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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.
| 451. |
Prove that the necessaryand sufficient condition for any four points in three-dimensional space to becoplanar is that there exists a liner relation connecting their positionvectors such that the algebraic sum of the coefficients (not all zero) in itis zero. |
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Answer» Let us assume that the points `A, B, C and D` whose position vectors are `veca, vecb, vecc and vecd`, respectively, are coplaner. In the case the lines AB and CD will intersect at some point P (it being assumed that AB and CD are not parallel, and if they are, then we will choose any other pair of non-parallel lines formed by the given points ). If P divides AB in the ratio `q:p` and CD in the ratio `n:m` , then the position vector of P written from AB and CD is `" "(pveca+qvecb)/(p+q)=(mvecc+nvecd)/(m+n)` or `" "(p)/(p+q)veca+(q)/(p+q)vecb-(m)/(m+n)vecc-(n)/(m+n)vecd=vec0` or `" "Lveca+Mvecb+Nvecc+Pvecd=vec0` where `" "L+M+N+P= (p)/(p+q)+(q)/(p+q)-(m)/(m+n)-(n)/(m+n)=1-1=0` Hence, the condition is necessary. |
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| 452. |
let us define , the length of a vector as `|a| + |b| +|c| `. this definition coincides with the usual definition of the length of a vector `ahati + bhatj + chatk ` ifA. `a= b= c=0`B. any two of a, b and c are zeroC. any one of a, b and c is zeroD. `a+ b+ c=0` |
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Answer» Correct Answer - B `|a| + |b| + |c| = sqrt(a^(2) + b^(2) + c^(2))` `hArr 2 |ab|+ 2|bc| + 2|ca|=0` `hArr ab = bc =ca =0 hArr ` any two of a, b and c are zero. |
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| 453. |
If ` vec a `is a non zero vecrtor iof modulus ` vec a a n d m`is a non zero scalar such that `m a`is a unit vector, write the value of `mdot`A. `m = pm 1 `B. `a = |m|`C. `a = 1//|m|`D. `a = 1/m` |
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Answer» Correct Answer - C `m veca` is a unit vector if and only if `|mveca| = 1 rArr |m| |veca| = 1 rArr |m| a =1 rArr a = (1)/(|m|)` |
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| 454. |
The vectors `veca and vecb ` are non collinear. Find for what value of x the vectors `vecc=(x-2)veca+vecb and vecd=(2x+1) veca-vecb` are collinear.? |
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Answer» Both the vectors `vecc and vecd` are non-zero as the cofficients of `vecb` in the both are non-zero. Two vectors `vecc and vecd` are collinear if one of them is a linear multiple of the other. Therefore, `" "vecd=lamdavecc` or `" "(2x+1)veca-vecb=lamda{(x-2)veca+vecb}" "`(i) or `" "{(2x+1)-lamda(x-2)}veca-(1+lamda)vecb=0` The above expression is of the form `pveca+qvecb=0,` where `veca and vecb` are non-collinear, and hence we have `p=0 and q=0`. Therefore, `" "2x+1-lamda(x-2)=0" "`(ii) and `" "1+lamda=0" "`(iii) From (iii), `lamda=-1`, and putting this value in (i), we get `x=(1)/(3)` Alternate method : `" "vecc=(x-2)veca+vecb and vecd = (2x+1)veca-vecb` are collinear. If `(x-2)/(2x+1)=(1)/(-1)`, then `x=(1)/(3)` |
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| 455. |
A, B, C and D have position vectors `veca, vecb, vecc and vecd`, repectively, such that `veca-vecb = 2(vecd-vecc)`. ThenA. AB and CD bisect each otherB. BD and AC bisect each otherC. AB and CD trisect each otherD. BD and AC trisect each other |
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Answer» Correct Answer - D `veca-vecb= 2(vecd-vecc)` `therefore (veca+ 2vecc)/(2+1) = (vecb+ 2vecd)/(2+1)` ltBrgt Hence, AC and BD trisect each other as L.H.S is the position vector of a point trisecting A an C, and R.H.S that of B and D. |
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| 456. |
If `veca, vecb, vecc, vecd` are the position vectors of points `A, B, C and D`, respectively referred to the same origin O such that no three of these points are collinear and `veca+vecc=vecb+vecd`, then prove that quadrilateral `ABCD` is a parallelogram. |
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Answer» Since `veca+vecc=vecb+vecd`, we have `" "(veca+vecc)/(2)=(vecb+vecd)/(2)` i.e., Midpoint of `AC` and `BD` coincide. Hence, quadrilateral `ABCD` is a parallelogram. |
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| 457. |
The position vectors of points A and B w.r.t. the origin are `veca=hati+3hatj-2hatk and vecb=3hati+hatj -2hatk`, respectively. Determine vector `vec(OP)` which bisects angle `AOB`, where P is a point on AB. |
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Answer» `|vec(OA)|= |vec(OB)| = sqrt(14)` `Delta AOB` is isosceles. Hence, the bisector of angle `AOB` will bisect the base `AB`. Hence, P is the midpoint `(2, 2, -2)` of AB. Therefore, `" "vec(OP)= 2(hati+hatj-hatk)` |
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| 458. |
The volume of a parallelopiped whose coterminous edges are `2veca , 2vecb , 2 vec c ` , isA. `2 [(a,b,c)]`B. `4 [(a,b,c)]`C. `8 [(a, b, c)]`D. `[(a,b,c)]` |
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Answer» Correct Answer - C Volume of parallelopiped=`["2a 2b 2c"]=8["abc"]` |
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| 459. |
If the volume of parallelopiped with coterminous edges `4hati+5hatj+hatk, - hatj+hatk and 3hati+9hatj+phatk` is 34 cu units, then p is equal toA. 4B. `-13`C. 13D. 6 |
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Answer» Correct Answer - B Since, volume of parallelopiped `=34` `:. |(4,5,1),(0,-1,1),(3,9,p)|=34` `implies 4(-p-9)-5(-3)+1(3)=34` `implies -4p-36+15+3=34` `implies 4p=-52` `implies p=-13` |
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| 460. |
If the volume of the parallelopiped with a, b and c as coterminous edges is 40 cubic units, then the volume of the parallelopiped having `b+c, c+a and a+b` as coterminous edges in cubic units isA. 80B. 120C. 160D. 40 |
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Answer» Correct Answer - A Given , volume of parallelopied `[abc]=40` `therefore` Volume of parallelopiped `=[b+c c+a a +b]=2[abc]` `2xx40=80` cu units. . |
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| 461. |
The position vectors of the points A, B and C are vector (2i + j - k),(3i - 2j + k) and (i+4j-3k) respectively. Show that the points A, B and C are collinear. |
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Answer» A = \(2\vec{i}+\vec{j}-\vec{k}\) B = \(3\vec{i}-2\vec{j}+\vec{k}\) C = \(\vec{i}+4\vec{j}-3\vec{k}\) \(\vec{AB}\) = \((3\vec{i}-2\vec{j}+\vec{k})\) - \((2\vec{i}+\vec{j}-\vec{k})\) = \(\vec{i}-3\vec{j}+2\vec{k}\) \(\vec{BC}\) \((\vec{i}+4\vec{j}-3\vec{k})\) - \((3\vec{i}-2\vec{j}+\vec{k})\) = \(-2\vec{i}+6\vec{j}-4\vec{k}\) (-3)\(\vec{AB}\) = \(\vec{BC}\) So, the points A, B and C are collinear. |
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| 462. |
Show that the points A, B and C having position vectors (i + 2j + 7k),(2i + 6j + 3k) and (3i + 10j - 3k) respectively, are collinear. |
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Answer» A = \(\vec{i}+2\vec{j}+7\vec{k}\) B = \(2\vec{i}+6\vec{j}+2\vec{k}\) C = \(3\vec{i}+10\vec{j}-3\vec{k}\) \(\vec{AB}\) = \((2\vec{i}+6\vec{j}+2\vec{k})\) - \((\vec{i}+2\vec{j}+7\vec{k})\) = \(\vec{i}+4\vec{j}-5\vec{k}\) \(\vec{BC}\) \((3\vec{i}+10\vec{j}-3\vec{k})\) - \((2\vec{i}+6\vec{j}+2\vec{k})\) = \(\vec{i}+4\vec{j}-5\vec{k}\) \(\vec{AB}\) = \(\vec{BC}\) So, the points A, B and C are collinear. |
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| 463. |
Two persons are raising a load pulling at an angle of each other. If they exert forces of `30 N` and ` 60 N` respectively and their effective pull is at right angles to the direction of the pull of the first person, what is the angle between their pulling forces ? What is the effective pull |
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Answer» Correct Answer - `120^(@), 30 sqrt(3)N` |
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| 464. |
Let `a=2hati+hatk, b=hati+hatj+hatk and c=4hati-3hatj+7hatk`. If r is a vector such that `rxxb=cxxb` and `r.a=0`, then value of r.b isA. 7B. `-7`C. `-5`D. 5 |
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Answer» Correct Answer - A Let `r=xhat(i)+yhat(j)+zhat(k)` Given `rxxb=cxxb` `rArr (xhat(i)+yhat(j)+zhat(j))xx(hat(i)+hat(j)+hat(k))` `=(4hat(i)-3hat(j)+7hat(k))xx(hat(i)+hat(j)+hat(k))` `rArr (y-z)hat(i)-(x-z)hat(j)+(x-y)hat(k)=-10hat(i)+3hat(j)+7hat(k)` `because|{:(hat(i),hat(j),hat(j)),(x,y,z),(1,1,1):}|=hat(i)(y-z)-hat(j)(x-z)+hat(k)(y-z)` `rArr|{:(hat(i),hat(j),hat(j)),(4,-3,7),(1,1,1):}|=hat(i)(-10)-hat(j)(-3)+hat(k)(7)` `rArr y-z=-10,-(x-z)=3,x-y=7` `rArr y-z=-10,-x+z=3,x-y=7`....(i) and `r.a=0` `rArr (xhat(i)+yhat(j)+zhat(k)).(2hat(i)+hat(k))` `rArr 2x+z=0` ......(ii) From Eqs.(i) and (ii) , we get `x=-1,y=-8,z=2` `therefore r.b=(-hat(i)-8hat(j)+2hat(k)).(hat(i)+hat(j)+hat(k))` `=-1-8+2=-7`. |
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| 465. |
When a force vector `vecF=(hati+2hatj+hatk)`N acts on a body and produces a displacement of `vecS=(4hati+hatj+7hatk)` m, then the work done isA. 9JB. 13JC. 5JD. 1J |
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Answer» Correct Answer - B `W=vecF.vecS.` |
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| 466. |
A force `vecF=3hati+chatj+2hatk` acting on a particle causes a displacement `vecd=-4hati+2hatj-3hatk`. If the work done is 6 J. then the value of c will beA. 12B. 6C. 1D. 0 |
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Answer» Correct Answer - A |
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| 467. |
A force `vecF=3hati+chatj+2hatkN` acting on a particle causes a displacement `vecS=-4hati+2hatj-3hatk` m. if the workdone is 6 joule, the value of of c is |
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Answer» Correct Answer - C `W=vecF.vecS` |
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| 468. |
If force `(vecF)=4hati+4hatj` and displacement `(vecs)=3hati+6hatk` then the work done isA. `4xx3`B. `5xx6`C. `6xx3`D. `4xx6` |
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Answer» Correct Answer - A |
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| 469. |
A force `vecF=5hati+6hatj+4hatk` acting on a body, produces a displacement `vecS=6hati-5hatk`. Work done by the force isA. 10 unitsB. 18 unitsC. 11 unitsD. 5 units |
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Answer» Correct Answer - A |
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| 470. |
Two constat forece `vecF_(1) = (2hati + 3hatj + 3hatk)` newton and `vecF_(2) = (5hati - 6hatj - 2hatk)` newton act toghter on a particle during its displacement from the position `(20hati + 15hatj)`m to `8vecKm`. Calculate the work done. |
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Answer» Correct Answer - `-87` joule `vecF_(1) = 2hati + 3hatj + 3hatk` `vecF_(2)= 5hati - 6hatj + hatk` `vecF_("net") = 7 hati - 3hatj + hatk` `vecr_(1) = 20hati + 15hatj" "vecr_(2) = 8hatk` Work done ` = (vecF_("net") ).(vecr_("net"))= - 87 "joule"` |
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| 471. |
If the resultant of three forces `vecF_1 = p hati + 3hatj -hatk, vecF_2 = 6hati-hatk and vecF_3 =-5hati +hatj +2hatk` acting on a particle has a magnitude equal to 5 units, then the value of `p` isA. `-6`B. `-4`C. 2D. 4 |
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Answer» Correct Answer - B::C Let `vecR` be the resultant. Then `vecR= vecF_1 + vecF_2 + vecF_3 = (p+1) veci + 4hatj` Given, `|vecR| =5`. Therefore, `(p+1)^(2)+ 16 = 25 or `p+1 = pm3` or `p=2, -4` |
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| 472. |
A paricle starting from the origin (0,0) moves in a straight line in `(x,y)` plane. Its coordinates at a later time are `(sqrt(3),3)`. The path of the particle makes with the x-axis an angle ofA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `0^(@)` |
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Answer» Correct Answer - C `vecd = vecAB=(sqrt(3)hati - 3hatj)` `costheta=(dhati)/(|vecd||hati|)=(sqrt(3))/(sqrt(12)xx1)=(1)/(2)` `" "theta=60^(@)` |
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| 473. |
The value of x and y for which vector `A= (6hati+xhatj-2hatk)and B=(5hati+6hatj-yhatk)` may be parallel areA. `x=0,y=(2)/(3)`B. `x=(36)/(5),y=(5)/(3)`C. `x=(15)/(3),y=(23)/(3)`D. `x=(36)/(5),y=(15)/(4)` |
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Answer» Correct Answer - B (b) For vector to be parallel ,ratio of coefficient should be same . `implies(6)/(5)=(x)/(6)=(-2)/(-y)impliesx=(36)/(5),y=(10)/(6)=(5)/(3)` |
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| 474. |
A and B are two vector given by `A= 2hati +3hatjand B=2hati+4hatj` The magnitude to the component of A along B isA. `(5)/(sqrt(2))`B. `(3)/(sqrt(2))`C. `(8)/(sqrt(5))`D. `(5)/(sqrt(13))` |
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Answer» Correct Answer - C (c ) Component of A along B=`(A.B)(B)` `=(4+12)/(sqrt((2)^(2)+(4)^(2)))=(16)/(2sqrt(5))=(8)/(sqrt(5))` |
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| 475. |
If ` vecalpha+ vecbeta+ vecgamma=a vecdeltaa n d vecbeta+ vecgamma+ vecdelta=b vecalpha, vecalphaa n d vecdelta`are non-colliner, then ` vecalpha+ vecbeta+ vecgamma+ vecdelta`equalsa. `a vecalpha`b. `b vecdelta`c. `0`d. `(a+b) vecgamma`A. `a vecalpha `B. `b vecdelta`C. `0`D. `(a+ b)vecgamma` |
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Answer» Correct Answer - C Given `vecalpha + vecbeta + vecgamma = a vecdelta" "` (i) `" "vecbeta + vecgamma + vecdelta = b vecalpha" "` (ii) From (i), ` vecalpha + vecbeta + vecdelta = (a+1) vecdelta" "` (iii) From (ii), `vecalpha + vecbeta + vecgamma + vecdelta = (b+1)vecalpha " "` (iv) From (iii) and (iv), we get `" "(a+1) vecdelta = (b+1) vecalpha " "` (v) Since `vecalpha` is not parallel to `vecdelta`, From (v), `a=1 =0 and b+1 =0` From (iii), `vecalpha + vecbeta + vecgamma + vecdelta =0` |
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| 476. |
If a and b are two vectors.then the value of `(a+b)xx(a-b)` isA. `2(bxxa)`B. `-2(bxxa)`C. `bxxa`D. `bxxa` |
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Answer» Correct Answer - A `(a+b)xx(a-b)=axxa-axxb+axxa-bxxb` `=2(bxxa)` |
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| 477. |
Three are N coplancar vectors each of magnitude V Each vector is inclined to the preceding vector atangle `(2pi)/(N)` What is the magnitude of their resultant ?A. `(V)/(N)`B. VC. zeroD. `(N)/(V)` |
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Answer» Correct Answer - C (c ) Since of N- coplancer vector is inclined at `(2pi)/(N)` to the preceding Hence ,they will form a closed polygon ,Therefore their resultant must be zero. |
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| 478. |
Find the vector sum of `N` coplanar forces, each of the magnitude `F`,when each force makes an angle of `2pi//N` with that preceding it.A. ZeroB. `1000 N`C. `500 N`D. `250 N` |
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Answer» Correct Answer - A Total angle `= 100xx(pi)/(50)= 2 pi` So all the forces will pass through one point and all forces will be balnced. i.e., their resultant will be zero. |
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| 479. |
The sum and difference of two vectors `vec(A)` and `vec(B)` are `vec(A) +vec(B) = 2hati +6 hatj + hatk` and `vec(A) - vec(B) = 4 hati +2 hatj - 11 hatk`. Find the magnitude of each vector and their scalar product `vec(A) vec(B)`. |
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Answer» Correct Answer - `sqrt(50),sqrt(41),-25` |
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| 480. |
A force `vec(F) = 5hati + 4hatj` newton displaces a body through `vec(S) = 3hati +4 hatk` metre in 3s. Find the power. |
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Answer» Correct Answer - 5W |
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| 481. |
The axes of coordinates arerotated about the z-axis though an angle of `pi//4`in the anticlockwise direction and thecomponents of a vector are 2`sqrt(2),`3`sqrt(2), 4.`Prove that the components of the same vectorin the original system are -1,5,4. |
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Answer» If `hati, hatj, hatk` are the new unit vectors along the coordinate axes, then `" " veca = 2sqrt2 hati + 2sqrt2hatj + 4hatk" "` (i) `hati, hatj, hatk` are obtained by rotating by `45^(@)` about the z-axis. Then `hati` is replaced by `hati cos 45^(@) + hatj sin 45^(@) = (hati + hatj)/(sqrt2)` and `hatj` is replaced by `" "-haticos45^(@)+hatj sin 45^(@) = (-hati + hatj)/(sqrt2)` ` " " hatk = hatk`, `" " veca = 2sqrt2 [(hati +hatj)/(sqrt2)] +3sqrt2[(-hati +hatj)/(sqrt2)] + 4hatk` `" " = (2-3)hati + (2+3)hatj + 4hatk` `" " = - hati +5hatj + 4hatk` |
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| 482. |
The vector ` vec a`has the components `2p`and 1 w.r.t. arectangular Cartesian system. This system is rotated through a certain angelabout the origin in the counterclockwise sense. If, with respect to a newsystem, ` vec a`has components `(p+1)a n d1`, then `p`is equal toa. `-4`b. `-1//3`c. `1`d. `2`A. `-1`B. `-1//3`C. `1`D. `2` |
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Answer» Correct Answer - B::C We have , `veca = 2phati+hatj` on rotaion, let `vecb` be the vector with components (`p+1`) and 1 so that `vecb= (p+1)hati+hatj`. Now, `|veca|= |vecb| or a^(2)= b^(2)` `rArr 4p^(2) +1 = (p+1)^(2) +1` or `4p^(2) = (p+1)^(2)` or `2p = pm (p+1)` `rArr 3p = -1 or p =1 ` `therefore p =-1//3 or p =1 ` |
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| 483. |
The components of a vector along the x- and y- directions are `(n+1)` and 1, respectively. If the coordinate system is rotated by an angle `theta=60^(@)`, then the components change to `n` and 3. The value of `n` isA. 2B. `cos 60^(@)`C. `sin 60^(@)`D. 3.5 |
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Answer» Correct Answer - D The length of the vector is not changed by the rotation of the coordinate axes. `sqrt((n+1)^(2)+1^(2))=sqrt(n^(2)+3^(2)` or `n^(2)+2n+2=n^(2)+9` or `2n=7` or `n=3.5` |
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| 484. |
Find the resultant of the three vectors `vec(OA), vec(OB)` and `vec(OC)` shown in figure. Radius of the circle is R. A. `2R`B. `R(1+sqrt(2))`C. `Rsqrt(2)`D. `R(sqrt(2)-1)` |
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Answer» Correct Answer - B `R_(net)=R+sqrt(R^(2)+R^(2))= R+sqrt(2)R= R(sqrt(2)+1)` |
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| 485. |
Find the resultant of the three vectors `vec(OA), vec(OB)` and `vec(OC)` shown in figure. Radius of the circle is R. A. 2RB. `R (1+sqrt(2))`C. `Rsqrt(2)`D. `R (sqrt(2)-1)` |
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Answer» Correct Answer - B `(b) R_("net") =R+sqrt(R^(2)+R^(2))=R+sqrt(2R)=R(sqrt(2)+1)` |
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| 486. |
If three vectors `2hat(i)-hat(j)-hat(k), hat(i)+2hat(j)-3hat(k)` and `3hat(i)+lambda hat(j)+5hat(k)` are coplanar, then the value of `lambda` isA. `-4`B. `-2`C. `-1`D. `-8` |
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Answer» Correct Answer - D Let `a=2 hati-hatj-hatk, ` `b=hati+2hatj-3hatkand c=3 hati+lamdahatj+5 hatk` It these vectors are coplanar, then [a b c] =0 `implies|{:(2,-1,-1),(1,2,-3),(3, lamda,5):}|=0` `implies2(10+3 lamda)+(5+9)-(lamda-6)=0` `=20+6 lamda+14-lamda +6=0` `implies5 lamda +40=0 implies lamda=-8` |
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| 487. |
Area of rhombus is ......., where diagonals are `a=2hat(i)-3hat(j)+5hat(k)` and `b=-hat(i)+hat(j)+hat(k)`A. `sqrt(21.5)`B. `sqrt(31.5)`C. `sqrt(28.5)`D. `sqrt(38.5)` |
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Answer» Correct Answer - C Given diaglonals of a rhombus are `a=2hati-3hatj+5hatk" and "b=-hati+hatj+hatk` `:." "` Area of rhombus `= 1/2|axxb|` `=1/2|-8hati-7hatj-hatk|=1/2sqrt(114)=sqrt(28.5)` |
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| 488. |
The angles with a vector `hat(i)+hat(j)+sqrt(2hat(k))` makes with X,Y and Z axes respectively areA. `60^(@),60^(@),60^(@)`B. `45^(@),45^(@),45^(@)`C. `60^(@),60^(@),45^(@)`D. `45^(@),45^(@),60^(@)` |
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Answer» Correct Answer - A |
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| 489. |
If vector `hat(i)+hat(j)+hat(k), hat(i)-hat(j)+hat(k)` and `2hat(i)+3hat(j)+lambda hat(k)` are coplanar, then `lambda` is equal toA. `-2`B. 3C. 2D. `-3` |
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Answer» Correct Answer - C Since, vectors `hati+hatj+hatk, hati-hatj+hatk" and "2hati+3hatj+lamdahatk` are coplanar, therefore `|{:(1,1,1),(1,-1,1),(2,3,lamda):}|=0` `rArr" "2(-lamda-3)-1(lamda-2)+1(3+2)=0` `rArr" "-lamda-3 -lamda+2+5=0` `rArr" "2lamda=4lamda=2` |
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| 490. |
The angles with a vector `hat(i)+hat(j)+sqrt(2hat(k))` makes with X,Y and Z axes respectively areA. `60^(@),60^(@),60^(@)`B. `45^(@), 45^(@), 45^(@)`C. `60^(@), 60^(@),45^(@)`D. `45^(@), 45^(@), 60^(@)` |
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Answer» Correct Answer - C `vec(R )= hat(i)+hat(j)+sqrt(2)hat(k)` Comparing the given vector `vec(R )= R_(x)hat(i)+R_(y)hat(j)+R_(s)hat(k)` `R_(x)= 1, R_(y)= 1, sqrt(2) and |vec(R )|= sqrt(R_(x)^(2)+R_(y)^(2)+R_(z)^(2))=2` `cos alpha= (R_(x))/R= 1/2implies alpha= 60^(@)` `cos beta =(R_(y))/R= 1/2 implies beta= 60^(@)` `cos gamma= (R_(z))/R= 1/(sqrt(2))implies gamma= 45^(@)` |
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| 491. |
A unit vector in the dirction of resultant vector of `vec(A)= -2hat(i)+3hat(j)+hat(k)` and `vec(B)= hat(i)+2hat(j)-4hat(k)` isA. `(-2hat(i)+3hat(j)+hat(k))/(sqrt(35))`B. `(-hat(i)+2hat(j)+4hat(k))/(sqrt(35))`C. `(-hat(i)+5hat(j)-3hat(k))/(sqrt(35))`D. `(-3hat(i)+hat(j)-5hat(k))/(sqrt(35))` |
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Answer» Correct Answer - C Here, `vec(A)= -2hat(i)+3hat(j)+hat(k)` `vec(B)= hat(i)+2hat(j)-4hat(k)` The resultant vector of `vec(A)` and `vec(B)` is `vec(R )= vec(A)+vec(B)= (2hat(i)+3hat(j)+hat(k))+(hat(i)+2hat(j)-4hat(k))` `= -hat(i)+5hat(j)-3hat(k)` `|vec(R )= sqrt((-1)^(2)+(5)^(2)+(-3)^(2))= sqrt(1+25+9)= sqrt(35)` Unit vector in the direction of resultant vector of `vec(A)` and `vec(B)` is `hat(R )= (vec(R ))/(|R|)= (-hat(i)+5hat(j)-3hat(k))/(sqrt(35))` |
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| 492. |
The vector `vec(P)= ahat(i)+ahat(j)+3hat(k)` and `vec(Q)= ahat(i)-2hat(j)-hat(k)` are perpendicular to each other. The positive value of `a` isA. 3B. 4C. 9D. 13 |
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Answer» Correct Answer - A `vec(P).vec(Q)=0 :. a^(2)-2a-3=0 implies a=3` |
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| 493. |
Given `vec(A)=2hat(i)+phat(j)+qhat(k)` and `vec(B)=5hat(i)+7hat(j)+3hat(k)`. If `vec(A)||vec(B)`, then the values of p and q are, respectively,A. `14/5and6/5`B. `14/3 and 6/5`C. `6/5 and 1/3`D. `3/4 and 1/4` |
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Answer» Correct Answer - A `vec(A)xxvec(B)=|(hat(i), hat(j), hat(k)) ,(2,p,q), (5,7 ,3)|=0` or `hat(i)(3p-7q)+hat(j)(5q-6)+hat(k)(14-5p)=0` `3p=7q,5q-6=0` or `q=6/5` |
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| 494. |
Given: `vec(A)=Acos theta hat(i)+Asin theta hat(j)`. A vector `vec(B)`, which is perpendicular to `vec(A)`,is given byA. `hat(i)B cos theta+ hat(j) Bsin theta`B. `hat(i) Bsin theta+hat(j) B cos theta`C. `hat(i) Bsin theta+ hat(j) B cos theta`D. `hat(i) B cos theta- hat(j) B sin theta` |
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Answer» Correct Answer - C Dot product of two perpendicular Vector will be zero. |
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| 495. |
If `vec(A)` and `vec(B)` are perpendicular Vectors and vector `vec(A)= 5hat(i)+7hat(j)-3hat(k)` and `vec(B)= 2hat(i)+2hat(j)-ahat(k)`. The value of `a` isA. -2B. 8C. 7D. -8 |
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Answer» Correct Answer - D For perpendicular vector `vec(A).vec(B)=0` `implies (5hat(i)+7hat(j)-3hat(k)).(2hat(i)+2hat(j)-ahat(k))=0` `implies 10+14+3a=0 implies a= -8` |
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| 496. |
You are given vector `vec(A)=5hat(i)-6.5hat(j)` and `vec(B)=10hat(i)-7hat(j)`. A third vector `vec(C )` lies in the `x-y` plane. Vector `(C )` is perpendicular to vector `vec(A)` and the scalar product of `vec(C )` with `vec(B)` is 15. From this information, find the component of `vec(C )` |
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Answer» Correct Answer - `y=3/4`, `x=39/40` Since `vec(C )` lies in x-y plane, let us assume `vec(C )=xhat(i)+yhat(j)`. Now `vec(C ).vec(A)=orArr5x-6.5y=0`..(i) and `vec(C ).vec(B)=15rArr10x+7y=15`..(ii) From Eqs.(i) and (ii),`y=3/4,x=39/40` |
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| 497. |
A cube is placed so that one corner is at the origin and three edges are along the x-,y-, and ,z-axes of a coordinate system (figure).Use vector to compute a.The angle between the edge along the z-axis (line ab) and the diagonal from the origin to the opposite corner (line ad). b. The angle between line ac (the diagonal of a face ) and line ad. |
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Answer» Correct Answer - a.`cos^(-1)(1/sqrt(3))`; b. cos^(-1)(sqrt(2)/sqrt(3)) a.Let side of the cube is `d`,then `vec(ab)=dhat(k)` and `vec(ad)=dhat(i)+dhat(j)+dhat(k)` `cos theta=(vec(ab).vec(ad))/(|vec(ab)||vec(ad)|)=(d^(2))/(dsqrt(3d))=1/sqrt(3)rArr theta =cos^(-1)(1/sqrt(3))` `vec(ac)=dhat(j)+dhat(k)` `cos theta=(vec(ac).vec(ad))/(|vec(ac)||vec(ad)|)=(2d^(2))/sqrt(2dsqrt(3d))=sqrt(2)/sqrt(3)` `rArrcos^(-1)((sqrt(2))/sqrt(2))` |
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| 498. |
If `vecA=3hati-4hatj` and `hatB=-hati-4hatj`, calculate the direction of `vecA+vecB`A. `tan^(-1)(4)` with positive X-axis in close wiseB. `tan^(-1)(4)` with X-axis in clock wiseC. `tan^(-1)(4)` with positive X-axis in anticlock wiseD. `tan^(-1)(4)` with negative X-axis in anticlock wise |
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Answer» Correct Answer - A `vecR=vecA+vecB,vecR=R_(x)hati+R_(y)hatj , Tan alpha=(R_(y))/(R_(x))` |
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| 499. |
A body is at rest under the action of three forces, two of which are `vecF_(1)=4hati, vecF_(2)=6hatj`, the third force isA. `4hati+6hatj`B. `4hati-6hatj`C. `-4hati+6hatj`D. `-4hati-6hatj` |
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Answer» Correct Answer - D |
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| 500. |
If vectors A and B be respectively equal to `3hati - 4hatj + 5hatk and 2hati + 3hatj - 4hatk.` Find the unit vector parallel t A + B |
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Answer» Correct Answer - A::B See the hint of Q-23 of objective type problems. |
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