn from B on AC, then `vec BM` is |
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Answer» `tantheta=|(m_1-m_2)/(1+m_1m_2)|` `tantheta=|((-3/4)+1)/(1+3/4)|` `tantheta=1/7` `tantheta=|(m+3/4)/(1-3/4m)|` `1/7=|(4m+3)/(4-3m)|` `1/7=(4m+3)/(4-3m)` `m=-17/31,-1` `3x+4y=5` `x+y=1` `3(1-y)+4y=5` `3-3y+4y=5` `3+y=5` `y=2,x=-1` `y-2=-17/31(x+1)` `31y+17x-45=0` `31y+17x=45` option b is correct. |
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| 62. |
Let `vec a= hat i` be a vector which makes an angle of `120^@` with a unit vector`vec b` in XY plane. then the unit vector `(vec a+ vec b)` is |
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Answer» `OA = hat i` `OB = vec b_1 = -(cos 60^@)hat i + sin 60^@` `OB = vec b_2= - cos 60^@ hat i - sin 60^@ hat j` `vec a + vec b_1 = hat i - hat i/2 + sqrt3/2 hat j` `= 1/2 hat i + sqrt3/2 hat j` `vec a + vec b_2 = hat i - hat i/2 - sqrt3/2 hat j` `= 1/2 hat i - sqrt3/2 hat j` Answer |
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| 63. |
Let `vec a= hat i` be a vector which makes an angle of `120^@` with a unit vector`vec b` in XY plane. then the unit vector `(vec a+ vec b)` isA. `-(1)/(2)hati+(sqrt(3))/(2)hatj`B. `-(sqrt(3))/(2)hati+(1)/(2)hatj`C. `(1)/(2)hati+(sqrt(3))/(2)hatj`D. `(sqrt(3))/(2)hati-(1)/(2)hatj` |
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Answer» Correct Answer - C `b=cos120^(@)hati+sin120^(@)hatj` or `b=-(1)/(2)hati+(sqrt(3))/(2)hatj` therefore, `a+b=hati-(1)/(2)hati+(sqrt(3))/(2)hatj=(1)/(2)hati+(sqrt(3))/(2)hatj` |
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| 64. |
.If ifference of two unit vectors is a unit vector, prove that magnitude of their sum is `sqrt3` |
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Answer» `| vec a| = 1, | vec b|=1` `| vec a - vec b | = 1` `sqrt( a^2 + b^2 - 2 vec a vec b cos theta) = 1` `sqrt(1 + 1 - 2*1*1*cos theta) = 1` `1 + 1 - 2 cos theta = 1` `2cos theta = 1` `cos theta = 1/2` `|vec a + vec b| = | vec a|^2 + |vec b|^2 + 2|vec a|| vec b| cos theta` `= sqrt( 1 + 1 + 2 xx1/2)` `= sqrt3` Hence proved |
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| 65. |
If `hati-3hatj+5hatk` bisects the angle between `hata and -hati+2hatj+2hatk`, where `hata` is a unit vector, thenA. `a=(1)/(105)(41hati+88hatj-40hatk)`B. `a=(1)/(105)(41hati+88hatj+40hatk)`C. `a=(1)/(105)(-41hati+88hatj-40hatk)`D. `a=(1)/(105)(41hati-88hatj-40hatk)` |
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Answer» Correct Answer - D We must have `lamda(hati-3hatj+5hatk)=a+(2hatk+2hatj-hati)/(3)` therefore, `3a=3lamda(hati-3hatj+5hatk)-(2hatk+2hatj-hati)` `=hati(3lamda+1)-hatj(2+9lamda)+hatk(15lamda-2)` or `3|a|=sqrt((3lamda+1)^(2)+(2+9lamda)^(2)+(15lamda-2)^(2))` or `9=(3lamda+1)^(2)+(2+9lamda)^(2)+(15lamda-2)^(2)` or `315lamda^(2)-18lamda=0 implies lamda=0, (2)/(35)`. if `lamda=0,a=hati-2hatj-2hatk` (not acceptable). for `lamda=(2)/(35),a=(41)/(105)hati-(88)/(105)hatj-(40)/(105)hatk` |
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| 66. |
If the resultannt of two forces of magnitudes P and Q acting at a point at an angle of `60^(@)` is `sqrt(7)Q`, then P/Q isA. 1B. `(3)/(2)`C. `2`D. 4 |
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Answer» Correct Answer - C `R^(2)=P^(2)+Q^(2)+2PQcostheta` `implies(sqrt(7)Q)^(2)=P^(2)+Q^(2)+2PQcos60^(@)` `implies7Q^(2)=P^(2)+Q^(2)+PQ` `impliesP^(2)+PQ-6Q^(2)=0` `impliesP^(2)+3PQ-2PQ-6Q^(2)=0` `impliesP(P+3Q)-2Q(P+3Q)=0` `implies(P-2Q)(P+3Q)=0` `impliesP-2Q=0` or `P+3Q=0` From `P-2Q=0 implies(P)/(Q)=2` |
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| 67. |
A unit vector `hata` makes an angle `pi/4` with z-axis, `if hata+hati+hatj` is a unit vector then `hata` is equal to (A) `hati+hatj+hatk/2` (B) `hati/2+hatj/2-hatk/sqrt(2)` (C) `-hati/2-hat/2+hatk/sqrt(2)` (D) `hati/2-hatj/2-hatk/sqrt(2)`A. `(hati)/(2)+(hatj)/(2)+(hatk)/(sqrt(2))`B. `(hati)/(2)+(hatj)/(2)-(hatk)/(sqrt(2))`C. `-(hati)/(2)-(hatj)/(2)+(hatk)/(sqrt(2))`D. none of these |
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Answer» Correct Answer - C Let `a=l hati+m hatj+n hatk`, where `l^(2)+m^(2)+n^(2)=1`. A makes an angle `(pi)/(4)` with Z-axis. `therefore n=(1)/(sqrt(2)),l^(2)+m^(2)=(1)/(2)` . . . (i) `therefore a=l hati+mhatj+(hatk)/(sqrt(2))` `a+hati+hatj=(l+1)hati+(m+1)hatj+(hatk)/(sqrt(2))` Its magnitude is 1, hence `(l+1)^(2)+(m+1)^(2)=(1)/(2)` . . . (ii) From eqs. (i) and (ii), we get `2lm=(1)/(2)impliesl=m=-(1)/(2)` hence, `a=-(hati)/(2)-(hatj)/(2)+(hatk)/(sqrt(2))`. |
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| 68. |
The non-zero vectors are `vec a,vec b and vec c` are related by `vec a= 8vec b and vec c = -7vec b`. Then the angle between `vec a and vec c` isA. `(pi)/(4)`B. `(pi)/(2)`C. `pi`D. 0 |
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Answer» Correct Answer - C a and b vectors are in the same direction. B and c are in the opposite direction. `implies`a and c are in opposite directions. `therefore` Angle between a and c is `pi`. |
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| 69. |
The non-zero vectors are `vec a,vec b and vec c` are related by `vec a= 8vec b and vec c = -7vec b`. Then the angle between `vec a and vec c` isA. `pi`B. `0`C. `pi/4`D. `pi/2` |
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Answer» Correct Answer - A Since, `a=8b and c=-7b` So, a is parallel to b and c is anti-parallel to b. `implies a and c` are anti-parallel. so, the angle between a and c is `pi`. |
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| 70. |
Let a,b and c be three non-zero vectors which are pairwise non-collinear. If a+3b is collinear with c and b+2c is collinear with a, then a+3b+6c isA. a+cB. aC. `c`D. 0 |
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Answer» Correct Answer - D As, a+3b is collinear with c. `thereforea+3b=lamdac` . . . (i) Also, b+2c is collinear with a. `implies b+2c=mua` . . (ii) From eq. (ii), we get `a+3b+6c=(lamda+6)c` . .. (iii) From eq. (ii), we get `a+3b+6c=(1+3mu)a` . . (iv) From eqs. (iii) and (iv), we get `therefore(lamda+6)c=(1+3mu)a` Since, a is not collinear with c. `implies lamda+6=1+3mu=0` from eq. (iv), we get `a+3b+6c=0`. |
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| 71. |
If `|veca|=|vecb|`, then necessarily it implies `veca=+-vecb`. |
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Answer» if `|veca|=|vecb|impliesveca=+-vecb` So, it is true statement. |
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| 72. |
If non-zero vectors `veca and vecb` are equally inclined to coplanar vector `vecc`, then `vecc` can beA. `(|a|)/(|a|=2|b|)a+(|b|)/(|a|+|b|)b`B. `|b|/(|a|+|b|)a+|a|/(|a|+|b|)b`C. `(|a|)/(|a|+|b|)a+(|b|)/(|a|+2|b|)b`D. `(|b|)/(2|a|+|b|)a+(|a|)/(2|a|+|b|)b` |
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Answer» Correct Answer - B::D Since, a and b are equally inclined to c, therefore c must be of the form `t((a)/(|a|)+(b)/(|b|))` Now, `(|b|)/(|a|+|b|)a+(|a|)/(|a|+|b|)b=(|a||a|)/(|a|+|a|)((a)/(|a|)+(b)/(|b|))` Also, `(|b|)/(2|a|+|b|)a+(|a|)/(2|a|+|b|)b=(|a||b|)/(2|a|+|b|)((a)/(|a|)+(b)/(|b|))` Other two vectors cannot be written in the form `t((a)/(|a|)+(b)/(|b|))`. |
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| 73. |
Show that the vector `i+j+k`is equally inclined with the axes `O X , O Y a n d O Zdot` |
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Answer» Let `a=hati+hatj+hatk` If a makes angles `alpha,beta,gamma` with X,Y and Z-axes respectively, then `cosalpha=(1)/(sqrt(1^(2)+1^(2)+1^(2)))=(1)/(sqrt(3))` `cos beta=(1)/(sqrt(3))` and `cos gamma=(1)/(sqrt(3))` Thus, we have `cos alpha=cosbeta=cosgamma,` i.e., `alpha=beta=gamma` Hence, a is equally inclined to the axes. |
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| 74. |
What should be added in vector `a=3hati+4hatj-2hatk` to get its resultant a unit vector `hati`?A. `-2hati-4hatj+2hatk`B. `-2hati+4hatj-2hatk`C. `2hati+4hatj-2hatk`D. none of these |
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Answer» Correct Answer - A Let b should be added, then `a+b=hati` `impliesb=hati-a=hati-(3hati+4hatj-2hatk)` `=-2hati-4hatj+2hatk` |
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| 75. |
The direction cosines of the vector `3hati-4hatj+5hatk` areA. `(3)/(5),(-4)/(5),(1)/(5)`B. `(3)/(5sqrt(2)),(-4)/(5sqrt(2)),(1)/(sqrt(2))`C. `(3)/(sqrt(2)),(-4)/(sqrt(2)),(1)/(sqrt(2))`D. `(3)/(5sqrt(2)),(4)/(5sqrt(2)),(1)/(sqrt(2))`. |
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Answer» Correct Answer - B `r=3hati-4hatj+5hatk` `implies |r|=sqrt(3^(2)+(-4)^(2)+5^(2))=5sqrt(2)` Hence, direction cosines are `(3)/(5sqrt(2)),(-4)/(5sqrt(2)),(5)/(5sqrt(2))` i.e., `(3)/(5sqrt(2)),(-4)/(5sqrt(2)),(1)/(sqrt(2))`. |
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| 76. |
Find a unit vector parallel to the vector `-3hati+4hatj`. |
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Answer» let `a=-3hati+4hatj` then, `|a|=sqrt((-3)^(2)+(4)^(2))=5` `therefore`Unit vector parallel to `a=hata=(1)/(|a|)*a` `=(-3hati+4hatj)/(5)=(-3)/(5)hati+(4)/(5) hatj` |
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| 77. |
In the following figure, which of the vectors are: (i) Collinear (ii) Equal (iii) Co-initial (iv) collinear but not equal . |
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Answer» (i) a,c and d are collinear vectors. (ii) a and c are equal vectors (iii) b,c and d are co-initial vectors. (iv) a and d are collinear but they are not equal, as their directions are not same. |
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| 78. |
Represent graphically a displacement of 40 km, `30o`east of north. |
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Answer» Please refer to the diagram in the video for the first question. Let P is the point. It will have a displacementof 40 km from origin O. OP will be 40 km and it will create an angle of `30^@` with north axis. Please refer to the diagram in video for graphical representation. |
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| 79. |
A girl walks 4 km towards west, then she walks 3 km in a direction `30o`east of north and stops. Determine the girls displacement from her initial point of departure. |
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Answer» `veca=4(-hati)km` `vecb=3(sin30^ohati+cos30^ohatj)km` =`3(1/2hati+sqrt3/2hatj)` =`(3/2hati+(3sqrt3)/2hatj)` distance between them =`veca+vecb` =`-4hati+3/2hati+(3sqrt3)/2hatj` =`-5/2hati+(3sqrt3)/3hatj` |
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| 80. |
Let ABC be a triangle, the position vectors of whose vertices are `7hatj+10hatk,-1hati+6hatj+6hatk` and `-4hati+9hatj+6hatk`. Then, `DeltaABC` isA. isoscelesB. equilateralC. right angledD. none of these |
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Answer» Correct Answer - A::C We have, `AB=-hati-hatj-4hatk,BC=-3hati+3hatj` and `CA=4hati-2hatj+4hatk`. therefore, `|AB|=|BC|=3sqrt(2) and |CA|=6`. Clearly, `|AB|^(2)+|BC|^(2)=|AC|^(2)` Hence, the triangle is right angled isoscels triangle. |
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| 81. |
If the vectors `4hati+11hatj+mhatk,7hati+2hatj+6hatk and hati+5hatj+4hatk` are coplanar, then m is equal toA. 38B. 0C. 10D. `-10` |
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Answer» Correct Answer - C Since the three vectors are coplanar, one will be a linear combination off the other two. `therefore4hati+11hatj+mhatk=x(7hati+2hatj+6hatk)+y(hati+5hatj+4hatk)` `implies4=7x+y` . . . (i) `11=2x+5y` . . . (ii) `m=6x+4y` . . . (iii) From eqs. (i) and (ii), we get `x=(3)/(11) and y=(23)/(11)` From eq. (iii), we get `m=6xx(3)/(11)+4xx(23)/(11)=10` Trick since, vectors `4hati+11hatj+mhatk,7hati+2hatj+6hatk and hati+5hatj+4hatk` are complanar. `therefore|(4,11,m),(7,2,6),(1,5,6)|=0` `implies4(8-30)-11(28-6)+m(35-2)=0` `implies-88-11xx22+33m=0` `implies-8-22+3m=0` `implies3m=30impliesm=10` |
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| 82. |
If `overset(to)(V) = 2oveset(to)(i) + overset(to)(j) - overset(to)(k) " and " overset(to)(W) = overset(to)(i) + 3overset(to)(k) .` if `overset(to)(U)` is a unit vectors then the maximum value of the scalar triple product `[overset(to)(U) , overset(to)(v) , overset(to)(W)]` isA. `(-1)/(sqrt(59))`B. `sqrt(10) + sqrt(6)`C. `sqrt(59)`D. `sqrt(60)` |
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Answer» Correct Answer - C Given `vec( V) = 2hat(i) + hat(j) - hat(k) " and " vec(W) = hat(i) + 3hat(k)` `[vec(U)vec(V) vec(W)] = vec(U) . [(2hat(i) + hat(j) -hat(k)) xx (hat(i) + 3hat(k))]` `=vec(U) .(3hat(i) -7hat(j) - hat(k)) = |vec(U)||3hat(i) - 7 hat(j) - hat(k)|` cos 0 Which is maximum if angle between `vec(U) " and " 3hat(i) - 7hat(j) - hat(k) ` is 0 and maximum value `+|3hat(i) -7hat(j) - hat(k)|= sqrt(59)` |
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| 83. |
If O is origin annd the position vector fo A is `4hati+5hatj`, then unit vector parallel to OA isA. `(4)/(sqrt(41))hati`B. `(5)/(sqrt(41))hati`C. `(1)/(sqrt(41))(4hati+5hatj)`D. `(1)/(sqrt(41))(4hati-5hatj)` |
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Answer» Correct Answer - C Unit vector parallel to `OA=(4hati+5hatj)/(sqrt(16+15))=(1)/(sqrt(41))(4hati+5hatj)` |
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| 84. |
The number of distinct real values of `lambda,` for the vector `-lambda^2hati+hatj+hatk,hati-lambda^2hatj+hatk and hati+hatj-lambda^2hatk` are coplanar, is |
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Answer» Correct Answer - C Since given vectors are coplanar `:. |{:(-lambda^(2),,1,,1),(1,,-lambda^(2),,1),(1,,1,,-lambda^(2)):}|=0` `rArr lambda^(6) - 3lambda^(2) - 2= 0 rArr (1+lambda^(2))^(2) (lambda^(2)-2) =0 rArr lambda =+- sqrt(2)` |
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| 85. |
If the position vectors of P and Q are `hati+2hatj-7hatk and 5hati-2hatj+4hatk` respectively, the cosine of the angle between PQ and Z-axis isA. `(4)/(sqrt(162))`B. `(11)/(sqrt(162))`C. `(5)/(sqrt(162))`D. `(-5)/(sqrt(162))` |
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Answer» Correct Answer - B `PQ=OQ-OP=4hati-5hatj+11hatk` `therefore(PQ)/(|PQ|)=(4)/(sqrt(162))hati-(5)/(sqrt(162))hatj+(11)/(sqrt(162))hatk` `therefore cos gamma=(11)/(sqrt(162))`, where `gamma` is the angle of PQ with Z-axis. |
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| 86. |
If the position vectors of P and Q are `(hati+3hatj-7hatk) and (5hati-2hatj+4hatk)`, then |PQ| isA. `sqrt(158)`B. `sqrt(160)`C. `sqrt(161)`D. `sqrt(162)` |
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Answer» Correct Answer - D `PQ=(5-1)hati+(-2-3)hatj+(4+7)hatk` `=4hati-5hatj+11hatk` `|PQ|=sqrt(16+25+121)=sqrt(162)` |
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| 87. |
The position vectors of the points A,B and C are `hati+2hatj-hatk,hati+hatj+hatk and 2hati+3hatj+2hatk`, respectively. If A is chosen as the origin, then the position vectors of B and C areA. `hati+2hatk,hati+hatj+3hatk`B. `hatj+2hatk,hati+hatj+3hatk`C. `-hatj+2hatk,hati--hatj+3hatk`D. `-hatj+2hatk,hati+hatj+3hatk` |
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Answer» Correct Answer - D `OA=hati+2hatj-hatk,OB=hati+hatj+hatk` and `OC=2hati+3hatj+2hatk` Position vector of B w.r.t. origin at A at `AB=OB-OA=-hatj+2hatk` Position vector of C w.r.t. origin at A is `AC=OC-OA=hati+hatj+3hatk`. |
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| 88. |
Let OABCD be a pentagon in which the sides OA and CB are parallel and the sides OD and AB are parallel as shown in figure. Also, OA:CB=2:1 and OD:AB=1:3. if the diagonals OC and AD meet at x, find OX:OC.A. `5//2`B. `6`C. 7//3`D. 4 |
| Answer» Correct Answer - B | |
| 89. |
Let OABCD be a pentagon in which the sides OA and CB are parallel and the sides OD and AB are parallel as shown in figure. Also, OA:CB=2:1 and OD:AB=1:3. if the diagonals OC and AD meet at x, find OX:OC.A. `3//4`B. `1//3`C. `2//5`D. `1//2` |
| Answer» Correct Answer - C | |
| 90. |
Let `vec a ` be vector parallel to line of intersection of planes `P_1 and P_2` through origin. If `P_1`is parallel to the vectors `2 bar j + 3 bar k and 4 bar j - 3 bar k` and `P_2` is parallel to `bar j - bar k` and ` 3 bar I + 3 bar j `, then the angle between `vec a` and `2 bar i +bar j - 2 bar k` is :A. `(pi)/(2)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(3pi)/(4)` |
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Answer» Correct Answer - B::D Let vector `vec(AO) ` be parallel to line of intersection of planes `P_(1) " and " P_(2) ` through origin . Normal to plane `P_(1)` is ` vec(n)_(1) =[(2hat(j) + 3hat(k)) xx (4hat(j) - 3hat(k))] =- 18hat(j)` So , `vec(OA) ` is parallel to `+- (vec(n)_(1) xx vec(n)_(2)) = 54 hat(j) - 54 hat(k)` `:. ` Angle between `54 (hat(j) - hat(k)) " and " (2hat(i) + hat(j) - 2hat(k)) ` is `" cos" 0 = +- ((54 + 108)/(3.54 "."sqrt(2))) =+- (1)/(sqrt(2))` ` :. 0 = (pi)/( 4) , ( 3pi)/(4)` Hence (b) and (d) are correct answers |
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| 91. |
If `vecx`and `vecy` are two unit vectors and `theta` is the angle between them, then `1/2|vecx -vecy|` is equal to |
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Answer» `(1/2|vecx-vecy|)^2=1/4(|vecx|^2+|vecy|^2-2|vecx|||vecy|costheta)` `1/4(1+1-2*1*1costheta)` `1/4(2-2costheta)` `1/4*2(1-costheta)` `1/2(2sin^2theta/2)` `(1/2|vecx-vecy|)^2=sin^2theta/2` `1/2|vecx-vecy|=|sintheta/2|` |
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| 92. |
Let ` vec a , vec b"and" vec c`be there unit vectors such that ` vec axx( vec bxx vec c)=(sqrt(3))/2( vec b+ vec c)`. If ` vec b`is not parallel to ` vec c`, then the angle between ` vec a"and" vec b`is:(1) `(3pi)/4`(2) `pi/2`(3) `(2pi)/3`(4) `(5pi)/6` |
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Answer» Let angle between `veca` and `vecb` is `theta`. We are given, `veca xx (vecb xx vecc) = sqrt3/2(vecb+vecc)` `=>(veca*vecc)vecb - (veca*vecb)vecc = sqrt3/2(vecb+vecc)` `=>veca*vecc = sqrt3/2 and veca*vecb = -sqrt3/2` `=>|veca||vecb|costheta = -sqrt3/2` `=>1*1*costheta = -sqrt3/2` ...(As a and b are unit vectors) `=>theta = pi-pi/6` `=>theta = (5pi)/6` |
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| 93. |
If the vectors ` bar A B=3 hat i+4 hat k`and ` bar A C=5 hat i-2 hat j+4 hat k`are the sidesof a triangle ABC, then the length of the median through A is(1) `sqrt(72)`(2) `sqrt(33)`(3) `sqrt(45)`(4) `sqrt(18)` |
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Answer» `vec( AM) = (vec(AB) + vec (AC))/2` `vec(AM) = ((3 hat i + 4 hat k)+ (5 hat i - 2 hat j + 4 hat k))/2` `= (8 hat i - 2 hat j + 8 hat k )/2` `vec (AM)= 4 hat i- hat j + 4 hat k ` `|vec(AM)| = sqrt(4^2 + 1^2 + 4^2)` `= sqrt(16 +1+16) = sqrt33` option 2 is correct |
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| 94. |
If `[ vec axx vec b"" vec bxx vec c"" vec cxx vec a]=lambda[ vec a vec b vec c]^2`, then l is equal to(1) 2(2) 3(3) 0(4) 1 |
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Answer» `[vec a xx vec b ; vec b xx vec c ; vec c xx vec a] = (vec a xx vec b)*[ (vec b xx vec c) xx (vec c xx vec a)]` `= (vec a xx vec b)[vec z xx (vec c xx vec a)]` `= (vec a xx vec b)[vec c(vec b xx vec c)]` Answer |
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| 95. |
The vectors `lamdahati+hatj+2hatk,hati+lamdahatj-hatkand2hati-hatj+lamdahatk` are coplanar, ifA. `lamda=-2`B. `lamda=0`C. `lamda=1`D. `lamda=-1` |
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Answer» Let `veca=lamdahati+hatj+2hatj,vecb=hati+lamdahatj-hatkandvecc=2hati-hatj+lamdahatk` `|[lamda,1,2],[1,lamda,-1],[2,-1,lamda]|=0` `implieslamda(lamda^(2)-1)-1(lamda+2)+2(-1-2lamda)=0` `implieslamda^(3)-lamda-lamda-2-2-4lamda=0` `implieslamda^(3)-6lamda-4=0` `implies(lamda+2)(lamda^(2)-2lamda-2)=0` `implieslamda=-2orlamda=(2+-sqrt12)/(2)` `implieslamda=-2orlamda(2+-2sqrt3)/(2)=1+-sqrt3` |
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| 96. |
The projection vector of `veca" on "vecb` isA. `((veca.vecb)/(|vecb|))vecb`B. `(veca.vecb)/(|vecb|)`C. `(veca.vecb)/(|veca|)`D. `((veca.vecb)/(|veca|^(2)))hatb` |
| Answer» Projection vector of `veca" on "vecb` is given by `=veca.(vecb)/(|vecb|)vecb=(veca.(vecb)/(|vecb|)).vecb` | |
| 97. |
If `overset(to)(a) , overset(to)(b) " and " overset(to)( c)` are unit coplanar vectors then the scalar triple product `[2 overset(to)(a) - overset(to)(b) 2 overset(to)(b) - overset(to)(c ) 2 overset(to)(c ) - overset(to)(a)]` is |
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Answer» Correct Answer - A If `vec(a) , vec(b) , vec(c ) ` are coplanar vectors then `2vec(a)-vec(b) ,2vec(b) - vec(c ) ` and `2vec( c) - vec(a) ` are also coplanar vectors. `I.e., [2vec(a) -vec(b) 2vec(b) -vec(c ) vec(c ) -vec(a)]=0` |
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| 98. |
(i) If C is a given non-zero scalar and `overset(to)(A) " and " overset(to)(B)` be given non-zero vectors such that `overset(to)(A) bot overset(to)(B)` then find the vectors `overset(to)(X)` which satisfies the equations `overset(to)(A) "."overset(to)(X) =c " and " overset(to)(A) xx overset(to)(X) = overset(to)(B)` (ii) `overset(to)(A)` vectors A has components `A_(1), A_(2) , A_(3) `in a right -handed rectangular cartesian coordinate system OXYZ. The coordinate system is rotated about the X-axis through an anlge `(pi)/(2)` . Find the components of A in the new coordinate system in terms of `A_(1),A_(2),A_(3)` |
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Answer» Correct Answer - `(i) vec(X) = ((vec(c))/(|vec(A)|^(2))) vec(A) - ((1)/(|vec(A)|^(2))) (vec(A) xx vec(B)) " " (ii) (A_(2) hat(i) -A_(1) hat(j) + A_(3) hat(k)) (i) Given `vec(A) bot vec(B) rArr vec(A) ". " vec(B)=0` and `vec(A) xx vec(X) =vec(B) rArr vec(A) ". " vec(B) =0 " and " vec(X) ". " vec(B)=0` Now `[vec(X) vec(A) vec(A) xx vec(B)] = vec(X) ". " {vec(A) xx (vec(A) xx vec(B))}` ` =vec(X) .{(vec(A) ". " vec(B)) vec(A) -(vec(A) ". " vec(B)) vec(B)}` ` = (vec(A) ". " vec(B))(vec(X) " ." vec(A)) - (vec(A) ". " vec(A)) (vec(X) " ." vec(B)) =0` `rArr vec(X) , vec(A) , vec(A) xx vec(B)` are coplanar So `vec(X)` can be represented as a linear combination of `vec(A) " and " vec(A) xx vec(B)` , Let us consider , `vec(X) = lvec(A) + m (vec(A) xx vec(B))` Since `vec(A) " . " vec(X) = c` `:. vec(A) " ." {(vec(A) +m (vec(A) xx vec(B)) }=c` ` rArr l(vec(A) xx vec(A)) +m {vec(A) xx (vec(A) xx vec(B))}= vec(B)` `rArr 0- m |vec(A)|^(2)vec(B) =vec(B)` `rArr m = -(1)/(|vec(A)|^(2))` `:. vec(X) =((C)/(|vec(A)|^(2)))vec(A) -((1)/(|vec(A)|^(2))) (vec(A) xx vec(B))` (ii) Since vector `vec(A)` has components `A_(1) , A_(2) , A_(3)` in the coordinate system OXYZ `:. vec(A) = A_(1) hat(i) + A_(2) hat(j) +A_(3) hat(k)` When the given system is rotated about an angle of `pi//2` the new X-axis is along old Y-axis and new Y-axis is along the old negative X - axis , whereas z remains same . Hence the components of A in the new system are `(A_(2) , -A_(1) , A_(3))` `:. vec(A) ` becomes `(A_(2) hat(i) - A_(2)hat(j) + A_(3) hat(k))` |
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| 99. |
If `overset(to)(a),overset(to)(b),overset(to)(c )` are three non-zero , non-coplanar vectors and `overset(to)(b)_(1)=overset(to)(b) -.(overset(to)b.overset(to)(a))/(|overset(to)(a)|) overset(to)(a) , overset(to)(b)_(2) +.(overset(to)(b).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a)` `overset(to)(c)_(1) =overset(to)(c)-.(overset(to)(c).overset(to)a)/(|overset(to)(a)|^(2))overset(to)(a)-.(overset(to)(c).overset(to)(b))/(|overset(to)(b)|^(2))overset(to)(b),overset(to)c_(2)=overset(to)(c) -.(overset(to)(c).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a)-.(overset(to)(c).overset(to)(b)_(1))/(|overset(to)(b)|^(2))overset(to)(b)_(1)` `overset(to)(c)_(3) =overset(to)(c) -.(overset(to)(c).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a)-.(overset(to)(c).overset(to)(b)_(2))/(|overset(to)(b)_(2)|^(2))overset(to)(b)_(2).overset(to)(c)_(4)=overset(to)(a) -.(overset(to)(c).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a)` Then which of the following is a set of mutually orthogonal vectors ?A. `{overset(to)(a),overset(to)(b)_(1),overset(to)(c)_(1)}`B. `{overset(to)(a),overset(to)(b)_(1),overset(to)(c)_(2)}`C. `{overset(to)(a),overset(to)(b)_(2),overset(to)(c)_(3)}`D. `{overset(to)(a),overset(to)(b)_(3),overset(to)(c)_(4)}` |
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Answer» Correct Answer - B Since `vec(b)_(1) = vec(b) - (vec(b).vec(a))/(|vec(a)|^(2)) vec(a) , vec(b)_(1) =vec(b) + (vec(b).vec(a))/(|vec(a)|^(2)) vec(a)` `" and " vec( c)_(1) =vec(c ) - (vec( c) "."vec( a))/(|vec(a)|^(2)) vec(a) - (vec(c )"."vec(b))/(|vec(b)|^(2)) vec(b) vec( c)_(2) = vec( c) - (vec( c). vec(a))/(|vec(a)|^(2)) vec(a) - (vec(c ) "."vec(b)_(1))/(|vec(b)|^(2)) vec(b)_(1)` `vec(c)_(3) =vec(c) -(vec(c ).vec(a))/(|vec(a)|^(2))vec(a) -(vec(c ).vec(b)_(2))/(|vec(b)_(2)|^(2)) vec(b)_(2), vec(c )_(4) =vec(a) -(vec(c ). vec(a))/(|vec(a)|^(2))vec(a)` Which shows `vec(a). vec(b)_(1) =0=vec(a) . vec(c )_(2)=vec(b)_(1).vec(c )_(2)` So `{vec(a),vec(b)_(1),vec(c )_(2)}` are mutually orthogonal vectors. |
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| 100. |
If `|veca|=4and-3lelamdale2`, then the range of `|lamdaveca|` isA. [8,0]B. [-12,8]C. [0,12]D. [8,12] |
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Answer» We have, `|veca|=4and-3lelamdale2` `:.|lamdaveca|=|lamda||veca|=lamda|4|` `implies|lamdaveca|=|-3|4=12,atlamda=-3` `|lamdaveca|=|0|4=0,atlamda=0` and `|lamdaveca|=|2|4=8,atlamda=2` So, the range of `|lamdaveca|" is " [0,12]` Alternate Method Since, `-3lelamdale2` `0le|lamda|le3` `implies0le4|lamda|le12` `|lamdaveca|in[0,12]` |
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