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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.
| 651. |
If the position vectors off A,B,C and D are `2hati+hatj,hati-3hatj,3hati+2hatj and hati+lamdahatj`, respectively and `AB||CD`, then `lamda` will beA. `-7`B. 7C. `-6`D. None of these |
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Answer» Correct Answer - C Now, `PQ= (hati - 3 hatj)-(2 hati + hatj) = - hati-4hatj` `and RS= - 2hati + (mu - 2)hatj` Given, `PQ||RS rArr PQ= lambda RS` `-hati - 4 hatj =[ lambda { - 2hati +(mu-2)hatj}]` `rArr lambda = 1//2 and mu = -6 ` |
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| 652. |
Statement 1: if threepoints `P ,Qa n dR`have position vectors ` vec a , vec b ,a n d vec c`, respectively, and `2 vec a+3 vec b-5 vec c=0,`then the points `P ,Q ,a n dR`must be collinear.Statement 2: If for threepoints `A ,B ,a n dC , vec A B=lambda vec A C ,`then points `A ,B ,a n dC`must be collinear.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
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Answer» Correct Answer - A `2veca + 3 vecb-5 vecc=0` `rArr 3(vecb- veca) = 5(vecc- veca) rArr vec(AB) = (5)/(3) vec(AC)` Hence, `vec(AB) and vec(AC)` must be parallel since there is a common point A. The points A, B and C must be collinear. |
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| 653. |
Two vector A and B have equal magnitudes. Then the vector `A+B` is perpendicular toA. `AxxB`B. `A-B`C. `3A-3B`D. All of these |
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Answer» Correct Answer - A |
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| 654. |
The vector projection of a vector `3hat(i)+4hat(k)` on y-axis isA. 5B. 4C. 3D. Zero |
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Answer» Correct Answer - D As the multiple of `hat(j)` in the given vector therfore this vector lies in XZ plane and projection of this vector on y-axis is zero. |
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| 655. |
Two vectors of equal magnitudes have a resultant equle to either of them, than the angel between them will beA. `30^@`B. `120^@`C. `60^@`D. `150^@` |
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Answer» Correct Answer - B `R = sqrt(A^2 +A^2 + 2AA cos theta)` ` theta =120^@.` |
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| 656. |
For the resultant of two vectors to be maximum , what must be the angle between them ?A. `0^(@)`B. `60^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - A Because ` cos 0^(@) = 1` (maximum). |
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| 657. |
Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angel between them if the magnitude of the resultant is (a). 1 unit (b). 5 unit and (c). 7 unit. |
| Answer» Correct Answer - A::B::C | |
| 658. |
What vector must be added to the two vectors `hati-2hatj+2hatk` and `2hati+hatj-hatk`, so that the resultant may be a unit vector along `x`- axisA. `2hati+hatj+k`B. `-2hati+hatj-hatk`C. `2hati-hatj+hatk`D. `-2hati-hatj-hatk` |
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Answer» Correct Answer - B |
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| 659. |
For the resultant of two vectors to be maximum , what must be the angle between them ? |
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Answer» Correct Answer - A (a) Resultant of two vector will be maxium when they are parallel, i.e,angle between them is zero. |
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| 660. |
Given that `P+Q+R=0`. Which of the following statement is true?A. `|P|+|Q|=|R|`B. `|P+Q|=|R|`C. `|P|-|Q|=|R|`D. `|P-Q|=|R|` |
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Answer» Correct Answer - B (b)`If P+Q=0` `then|P+Q|=|R|` |
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| 661. |
For the resultant of two vectors to be maximum , what must be the angle between them ? |
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Answer» Correct Answer - A (a) Resultant of two vector will be maxium when they are parallel, i.e,angle between them is zero. |
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| 662. |
Two vector having magnitude 8 and 10 can maximum and minium value of magnitude of their resultant asA. 12,6B. 10,3C. 18,2D. none of these |
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Answer» Correct Answer - C (C ) Maximum resultant `A=hati+hatj` `|A|=sqrt((1)^(2)+(1)^(2))=sqrt(2)` |
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| 663. |
Find the projection of `veca=2hati-hatj+hatk and vecb=hati-2hatj+hatk.` |
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Answer» projectin of A on B=A`cos theta(theta=`angle between Aand B ) `=(A.B)/(B)=(2hati-hatj+hatk).((hati-2hatj+hatk))/(sqrt((1)^(2)+(-2)^(2)+(1)^(2)))=(2+2+1)/(sqrt(6))=(5)/(sqrt(6))` |
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| 664. |
The angle between Vectors `(vec(A)xxvec(B))` and `(vec(B)xxvec(A))` isA. `pi//2`B. `pi//3`C. `pi`D. `pi//4` |
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Answer» Correct Answer - C We know that `vec(A)xxvec(B)= -(vec(B)xxvec(A))` because the angle between these two is always `90^(@)` But if the angle between `vec(A)` and `vec(B)` is `0 or pi`. Then `vec(A)xxvec(B)= vec(B)xxvec(A)=0` |
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| 665. |
The angle between Vectors `(vec(A)xxvec(B))` and `(vec(B)xxvec(A))` isA. ZeroB. `pi`C. `pi//4`D. `pi//2` |
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Answer» Correct Answer - B `vec(A)xxvec(B)` and `vec(B)xxvec(A)` are parallel and opposite to each other. So the angle will be `pi` |
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| 666. |
Two vector `vec(A)` and `vec(B)` have equal magnitudes. Then the vector `vec(A) + vec(B)` is perpendicular :A. `AxxB`B. `A-B`C. `3A-3B`D. All of these |
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Answer» Correct Answer - A `vec(A)xxvec(B)` is a vector perpendicular to plane `vec(A)+vec(B)` and hence perpendicular to `vec(A)+vec(B)`. |
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| 667. |
The value of `(vec(A)+vec(B))xx(vec(A)-vec(B))` is |
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Answer» Correct Answer - D `(vec(A)xxvec(B))xx(vec(A)-vec(B))=vec(A)xxvec(A)-vec(A)xxvec(B)+vec(B)xxvec(A)-vec(B)xxvec(B)` `=0-vec(A)xxvec(B)+vec(B)+vec(B)xxvec(A)-0` `=vec(B)xxvec(A)+vec(B)xxvec(A)= 2(vec(B)xxvec(A))` |
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| 668. |
For two vectors `vecA` and `vecB` which of the following relations are not commutative ?A. `vecA + vecB`B. `vecA- vecB`C. `vecA xx vecB`D. `vecA . vecB` |
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Answer» Correct Answer - B::C Vector difference and cross product are not commutative . |
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| 669. |
Assertion: Vector addition is commutative. Reason: `(vecA+vecB)!=(vecB+vecA)`A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 670. |
Assertion: Vector addition is commutative. Reason: Two vectors may be added graphically using head- to-tail method or parallelogram method.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 Vector addition is commutative i.e., `vec(A)+vec(B)= vec(B)+vec(A)` Where `vec(A)` and `vec(B)` are two vectors Two vectors `vec(A)` and `vec(B)` may be added graphically using head-to-tail method or parallelogram methdo. |
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| 671. |
Assertion: Two vectors are said to be like vectors if they have same direction but different magnitude. Reason: Vector quantities do not have specific direction.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 672. |
Assertion: Two vectors are said to be like vectors if they have same direction but different magnitude. Reason: Vector quantities do not have specific direction.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 - C If two vectors are in opposite direction, then they cannot be like vectors. |
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| 673. |
Assertion: Two vectors are said to be equal if , and only if, they have the same magnitude and the same dirction. Reason: Addition and subtraction of scalars make sense only for quantities with same units.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 Addition and subtraction of scalars make sense only for quantities with same unit. But you can multiply and divide sacalrs of different units. |
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| 674. |
An aeroplane moving in a circular path with a speed `250 km//h`. The change in velocity in half of the revolution is.A. 500 km/hrB. 250 km/hrC. 125 km/hrD. zero |
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Answer» Correct Answer - A `v_(i) = 250km//hr` `v_(r) = 2 sin .(theta)/(2)` `theta= pi` Then `v_(F) = 2 xx 250 xx sin .(pi)/(2)km//hr` `" "v_(F) = 500 km//hr` |
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| 675. |
Given : `vec A = hati + hatj +hatk and vec B =-hati-hatj-hatk` What is the angle between `(vec A - vec B) and vec A` ?A. `0^(@)`B. `180^(@)`C. `90^(@)`D. `60^(@)` |
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Answer» Correct Answer - A `(a) A-B=(hati+hatj+hatk)-(-hati-hatj-hatk)` `=2hatj+2hatj+2hatk=2A` `i.e A-B and A" parallel".` |
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| 676. |
Obtain the magnitude of 2A - 3B if `A = hati +hatJ- 2 hat k and b = 2 hat I - hat j+hat k` . |
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Answer» `2A-3B=(hati+hatj-2hatk)-3(2hatj-hatj+hatk)=-4hati+5hatj-7hatk` ` therefore` Magnitude of `2A-3B=sqrt((-4)^(2)+(5)^(2)+(-7)^(2))` `=sqrt(16+25+49)=sqrt(90)` |
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| 677. |
Assertion - Vector addition of two vector is always greater then their vector subtraction. Reason At `theta =90^(@)` , addition and subreaction of two vector are equal .A. If both Assetion and Reason are correct but Reason is the correct explanation of Assertion.B. If both Assetion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Asserion is true but Reason is falseD. If Asserion is false but Reason is true |
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Answer» Correct Answer - D (d) `|AxxB|=sqrt(A^(2)+B^(2)+2AB costheta)` `and |A-B|=sqrt(A^(2)+B^(2)-2AB cos theta)` `theta=90^(@)` `|AxxB|=|A-B|=sqrt(A^(2)+B^(2))` |
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| 678. |
Assertion - Finite angular disacement is not a vector quantiy. Reason It does not obey vector laws.A. If both Assetion and Reason are correct but Reason is the correct explanation of Assertion.B. If both Assetion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Asserion is true but Reason is falseD. If Asserion is false but Reason is true |
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Answer» Correct Answer - A (a) If a particle is roting about on axis on its elf the orientain of the particle is not same under a different order of addition of angular displacement ,so it cannot be regarded as vector quantity . `d theta+d theta_(2)=d theta_(2)+d theta_(1)` `but theta_(1)+theta_(2) netheta_(2)+theta_(1)` |
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| 679. |
Which of the following is the unit vector perpendicular to `vec(A)` and `vec(B)`?A. `(hat(A)xxhat(B))/(AB sin theta)`B. `(hat(A)xxhat(B))/(AB cos theta)`C. `(vec(A)xxvec(B))/(AB sin theta)`D. `(vec(A)xxvec(B))/(AB cos theta)` |
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Answer» Correct Answer - C Vector perpendicular to A and B, `vec(A)xxvec(B)= AB sin theta hat(n)` `:.` Unit vector perpendicular to A and B `hat(n)=(vec(A)xxvec(B))/(|vec(A)|xx|vec(B)|sin theta)` |
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| 680. |
Writedown a unit vector in XY-plane, making an angle of 30 with the positivedirection of x-axis. |
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Answer» Let, in XY-plane, OP makes `30^(@)` angle from X-axis, `60^(@)` angle from Y-axis and `90^(@)` from Z-axis. Therefore, direction cosines of vector OP are cos `30^(@)`, cos `60^(@)` and `90^(@)` i.e., `(sqrt(3))/(2),(1)/(2),0`. Therefore, OP `=(sqrt(3))/(2)hat(i)+(1)/(2)hat(j)` `|vec(OP)|-sqrt(((sqrt(3))/(2))^(2)+((1)/(2))^(2))` `= sqrt((3)/(4)+(1)/(4))=sqrt(1)=1` Therefore required unit vector in XY-plane is `(sqrt(3))/(2)hat(i) + (1)/(2)hat(j)`. |
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| 681. |
Find the value of `x`for which `x ( hat i+ hat j+ hat k)`is a unit vector. |
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Answer» Given that `x(hat(i) + hat(j) + hat(k))` is a unit vector, so `|xhat(i) + xhat(j) + xhat(k)|=1 implies sqrt(x^(2) + x^(2) +x^(2)) =1` `implies sqrt(3x^(2)) = 1 implies pm sqrt(3)x =1 implies x = pm (1)/(sqrt(3))` `therefore x = pm (1)/(sqrt(3))` |
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| 682. |
Find the scalar components and magnitude of the vector joining the points `P(x_1,y_1,z_1)`and `Q(x_2,y_2,z_2)` |
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Answer» Vector of joining `P(x_(1),y_(1),z_(1))` and `P(x_(1),y_(1),z_(1))` is : PQ = Position vector of Q - Position vector of P `=(x_(2)hat(i) + y_(2)hat(j) + z_(2)hat(k)) - (x_(1)hat(i) + y_(1)hat(j) + z_(1)hat(k))` `=(x_(2) - x_(1)) hat(i) + (y_(2) - y_(1))hat(j)+ (z_(2) - z_(1)) hat(k)` Scalar component of PQ are `x_(2)-x_(1), y_(2) -y_(1), z_(2)-z_(1)`. Magnitude of `vec(PQ)` `|vec(PQ)| = sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)+(z_(2)-z_(1))^(2))` |
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| 683. |
Let a, b, c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is(A) The arithmetic mean of a and b (B) The geometric mean of a and b (C) The harmonic man of a and b (D) 0 |
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Answer» Correct answer is (B) The geometric mean of a and b |
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| 684. |
Find the direction cosines of the vector joining the points `A(1, 2, -3) and B(-1, -2, 1)` directed from `A` to `B`. |
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Answer» The given points are `A(1, 2, -3) and B(-1, -2, 1)`. Therefore, `" "vec(AB)=(-1-1)hati+(-2-2)hatj+{1-(-3)}hatk` `" "=-2hati-4hatj+4hatk` `therfore" "|vec(AB)|=sqrt((-2)^(2)+(-4)^(2) + 4^(2)) =sqrt(4+ 16 +16) ` `" "=sqrt(36) = 6` Hence, the direction cosines of `vec(AB)` are `" "(-(2)/(6), -(4)/(6), (4)/(6))=(-(1)/(3), -(2)/(3), (2)/(3))`. |
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| 685. |
If ` vec A O+ vec O B= vec B O+ vec O C`, then `A ,Bn a dC`are (where `O`is the origin)a. coplanar b. collinear c.non-collinear d. none of these |
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Answer» We have, `vec(AO) + vec(OB) = vec(BO ) + vec(OC)` `rArr vec(AB) = vec(BC)` Since the initial point of `vec(BC)` is the terminal point of `vec(AB), A, B and C` are collinear. |
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| 686. |
The position vectors of `P and Q` are `5hati+4hatj+ahatk and -hati+2hatj-2hatk`, respectively. If the distance between them is 7, then find the value of `a`. |
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Answer» Given points are `P(5hati+4hatj+ahatk) and Q(-hati+2hatj-2hatk)` Now `" "vec(PQ)= (-hati+2hatj-2hatk)-(5hati+4hatj+ahatk)` `" "=(-6hati-2hatj-(2+a)hatk)` Given `|vec(PQ)|=7` `therefore" "sqrt(36+4+(a+2)^(2))=7` or `" "(a+2)^(2)=9 ` or `" "a+2=pm3` or `" "a=-5, 1`. |
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| 687. |
Given three points are `A(-3,-2,0),B(3,-3,1)a n dC(5,0,2)dot`Then find a vector having the same directionas that of ` vec A B`and magnitude equal to `| vec A C|dot` |
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Answer» Given points are `A(-3, -2, 0), B(3, -3, 1) and C(5, 0, 2)`. Therefore, `" "vec(AB)=6hati-hatj+hatk` and `" "vec(AC)=8hati+2hatj+2hatk` `" "|vec(AC)|=sqrt(64+4+4)=sqrt(72)` Unit vector in direction of `vec(AB)` is `(6hati-hatj+hatk)/(sqrt(38))` `therefore" "`Required vector `" "=(sqrt(72))/(sqrt(38))(6hati-hatj+hatk)= (6)/(sqrt(19))(6hati-hatj+hatk)` |
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| 688. |
Find the vector of magnitude 3, bisecting the angle between the vectors `veca=2hati+hatj-hatk and vecb=hati-2hatj+hatk`. |
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Answer» A vector bisecting the angle between `veca and vecb` is `(veca)/(|veca|)pm (vecb)/(|vecb|)`. Here `" "(2hati+hatj-hatk)/(sqrt6)pm (hati-2hatj+hatk)/(sqrt6)` i.e., `" "(3hati-hatj)/(sqrt6) or (hati+3hatj-2hatk)/(sqrt6)` A vector of magnitude 3 along these vectors is `" "(3(3hati-hatj))/(sqrt(10)) or (3(hati+3hatj-2hatk))/(sqrt(14))` |
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| 689. |
Show that the points `A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7)` are collinear, and find the ratio in which B divides AC. |
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Answer» The given point are `A(1, -2, -8), B(5, 0, -2) and C(11, 3, 7)`. Therefore, `vec(AB) = (5-1)hati + (0+2)hatj + (-2 + 8)hatk` `" " = 4hati + 2hatj + 6hatk` `vec(BC) = (11 -5)hati + (3-0)hatj + (7+2)hatk` `= 6hati + 3hatj + 9hatk` `vec(AC) = (11-1)hati + (3+1)hatj + (7+8)hatk` `" " = 10hati + 5hatj + 15hatk` `" "|vec(AB)| = sqrt(4^(2) + 2^(2) + 6^(2))= sqrt(16 + 4+ 36)` `" " = sqrt(56) = 2 sqrt(14)` `|vec(BC)| = sqrt(6^(2)+3^(2)+9^(2)) = sqrt(36 + 9 + 81)` `" "= sqrt(126) = 3sqrt(14)` `" "|vec(AC)| = sqrt(10^(2) + 5^(2) + 15^(2)) = sqrt(100 + 25 + 225)` `" " = sqrt(350) = 5sqrt(14)` `therefore " "|vec(AC)| = |vec(AB)| + |vec(BC)|` Thus, the given points A, B, and C are collinear, Also `2 |vec(BC)| = 3|vec(AB)|` Hence, point B divides AC in the ratio `2:3`. |
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| 690. |
If the vectors `a + lamda b + 3c, -2a + 3b - 4c` and `a - 3b + 5c` are coplanar and `a, b, c` are non-coplanar, then the value of `lambda` isA. 2B. `-1`C. 1D. `-2` |
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Answer» Correct Answer - D Since, the given three vectors are coplanar, therefore one of them shold be expressblle as a llinear Scalars x and y such that `a+lamda b+3c=x(-2a+3b-4c)+y(a-3b+5c)` On comparing the coefficient of a, b anc c on both sides, we get `-2x+y=1, 3x=lamda` and `-4x+5y=3` On sloving first and third equations, we get `x=-1/3, y=1/3` Since, the vectors are coplanar, therefore these values of x and y, also satisty the second equation i.e., `-1-1=lamda` `:." "lamda=-2` |
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| 691. |
Find a vector of magnitude 10 units and parallel to the resultant of the vectors p = 3i -5j +7k and vector q =2i +5j -5k. |
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Answer» Resultant vector \(\vec p\times\vec q\)\(=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-5&7\\2&5&-5\end{vmatrix}\) = \(\hat i(25-35)-\hat j(-15-14)+\hat k(15+10)\) = -10\(\hat i\) + 29\(\hat j\) + 25\(\hat k\) vector of magnitude 10 and parallel to resultant vector is \(\vec n=\frac{\pm10(-10\hat i+29\hat i+25\hat k)}{1-10\hat i+29\hat j+25\hat k }\) \(=\frac{\pm10(-10\hat i+29\hat j+25\hat k)}{\sqrt{(-10)^2+(29)^2+(25)^2}}\) \(=\frac{\pm10}{\sqrt{1566}}(-10\hat i+29\hat j+25\hat k)\) |
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| 692. |
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors `" "veca=2hati+3hatj-hatk and vecb=hati-2hatj+hatk`. |
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Answer» We have, `" "veca=2hati+3hatj-hatk and vecb=hati-2hatj+hatk` Let `vecc` be the resultant of `veca and vecb`. Then `" "vecc=veca+vecb=(2+1)hati+(3-2)hatj+(-1+1)hatk` `" "=3hati+hatj` `therefore" "|vecc|=sqrt(3^(2)+1^(2))=sqrt(9+1)=sqrt(10)` `therefore" "hatc=(vecc)/(|vecc|)=((3hati+hatj))/(sqrt(10))` Hence, the vector of magnitude 5 units and parallel to the resultant to vectors `veca and vecb` is `" "pm5*vecc=pm5*(1)/(sqrt(10))(3hati+hatj)` `" "=pm(3sqrt(10)hati)/(2)pm(sqrt(10))/(2) hatj`. |
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| 693. |
If a,b,c ar enon-coplanar vectors and `lamda` is a real number, then the vectors `a+2b+3c,lamdab+4c and (2lamda-1)c` are non-coplanar forA. no value of `lambda`B. all except one value of `lambda`C. all except two value of `lambda`D. all values of `lambda` |
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Answer» Correct Answer - C Let `alpha=a+2b+3c, beta=lambdab+4c` and `gamma (2lambda-1)c` Then, `[ alpha, beta, gamma] = | (1, 2 , 3),(0 ,lambda, 4),(0,0,(2lambda-1))|[ "a b c"]` `(alpha, beta, gamma] = lambda (2 lambda -1) [abc] ` Now, consider `lambda (2 lambda -1) =0` `lambda =0,(1)/(2) " " [ because [abc] ne 0]` Hence, `alpah, beta and gamma ` are non-coplanar for all value of `lambda` expcept two values `0 and (1)/(2)`. |
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| 694. |
If sum of two unit vectors is itself a unit vector, then the magnitude of their difference is(A) √2(B) √3(C) 1(D) 2 |
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Answer» Correct answer is (B) √3 |
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| 695. |
If vectors |a| = 3, |b| = 4, then the value of λ for which a + λb is perpendicular to a - λb, is(A) 9/16(B) 3/4(C) 3/2(D) 4/3 |
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Answer» Correct answer is (b) 3/4 |
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| 696. |
`ABCD` is a parallelogram whose diagonals meet at P. If O is a fixed point, then `bar(OA)+bar(OB)+bar(OC)+bar(OD)` equals :A. AB + ACB. 0C. 2 (AB + BC)D. AC + BD |
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Answer» Correct Answer - B since, the diagonals of a rhombus bisect each other OA = - OC and OB =- OD OA + OB +OC + OD = - OC - OD + OC + OD = 0 |
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| 697. |
Show that the line segment joining the origin to the point A(2, 1, 1) is perpendicular to the line segment joining the points B(3, 5, -1) and C(4,3, -1). |
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Answer» Given: O(0, 0, 0) and A(2, 1, 1) The line joining these two points is given by, OA = 2i + j +k B(3, 5, -1) and D(4, 3, -1), The line joining these two points, BC = i - 2j +0k To prove that the two lines are perpendicular we need to show that the angle between these two lines is So, OA.BC = 0 (dot product) Thus, (2i + j +k). (i - 2j +0k) = 2 - 2 + 0 = 0. Thus, the two lines are perpendicular |
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| 698. |
Three forces are acting on a particle as shown in the figure. To have the resultant force only along the Y-direction, the magnitude of the maximum additional force needed is A. `0.866 N`B. `1.732 N`C. `0.5 N`D. `4 N` |
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Answer» Correct Answer - C Resulant is along Y-axis. So, the component of resultant force along X-axis should be zero. `:. SigmaF_(x)=0` or `1 sin 30^(@)+ 2sin30^(@)- 4cos60^(@)+F_(0)=0` `:. F_(0)= -1/2-1+2= 1/2= 0.5N` |
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| 699. |
A moves with `65km//h` while `B` is coming back of `A` with `80km//h`. The relative velocity of `B` with respect to `A` isA. `80km//h`B. `60km//h`C. `15km//h`D. `145km//h` |
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Answer» Correct Answer - C |
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| 700. |
The sum of two forces at a point is 16N. if their resultant is normal to the smaller force and has a magnitude of 8N, then two forces areA. `6 N and 10 N`B. `8 N and 8 N`C. `4 N and 2 N`D. `2 N and 14 N` |
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Answer» Correct Answer - A `A+B= 16` (given)….(i) `tan alpha=(B sin theta)/(A+B cos theta)= tan 90^(@)` `:. A+B cos theta= 0impliescos theta= (-A)/B`.....(ii) `8= sqrt(A^(2)+B^(2)2AB cos theta)`....(iii) By solving equ.(i),(ii),(iii) we get `A=6 N,B= 10 N` |
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