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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.
| 101. |
If `|AxxB| = sqrt3 A.B,` then the value of |A+B| isA. `(A^2+B^2 +AB)^(1//2)`B. `(A^2 +B^2 + (AB)/(sqrt3))^(1//2)`C. `(A+B)`D. `(A^2 +B^2 +sqrt3 AB)^(1//2)` |
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Answer» Correct Answer - A `|AxxB|= sqrt3 A.B` `rArr AB sin theta = sqrt3 AB cos theta` or `tan theta = sqrt3` rArr `theta = 60^@` `|A + B|=sqrt(A^2 + B^2 + 2AB cos 60^@)` `=sqrt(A^2 +B^2 +AB)` |
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| 102. |
The position vector of a particle is `vec( r) = a cos omega t i + a sin omega t j`, the velocity of the particle isA. parallel to the position vectorB. perpendicular to the position vectorC. directed towards the originD. directed away from the origin |
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Answer» Correct Answer - B `vec( r) = a cos theta omega t hat (i) + alpha sin omega t hat(j)` `vec(v ) = ( dvec(r ))/(dt) = -a sin omega t + a cos omega t . omega hat(j)` `vec( r) . vec(v ) = 0 , vec(r )` and `vec(v) are_|_^(ar)`. |
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| 103. |
Arrange the magnitude of cross products in the decreasing order. (a) `vecA` and `vecB` making angle zero (b) `vecA` and `vecB` making angle `30^(@)` (c) `vecA` and `vecB` making angle `120^(@)`A. a,b,cB. b,c,aC. c,a,bD. c,b,a |
| Answer» Correct Answer - D | |
| 104. |
A particle moves so that its position vector varies with time as `vec(r )= A cos omegathat(i)+A sin omega t hai(j)`. The initial velocity of the particel the particle isA. `A omega hat(i)`B. `A omega hat(j)`C. `A omega(hat(i)+hat(j))`D. `A omega (hat(i)-hat(j))` |
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Answer» Correct Answer - B Substituting `vec(r )=(A cos omega t hat(i)+A sin omega t hat(j))` in `vec(v)= (dvec(r ))/(dt)` we have `vec(v) = A(d)/(dt)(cos omega t)hat(i)+A(d)/(dt)(sin omega t)hat(j)` `= - A omega sin omega t hat(i)+A omega cos omega t hat(j)` At `t = 0 , v= - A omega sin 0 hat(i)+A omega cos 0hat(j)= A omegahat(j)` |
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| 105. |
Arrange the vector subtractions so that their magnitude are in decreasing order. If the two vectors `vecA` and `vecB` are acting at an angle `(|vecA|gt|vecB|)`. (a) `60^(@)` (b) `90^(@)` (c) `180^(@)` (d) `120^(@)`A. a,b,cB. c,b,aC. c,d,b,aD. a,c,b |
| Answer» Correct Answer - C | |
| 106. |
The work done by a force `vecF` during a displacement `vecr` is given by `vecF.vecr`. Suppose a force of 12 N acts on a particle in vertically upward directionand the particle is displaced through 2.0 m in vertically downward direction. Find the work done by the force during this displacement. |
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Answer» Correct Answer - B::D The angle between the force F and the displacement r is `180^(@)`. Thus, the work done is ` W = F.r` `=Fr cos theta` `= (12 N)(2.0m)(cos 180^@)` `=- 24 N-m =- 24 J`. |
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| 107. |
Two forces `vecF_(1)=5hati+10hatj-20hatk and vecF_(2)=10hati-5hatj-15hatk ` act on a single point. The angle between `vecF_(1) and vecF_(2)` is nearlyA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - B |
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| 108. |
The unit vector parallel to the resultant of the vectors `vec(A) = hat(i) + 2 hat(j) - hat(k)` and `vec(B) = 2 hat(i) + 4 hat(j) - hat(k)` isA. `(1)/(49) (7 hat(i) + 6 hat(j) - 2 hat(k))`B. `(1)/(7) (3 hat(i) + 6 hat(j) - 2 hat(k))`C. `(1)/(49) (3 hat(i) + 6 hat(j) - 2 hat(k))`D. `(1)/(7) (7 hat(i) + 6 hat(j) - 2 hat(k))` |
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Answer» Correct Answer - B `vec(A) + vec(B) = 3i + 6 j - 2k` `|vec(A) + vec(B)| = sqrt((3)^(2) + (6)^(2) + (-2)^(2)) = 7` Unit vector along `(vec(A) + vec(B)) = (vec(A) + vec(B))/(|vec(A) + vec(B)|) = (1)/(7) ( 3i + 6j - 2k)` |
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| 109. |
Three forces acting on a body are shown in figure. To have the resultant force only along the y-directon, the magnitude of the minimum additional force needed si A. `(sqrt(3))/(4)N`B. `sqrt(3)N`C. `0.5N`D. `1.5N` |
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Answer» Correct Answer - C (c ) According to given problem there should be `sumF_(x)=0` In the given figure net force along X-axis . `(1)/(2)hati+1hati-2hati` `impliesF=-(1)/(2) hati=- 0.5 hati` |
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| 110. |
The value of the` lambda` so that P, Q, R, S on the sides OA, OB, OC and AB of a regular tetrahedron are coplanar. When `(OP)/(OA)=1/3 ;(OQ)/(OB)=1/2` and `(OS)/(AB)=lambda` is (A)`lamda=1/2` (B) `lamda=-1` (C) `lamda=0` (D) `lamda=2`A. `lamda = (1)/(2)`B. `lamda =-1`C. `lamda =0`D. for no value of `lamda` |
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Answer» Correct Answer - B Let `vec(OA) = veca, vec(OB) = vecb and vec(OC) = vecc`, then `vec(AB) = vecb - veca and vec(OP) =(1)/(3) veca`, `" "vec(OQ) = (1)/(2) vecb, vec(OR) = (1)/(3) vecc`. Since P, Q, R and S are coplanar, then `vec(PS) = alpha vec(PQ) + betavec(PR) (vec(PS) ` can be written as a linear combination of `vec(PQ) and vec(PR))` `= alpha (vec(OQ) - vec(OP))+ beta(vec(OR) - vec(OP))` i.e., `vec(OS) - vec(OP) = -(alpha + beta)( veca)/(3) + (alpha )/(2) vecb + (beta)/(3) vecc` `rArr vec(OS) = (1-alpha- beta) (veca)/(3) + (alpha)/(2) vecb+ (beta)/(3) vecc" "` (i) Given `vec(OS) =lamda (vecb -veca) " "` (ii) From (i) and (ii), `beta =0, (1-alpha)/(3) = -lamda and (alpha)/(2) =lamda` `rArr 2 lamda = 1+ 3lamda` or `lamda =-1 ` |
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| 111. |
Choose th false statementA. A vector having zero magnitude can have a directionB. If `vecAxxvecB=vec0`, then either `vecA` or `vecB` or both must have zero magnitudeC. The component of a vector is a vectorD. All the above |
| Answer» Correct Answer - D | |
| 112. |
If `vecA,vecB` and `vecC` are coplanar vectors, thenA. `(vecA.vecB)xxvecC=0`B. `(vecAxxvecB)vecC=0`C. `(vecA.vecB)vecC=0`D. all the above are true |
| Answer» Correct Answer - B | |
| 113. |
The resultant of P and Q is perpendicular to P. what is the angle between P & Q?(a) cos-1 (P/Q)(b) cos-1 (-P/Q)(c) sin-1 (P/Q)(d) sin-1 (-P/Q) |
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Answer» Answer is (b) cos-1 (-P/Q) tan β = Q sinθ/(P + Q cosθ), here β = 90°. Hence P + Q cosθ = 0, Hence θ = cos-1 (-P/Q) |
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| 114. |
If vectors `veca =hati +2hatj -hatk, vecb = 2hati -hatj +hatk and vecc = lamdahati +hatj +2hatk` are coplanar, then find the value of `(lamda -4)`. |
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Answer» Correct Answer - 9 Vector `veca = hati + 2hatj -hatk, vecb = 2hati -hatj +hatk, vecc = lamda hati + hatj + 2hatk ` are coplanar `rArr |{:(1,,2,,-1),(2,,-1,,1),(lamda,,1,,2):}|=0` or `lamda -3 + 2(-5)=0` or `lamda = 13` |
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| 115. |
Out of the following forces, the resultant of which cannot be `10N`?A. `15N` and `20N`B. `10N` and `10N`C. `5N` and `12N`D. `12N` and `1N` |
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Answer» Correct Answer - D The resultant of two forces can lie between `A-B` and `A+B, i.e., 12-1 = 11N and 12+1=13N`. |
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| 116. |
Two forces, while acting on particle in opposite directions,have the resultant of 10N. If they act at right angles to each other, the resultant is found to be 50N. Find the two forces?A. `40 N , 30 N`B. `50 N , 40 N`C. `30 N , 20 N`D. `35 N , 25 N` |
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Answer» Correct Answer - A `P - Q = 10` (i) `P^(2) + Q^(2) = (50)^(2)` (ii) `(10 + Q)^(2) + Q^(2) = 2500` `100 + Q^(2) + 20 Q + Q^(2) = 2500` `Q^(2) + 10 Q - 1200 = 0` `(Q + 40) (Q - 30) = 0` `Q = -40 , P = -30` `Q = 30 , P = 40` Magnitude of forces : `30 N , 40 N` |
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| 117. |
`hata, hatb and hatc` are unit vectors. If the `hata +hatb = hatc`, them the magnitude of `hata -hatb` is :A. `(1)/(sqrt(3))`B. `(1)/(sqrt(2))`C. `sqrt(2)`D. `sqrt(3)` |
| Answer» Correct Answer - D | |
| 118. |
If `vecA=9hati-7hatj+5hatk` and `vecB=3hati-2hatj-6hatk` then the value of `(vecA+vecB),(vecA-vecB)` isA. 206B. 128C. 106D. `-17` |
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Answer» Correct Answer - C `(vecA+vecB).(vecA-vecB)=A^(2)-B^(2)` |
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| 119. |
The work done by a force `2hati-hatj+5hatk` when it displaces the body from a point (7.2.5) isA. 5unitsB. 7unitsC. 1unitsD. 15 units |
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Answer» Correct Answer - A `W=vecF.vecS=vecF.(vecr_(2)-vecr_(1))` |
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| 120. |
Find the vector that must be added to the vector `hat(i)-3hat(j)+2hat(k)` and `3hat(i)+6hat(j)-7hat(k)` so that the resultant vector is a unit vector along the y-axis. |
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Answer» Unit vector along y-axis `=hat(j).` so the required vector, `hat(j)-[(hat(i)-3hat(j)+2hat(k))+(3hat(i)+6hat(j)-7hat(k))]=-4hat(i)-2hat(j)+5hat(k)` |
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| 121. |
A particle moves from position `3hat(i)+2hat(j)-6hat(k)` to `14hat(i)+13hat(j)+9hat(k)` due to a uniform force of `4hat(i)+hat(j)+3hat(k)N`. If the displacement is in meters, then find the work done by the force. |
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Answer» `S=vec(r )_(2)-vec(r )_(1), W=vec(F).vec(S)=(4hat(i)+hat(j)+3hat(k)).(11hat(i)+11hat(j)+15hat(k))` `=(4xx11+1xx11+3xx15)=100J` |
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| 122. |
If two vectors `2hat(i)+3hat(j)-hat(k)` and `-4hat(i)-6hat(j)-lambda hat(k)` are parallel to each other,then find the value of `lambda` |
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Answer» Correct Answer - B Let `vec(A)= 2hat(i)+3hat(j)-hat(k)` and `vec(B)= -4hat(i)-6hat(j)+lambdahat(k)` `vec(A)` and `vec(B)` are parallel to each other `(a_(1))/(b_(1))= (a_(2))/(b_(2))=(a_(3))/(b_(3)) i.e., 2/(-4)= 3/(-6)= (-1)/(lambda)implies lambda= 2` |
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| 123. |
The vector from origion to the point A and B are `vec(A)=3hat(i)-6hat(j)+2hat(k)` and `vec(B)=2hat(i)+hat(j)-2hat(k)`,respectively. Find the area of the triangle OAB.A. `5/2sqrt(17) sq.unit`B. `2/5sqrt(17)sq.unit`C. `3/5sqrt(17)sq.unit`D. `5/3sqrt(17)sq.unit` |
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Answer» Correct Answer - A Given `vec(OA)= vec(a)= 3hat(i)-6hat(j)+2hat(k)` and `vec(OB)= vec(b)= 2hat(i)+hat(j)-2hat(k)` `:. (vec(a)xxvec(b))=|(hat(i),hat(j),hat(k)),(3,-6,2),(2,1,-2)|` `=(12-2)hat(i)+(4+6)hat(j)+(3+12)hat(k)` `=10hat(i)+10hat(j)+15hat(k)implies |vec(a)xxvec(b)|` `=sqrt(10^(2)+10^(2)+15^(2))` `=sqrt(425)=5sqrt(17)` Area of `DeltaOAB=1/2|vec(a)xxvec(b)|=(5sqrt(17))/2sq`.unit. |
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| 124. |
Given `vec(A)=4hat(i)+6hat(j)` and `vec(B)=2hat(i)+3hat(j)`. Which of the followingA. `vec(A)xxvec(B)=vec(0)`B. `vec(A).vec(B)=24`C. `(|vec(A)|)/(|vec(B)|)=1/2`D. `vec(A)` and `vec(B)` are antiparallel |
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Answer» Correct Answer - A `vec(A)xxvec(B)=(4hat(i)+6hat(j))xx(2hat(i)+3hat(j))=12(hat(i)xxhat(j))+12(hat(j)xxhat(i))` `=12(hat(i)xxhat(j))-12(hat(i)xxhat(j))=0` Again, `vec(A).vec(B)=(4hat(i)+6hat(j)).(2hat(i)+3hat(j))=8+18=26` Again`(|vec(A)|)/(|vec(B)|)=(sqrt(16+36))/(sqrt(4+9))!=1/2` Also, `vec(B)=1/2vec(A) rArr vec(A)` and `vec(B)` are parallel and not antiparallel. |
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| 125. |
a. Calculate `vec(r )=vec(a)-vec(b)+vec(c )`, where `vec(a)=5hat(i)+4hat(j)-6hat(k)`, `vec(b)= -2hat(i)+2hat(j)+3hat(k)`, and `vec(c )= 4hat(i)+3hat(j)+2hat(k)`. b. Calculate the angle between `vec(r )` and the `z`-axis. |
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Answer» Correct Answer - a.`11hat(i)+5hat(j)-7hat(k)`; b. `cos^(-1)-7/sqrt(195)` a. `vec(r )= vec(a)-vec(b)+vec(c )` `=(5hat(i)+4hat(j)-6hat(k))-(-2hat(i)+2hat(j)+3hat(k))+(4hat(i)+3hat(j)+2hat(k))` `5hat(i)+2hat(i)+4hat(i)+4hat(j)-2hat(j)+3hat(j)-6hat(k)-3hat(k)-2hat(k)` `11hat(i)+5hat(j)-7hat(k)` b. `cos r` or angle between `vec(r )` and z-axis `cos r =-(-7)/(sqrt((11)^(2)+(5)^(2)+(7)^(2)))=(-7)/sqrt(195)` or angle `=cos^(-1) (-7)/sqrt(195)` |
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| 126. |
Calculate the area of the parallelogram when adjacent sides are given by the vectors `vec(A)=hat(i)+2hat(j)+3hat(k)` and `vec(B)=2hat(i)-3hat(j)+hat(k)`. |
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Answer» We know that area of the parallelogram is equal to magnitude of the cross product of given vectors.,Now `vec(A)xxvec(B)=|(hat(i), hat(j), hat(k)) ,(1,2,3), (2 ,-3 ,1)|` `=hat(i)(2+9)+hat(j)(6-1)+hat(k)(-3-4)=11hat(i)+5hat(j)-7hat(k)` So area of parallelogram: `|vec(A)xxvec(B)|=sqrt(11^(2)+5^(2)+(-7)^(2))` `=sqrt(195)sq.unit` |
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| 127. |
Determine a vector which when added to the resultant of `vec(A)= 2hat(i)+5hat(j)-hat(k)` and `vec(B)= 3hat(i)-4hat(j)-hat(k)` gives unit Vector along negative y direction.A. `-5hat(i)-2hat(j)+2hat(k)`B. `-5hat(i)-hat(j)+hat(k)`C. `5hat(i)-hat(j)+2hat(k)`D. `-5hat(i)-hat(j)+2hat(k)` |
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Answer» Correct Answer - A Let `vec(C )`is required Vector, then `vec(C )+vec(A)+vec(B)= -hat(j)` `implies vec(C )+2hat(i)+5hat(j)-hat(k)+3hat(i)-4hat(j)-hat(k)= -hat(j)` `implies vec(C )= -5hat(i)-2hat(j)+2hat(k)` |
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| 128. |
The vector `vecP=ahati+ahatj+3hatj and vecQ=ahati-2hatj-hatk`, are perpendicular to each other. The positive value of a isA. 3B. 4C. 9D. 13 |
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Answer» Correct Answer - A |
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| 129. |
The vector sum of two forces is perpendicular to their vector differences. In that case, the forcesA. Are equal to each other in magnitudeB. Are not equal to each other in magnitudeC. Cannot be predictedD. Are equal to each other |
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Answer» Correct Answer - A If two vector are perpendicular then their dot product must be equal to zero. According to problem `(vec(A)+vec(B)).(vec(A)-vec(B))= 0 implies vec(A).vec(A)-vec(A).vec(B)+vec(B).vec(A)-vec(B).vec(B)=0` `implies A^(2)-B^(2)= 0 implies A^(2)= B^(2)` i.e., two vector are equal to each other in magnitude. |
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| 130. |
Two vectors A and B have magnitudes 6 units and 8 units respectively. Find |A-B|, if the angle between two vectors is .(a) `0^(@)` (b) `180^(@)` (c) `180^(@)` (d) `120^(@)` . |
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Answer» Correct Answer - A::B::C::D Apply `S = sqrt(A^2 + B^2 - 2AB cos theta)` |
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| 131. |
A body, under the action of a force `vec(F)=6hati -8hatj+10hatk` , acquires an acceleration of `1ms^(-2)` . The mass of this body must be.A. 200 kgB. 20 kgC. `10sqrt(2)kg`D. `6 sqrt(2) kg` |
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Answer» Correct Answer - C `m=sqrt(6^(2) + (-8)^(2) + 10^(2))` `m=sqrt(36+64+100)=10sqrt(2)` |
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| 132. |
When three forces of `50 N, 30 N and 15 N` act on body, then the boy isA. At restB. Moving with a uniform velocityC. In equilibriumD. Moving with an acceleration |
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Answer» Correct Answer - D Here all the three force will not keep the particle in equlibrium so the net force will not force will not be zero and the particle will move with an acceleration. |
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| 133. |
If the sum of two unit vectors is a unit vector,then find the magnitude of their differences.A. `sqrt(2)`B. `sqrt(3)`C. `1/sqrt(2)`D. `sqrt(5)` |
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Answer» Correct Answer - B |
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| 134. |
If the line \(\frac{x + 1}{1}=\frac{y - m}{3}=\frac{z - 4}{6}\) lies in the plane 3x – 14y + 6z + 49 = 0, then the value of m is :(x + 1)/1 = (y - m)/3 = (z - 4)/6(A) 5 (B) 3 (C) 2 (D) -5 |
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Answer» Correct answer is (A) 5 |
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| 135. |
The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is(A) 4x + y + 5z = 14 (B) 4x – 2y – 5z = 45 (C) x – 2y – 5z = 10 (D) 4x + y + 6z = 11 |
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Answer» Correct answer is (B) 4x – 2y – 5z = 45 |
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| 136. |
Let A,B and C be the unit vectors . Suppose that A.B=A.C =0 and the angle between B and C is `(pi)/(6)` then prove that `A = +-2(BxxC)` |
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Answer» `since ,A.B=O A.C=O` ` Hence , (B+C).A=O` so ,A is perpendicular to (B+C)and A is a unit vector perpendicular to the polane of vector BandC. `A=(BxxC)/(|BxxC|)` `where ,|BxxC|=|B||C|sintheta` ` =|B||c|"sin"(pi)/(6) (therefore sintheta=(pi)/(6))` `=1xx1xx(1)/(2)=(1)/(2)` `A=(BxxC)/(|BxxC|)=+-2(BxxC)` |
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| 137. |
A, B, C, D, E, F in that order, are the vertices of a regular hexagon with centre orgin. If the position vectors of the vertices A and B are respectively, `4hati+3hatj-hatk and -3hati+hatj+hatk`, then DE is equal toA. `7hati+2hatj-2hatk`B. `-7hati-2hatj+2hatk`C. `3hati-hatj-hatk`D. `-4hati-3hatj+2hatk` |
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Answer» Correct Answer - A Given, `OA=4hati+3hatj-hatk` `OB=-3hati+hatj+hatk` Now, `AB=OB-OA=-7hati-2hatj+2hatk` `therefore DE=-AB=7hati+2hatj-2hatk` |
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| 138. |
Three vectors `a=hati+hatj-hatk, b=-hati+2hatj+hatk and c=-hati+2hatj-hatk`, then the unit vector perpendicular to both `a+b` and `b+c` isA. `(hati)/(sqrt3)`B. `6hatk`C. `(hatk)/(sqrt3)`D. `(hati+hatj+hatk)/(sqrt3)` |
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Answer» Correct Answer - B `(a+b) xx (b+c)=3hatj xx (-2hati+4hatj)=6hatk` So, required unit vector is `6hatk`. |
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| 139. |
Assertion: If `hat(i)` and `hat(j)` are unit Vectors along x-axis and y-axis respectively, the magnitude of Vector `hat(i)+hat(j)` will be `sqrt(2)` Reason: Unit vectors are used to indicate a direction only.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both aseertion and reason are false. |
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Answer» Correct Answer - B Since `hat(i)` and `hat(j)` are unit Vectors, their magnitudes are `|vec(i)|=1 and |hat(i)|=1`. Magnitude of resulant vector is equal to `sqrt(|hat(i)|^(2)+|hat(j)|^(2))=sqrt((1)^(2)+(1)^(2))=sqrt(2)` |
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| 140. |
When a force F acs on a body of mass m the acceleration product in the body is a . If htree equal forces `F_(1)=F_(2)=F_(3)=F` act on the same body as shown in figure the accleration produced is A. `(sqrt(2)-1)a`B. `(sqrt(2)+1)a`C. `sqrt(2)a`D. a |
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Answer» Correct Answer - A (a) Accelertion `a=(F)/(m)` Resultant of `F_(1)andF_(2)`will be `F_(12)=sqrt(2F)`(in opposite directiomn o f`F_(3)`) Now ,resultant of`F_(12)and F_(3)`will be , `F_("net")=(sqrt(2)-1)F` `thereforea=(F_("net"))/(m)=(sqrt(2)-1)(F)/(m)=(sqrt(2)-1)a` |
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| 141. |
`hat(e )_(r)` is unit Vector along radius of a circle shown in figure `hat(e )_(r)` can be represented as ` A. `cos theta hat(i)+sin theta hat(j)`B. `sin theta hat(i)+cos theta hat(j)`C. `cos theta hat(i)-sin theta hat(j)`D. `-cos theta hat(i)+sin theta hat(i)` |
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Answer» Correct Answer - A The component of `hat(e )_(r )` along X-axis is `hat(e )_(r ) cos theta= 1xxcos theta= cos theta` The component of `hat(e )_(r )` along Y-axis is `hat(e )_(r ) sin theta= 1xx sin theta= sin theta` `:. hat(e )_(r )= cos theta hat(i)+sin theta hat(j)` |
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| 142. |
The diagonals of a parallelogram are `vecA=2hati-3hatj+hatk` and `vecB=-2hati+4hatj-hatk` what is the area of the paralleogram?A. 4 unitsB. 7 unitsC. `sqrt(5)` unitsD. `sqrt(8)` units |
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Answer» Correct Answer - C Area of parallelogram `=|vecAxxvecB|` |
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| 143. |
If `P+Q=0`. Then which of the following is necessarily ture?A. P=0B. P=-QC. Q=0D. P=Q |
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Answer» Correct Answer - B (b ) The sum of two vector is zero only when one vector is equal in magnitude but opposite in direction to that of other vector. |
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| 144. |
True and False:Position vector of a point P is a vector whose initial point is origin. |
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Answer» True Explanation: Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector \(\vec {OP}\) having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. |
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| 145. |
If `hat(i)+hat(j), hat(j)+hat(k), hat(i)+hat(k)` are the position vectors of the vertices of a `Delta ABC` taken in order, then `angleA` is equal toA. `pi/2`B. `pi/5`C. `pi/6`D. `pi/3` |
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Answer» Correct Answer - D Let position vector of the vertices are `OA = hati +hatj , OB = hatj + hatk` and `OC = hati + hatk` Now, `AB = - hati + hatk` and `AC = hatk - hati` `:.cos theta = ((AB).(AC))/(|AB||AC|) = ((-hati + hatk).(hatk -hatj))/(sqrt(1^(2) + 1^(2))sqrt(1^(2) +1^(2))) = (1)/(sqrt(2)sqrt(2)) = 1/2` `rArr theta= (pi)/(3)` |
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| 146. |
Let ` vec a= hat i+ hat j+ hat k , vec b= hat i- hat j+ hat ka n d vec c= hat i- hat j- hat k`be three vectors. A vector ` vec v`in the plane of ` vec aa n d vec b ,`whose projection on ` vec c`is `1/(sqrt(3))`is given by` hat i-3 hat j+3 hat k`b. `-3 hat i-3 hat j+3 hat k`c. `3 hat i- hat j+3 hat k`d. ` hat i+3 hat j-3 hat k`A. `3 hat(i)+hat(j)-3hat(k)`B. `3hat(i)-hat(j)+3hat(k)`C. `3hat(i)+hat(j)+3hat(k)`D. `-3hat(i)+hat(j)+3hat(k)` |
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Answer» Correct Answer - B Here, `(v.c)/(|c|) = 1/(sqrt(3))` `rArr (v.c)/(sqrt(1+1+1)) = 1/(sqrt(3))` `rArr (v.c)/(sqrt(3)) = 1/(sqrt(3)) rArr v.c = 1` `rArr v.(hati-hatj -hatk) = 1 "….."(i)` we get from option (b) `v=3 hati-hatj+3hatk` ` v.c (3hati -hatj+3hatk)` `=3+1-3=1` which satisfies Eq. 9i) `v=3hati-hatj+3hatk` |
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| 147. |
If `hati-3hatj+5hatk` bisects the angle between `hata and -hati+2hatj+2hatk`, where `hata` is a unit vector, thenA. `hata = (1)/(150)(41hati+88hatj-40hatk)`B. `hata=(1)/(105)(41hati + 88hatj + 40hatk)`C. `hata = (1)/(105) (-41hati + 88hatj-40hatk)`D. `hata= (1)/(105) (41hati - 88hatj - 40 hatk )` |
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Answer» Correct Answer - D We must have `lamda (hati-3hatj +5hatk) = hata + (2hatk+2hatj-hati)/(3)` Therefore `3hata = 3lamda = (hati-3hatj +5hatj) - (2hatk +2hatj -hati)` `" "= hati (3lamda +1) -hatj( 2+ 9lamda) + hatk ( 15lamda -2)` or `" "3|hata| = sqrt((3lamda +1)^(2) + (2+ 9lamda)^(2) + (15lamda -)^(2))` or ` " "9 = (3lamda +1)^(2) + (2 + 9lamda )^(2) + (15lamda -2)^(2)` or `" "315lamda ^(2) - 18lamda =0 rArr lamda =0, (2)/(35)` if `lamda =0, veca =hati-2hatj -hatk` ( not acceptable) For `lamda = (2)/(35) , veca = (41)/(105)hati - (88)/(105) hatj - (40/(105)hatk` |
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| 148. |
If `a=hat(i)+hat(j)+hat(k), b=2hat(i)+lambda hat(j)+hat(k), c=hat(i)-hat(j)+4hat(k)` and `a.(b xx c)=10`, then `lambda` is equal toA. 6B. 7C. 9D. 10 |
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Answer» Correct Answer - A Given `a=hati+hatj+hatk, b=2hati+lamda hatj+hatk and c= hati-hatj+4 hatk` Aslo, given `a*(bxx c)=10` `implies|{:(1,1,1),(2 , lamda,1),(1,-1,4):}|=10` `implies1(4lamda+1)-1(8-1)+1(-2-lamda)=10` `implies 4 lamda +1-7-2-lamda=10` `implies3 lamda =18 implieslamda =6` |
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| 149. |
`vec(A)=2hat(i)+4hat(j)+4hat(k)` and `vec(B)=4hat(i)+2hat(j)-4hat(k)` are two vectors. Find the angles between them. |
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Answer» `costheta=(vec(A).vec(B))/(|vec(A)|.|vec(B)|)` `=(a_(1) b_(1)+a_(2)b_(2)+a_(3)b_(3))/(|vec(A)|.|vec(B)|)=(2xx4+4xx2-4xx4)/(|vec(A)|.|vec(B)|)=0` `:. theta=cos^(-1)(0^(@))rArrtheta=90^(@)` |
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| 150. |
The angle between the two vectors `vecA=hati+2hatj-hatk` and `vecBi=-hati+hatj-2hatk` isA. `90^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - D `cos theta=(vecA.vecB)/(|vecA||vecB|)` |
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