704 + Interview Questions in VECTORS IN GENERAL AWARENESS Page 9 InterviewSolution

401.	Statement 1: In `"Delta"A B C , vec A B+ vec A B+ vec C A=0`Statement 2: If ` vec O A= vec a , vec O B= vec b ,t h e n vec A B= vec a+ vec b`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.
Answer» Correct Answer - C In `Delta ABC`,` vec(AB) + vec(BC) = vec(AC) =- vec(CA)` or `" " vec(AB) + vec(BC) + vec(CA) = vecO` `vec(OA) + vec(AB) = vec(OB)` is the triangle law of addition. Hence statement 1 is true and Statement 2 is false.

Discussion

402.	Statement 1: If `cosalpha,cosbeta,a n dcosgamma`are the direction cosinesof any line segment, then `cos^2alpha+cos^2beta+cos^2gamma=1.`Statement 2: If `cosalpha,cosbeta,a n dcosgamma`are the direction cosinesof any line segment, then `cos^2alpha+cos^2beta+cos^2gamma=1.`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.
Answer» Correct Answer - B Obviously, Statement 1 is true. `cos 2alpha + cos 2 beta + cos 2gamma ` `" " = 2cos^(2) alpha -1 = 2cos^(2) beta -1 + 2cos^(2)gamma -1 ` `" " = 2(cos^(2)alpha = cos^(2) beta + cos^(2)gamma) -3 =2 -3 =-1 ` Hence, Statement 2 is true but does not explain Statement 1 as it is resul derived usint the result in the statement.

Discussion

403.	Which of the following relation are worng ?A. `vecA + vecB= C`B. `vecA + vecB= C`C. `vecA + vecB= vecC`D. `A + vecB= vecC`
Answer» Correct Answer - A::B::D Vectors can be added with vectors only to give resultant vector.

Discussion

404.	Area of a parallelogram formed by vectors `(3hati-2hatj+hatk)m` and `(hati+2hatj+3hatk)` m as adjacent sides isA. 8B. `8sqrt(3)`C. `3sqrt(8)`D. 192
Answer» Correct Answer - B

Discussion

405.	Area of a parallelogram formed by vectors `(3hati-2hatj+hatk)m` and `(hati+2hatj+3hatk)` m as adjacent sides isA. `3sqrt(8)m^(2)`B. `24m^(2)`C. `8sqrt(3)m^(2)`D. `4sqrt(3)m^(2)`
Answer» Correct Answer - C Area of parallelogram `=\|vecAxxvecB\|`

Discussion

406.	The magnitude of scalar and vector products of two vectors are 144 and `48sqrt(3)` respectivley. What is the angle between the two vectors?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`
Answer» Correct Answer - A `ab cos theta=144, ab sin theta=48 sqrt(3)`

Discussion

407.	The magnitude of scalar and vector products of two vectors are `48sqrt(3)` and 144 respectively. What is the angle between the two vectors?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`
Answer» Correct Answer - C `ab cos theta=48sqrt(3), ab sin theta=144`

Discussion

408.	The modulus of the vector product of two vector is `(1)/(sqrt(3)` times their scalar product . The angle between vectors isA. `(pi)/(6)`B. `(pi)/(2)`C. `(pi)/(4)`D. `(pi)/(3)`
Answer» Correct Answer - A `(a) AB sintheta=(1)/(sqrt(3))ABcos theta` ` therefore tantheta=(1)/(sqrt(3))or theta=30^(@)`

Discussion

409.	The value of the sum of two vectors `vecA` and `vecB` with `theta` as the angle between them isA. `sqrt(A^(2)+B^(2)+2ABcos theta)`B. `sqrt(A^(@)-B^(2)+2ABcostheta)`C. `sqrt(A^(@)+B^(2)-2ABsintheta)`D. `sqrt(A^(2)+B^(2)+2ABsintheta)`
Answer» Correct Answer - A

Discussion

410.	Column-I gives operation of vectors `vecA and vecB` and column- II gives the angle `(theta)` between `vecA` and `vecB` (`vecA` and `vecB` are not zero vectors ).
Answer» Correct Answer - `a to r,s; b to p; c to q; d to p` `\|vecA xx vecB\| = vecA.vecB` `AB sin theta = AB cos theta ` `tan theta = 1 rArr theta = pi // 4 or 3pi//4` (b) `vecA xx vecB = vecB xx vecA ` only if `theta = 0^(@)` (c) `\|vecA + vecB\|= \|vecA - vecB\|` `sqrt(A^(2) +B^(2) + 2AB cos theta )= sqrt(A^(2) + B^(2) - 2AB cos theta)` `" "4 AB cos theta =0` `" "theta = pi//2` (d) `vecA + vecB = vecC` with A+B= C if `theta = 0^(@)`.

Discussion

411.	Find the torque of a force `vecF=3hati+2hatj+hatk` acting at the point `vecr=8hati+2hatj+3hatk` about originA. `14hati-38hatj+3hatk`B. `4hati+4hatj+6hatk`C. `-14hati+38hatj-16hatk`D. `-4hati-17hatj+22hatk`
Answer» Correct Answer - D Torque of the force, `vectau=vecrxxvecF`

Discussion

412.	If `theta` is the angle between unit vectors `vecA` and `vecB`, then `((1-vecA.vecB))/(1+vecA.vecB))` is equal toA. `tan^(2)(theta//2)`B. `sin^(2)(theta//2)`C. `cot^(2)(theta//2)`D. `cos^(2)(theta//2)`
Answer» Correct Answer - A `(1-vecA.vecB)/(1+vecA.vecB)=(1-cos theta)/(1+cos theta)`

Discussion

413.	A force of `2i+3j+2k N` acts on a body for 4 and produces a displacement of `3hati+4hatj+5hatk` m calculate the power?A. 5WB. 6WC. 7WD. 9W
Answer» Correct Answer - C `P=vecF.vecV=(vecF.vecS)/(t)`

Discussion

414.	What is the unit vector perpendicular to the following Vector `2hat(i)+2hat(j)-hat(k)` and `6hat(i)-3hat(j)+2hat(k)?`A. `(hat(i)+10hat(j)-18hat(k))/(5sqrt(17))`B. `(hat(i)-10hat(j)+18hat(k))/(5sqrt(17))`C. `(hat(i)-10hat(j)-18hat(k))/(5sqrt(17))`D. `(hat(i)+10hat(j)+18hat(k))/(5sqrt(17))`
Answer» Correct Answer - C `vec(A)= 2hat(i)+2hat(j)-hat(k)` and `vec(B)= 6hat(i)-3hat(j)+2hat(k)` `vec(C )= vec(A)xxvec(B)=(2hat(i)+2hat(j)-hat(k))xx(6hat(i)-3hat(j)+2hat(k))` `=\|(hat(i),hat(j),hat(k)),(2,2,-1),(6,-3,2)\|= hat(i)-10hat(j)-18hat(k)` Unit vector perpendicular to both `vec(A)` and vec(B)` =(hat(i)-10hat(j)-18hat(k))/(sqrt(1^(2)+10^(2)+18^(2)))=(hat(i)-10hat(j)-18hat(k))/(5sqrt(17))`

Discussion

415.	What is the angle between the following pair of vectors? `vec(A)=hat(i)+hat(j)+hat(k)` and `vec(B)=-2hat(i)-2hat(j)-2hat(k)`.
Answer» `vec(A).vec(B)=ABcos theta` or `cos theta=(vec(A)vec(B))/(AB)`….(i) But, `vec(A).vec(B)=(hat(i)+hat(j)+hat(k)).(-2hat(i)-2hat(j)-2hat(k))` `vec(A).vec(B)=-2-2-2=-6` Again `A=\|vec(A)\|=sqrt((1)^(2)+(1)^(2)+(1)^(2))=sqrt(3),` `B=\|vec(B)\|=sqrt((-2)^(2)+(-2)^(2)+(-2)^(2))=sqrt(12)=2sqrt(3)` Now, `cos theta=(-6)/(sqrt(3)xx2sqrt(3))=-1 rArr theta=180^(@)`

Discussion

416.	Find the value of m so that the vector `3hat(i)-2hat(j)+hat(k)` may be perpendicular to the vector `2hat(i)+6hat(j)+mhat(k)`.
Answer» The given vectors will be perpendicular of their dot product is zero. `(3hat(i)-2hat(j)+hat(k)).(2hat(i)+6hat(j)+mhat(k))=0` `6(hat(i).hat(i))-12(hat(j).hat(j))+m(hat(k).hat(k))=0` or `6-12+m=0` or `m-6=0` or `m=6`

Discussion

417.	A boat crosses a river with a velocity of `8(km)/(h)`. If the resulting velocity of boat is `10(km)/(h)` then the velocity of river water isA. `4km//h`B. `6km//h`C. `8km//h`D. `10km//h`
Answer» Correct Answer - B

Discussion

418.	A man walk 20 m at an angle` 60^(@)` north of east . How far towards east has he he travellled ?A. 10 mB. 20 mC. `20sqrt(3)` mD. `10sqrt(3)`m
Answer» Correct Answer - A (a) `20cos l60^(@)=10m`

Discussion

419.	The diagonals of a parallelogram are `vecA=2hati-3hatj+hatk` and `vecB=-2hati+4hatj-hatk` what is the area of the paralleogram?A. 2 unitsB. 4 unitsC. `sqrt(5)` unitsD. `sqrt(5)/(2)` units
Answer» Correct Answer - D area of parallelogram `=(1)/(2)\|vecd_(1)xxvecd_(2)\|`

Discussion

420.	The area of the parallelogram represented by the vectors `vecA=2hati+3hatj` and `vecB=hati+4hatj` isA. 14 unitB. 7.5 unitsC. 10 unitD. 5 units
Answer» Correct Answer - D (d) Aera`=\|Axxb\|=5` units `Here AxxB=(2hati+3hatj)xx(hati+4hatj)` `=8hatk-3hatk=5hatk`

Discussion

421.	A unit vector of modulus 2is equally inclined to `x`- and `y`-axes angle at an angle `pi//3`. Find the length ofprojection of the vector on the `z`-axis.
Answer» Given that the vector is inclined at an angle `pi//3` with both x- and y-axes. Then `" "cosalpha=cosbeta=(1)/(2)` Also we know that `cos^(2)alpha +cos^(2)beta+cos^(2)gamma=1`. Therefore, `" "cos^(2)gamma=(1)/(2)` or `" "cosgamma=pm(1)/(sqrt(2))` Thus, the given vector is `" "2(cosalphahati+cosbetahatj+cosgammahatk)=2((hati)/(2)+(hatj)/(2)pm(hatk)/(sqrt(2)))=hati+hatjpmsqrt(2)hatk` Hence, the length of projection of vector on the z-axis is `sqrt(2)` units.

Discussion

422.	Assertion: `vecAxxvecB` is perpendicular to both `vecA+vecB` as well as `vecA-vecB`. Reason: `vecA+vecB` as well as `vecA-vecB` lie in the plane containing `vecA` and `vecB`, but `vecAxxvecB` lies perpendicular to the plane containing `vecA` and `vecB`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.
Answer» Correct Answer - A

Discussion

423.	Assertion: Angle between `hat(i)+hat(j)` and `hat(i)` is `45^(@)`. Reason: `hat(i)+hat(j)` is equally inclined to both `hat(i)` and `hat(j)` and the angle between `hat(i)` and `hat(j)` is `90^(@)`.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.
Answer» Correct Answer - A `cos theta=((hat(i).hat(j)).(hat(i)))/(\|hat(i)+hat(j)\|\|hat(i)\|)=1/(sqrt(2)). Hence theta = 45^(@)`.

Discussion

424.	Assertion: If `theta` be the angle between `vecA` and `vecB` then `tan theta=(vecAxxvecB)/(vecA.vecB)` Reason: `vecAxxvecB` is perpendicular to `vecA.vecB`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.
Answer» Correct Answer - D

Discussion

425.	Assertion: Angle between `hati+hatj` and `hati` is `45^(@)`. Reason: `hati+hatj` is equally inclined to both `hati` and `hatj` and the angle between `hati` and `hatj` is `90^(@)`.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.
Answer» Correct Answer - A

Discussion

426.	The unit vector parallel to the resultant of the vectors `2hati+3hatj-hatk` and `4hati-3hatj+2hatk` isA. `1/sqrt(37)(6hati+hatk)`B. `1/sqrt(37)(6hati+hatj)`C. `1/sqrt(37)(6hati+hatk)`D. none of these
Answer» Correct Answer - A Let `vecR` be the resultant of the given vectors. Then, `vecR=(2hati+3hatj-hatk)+4(hati-3hatj+2hatk)` `=6hati+0hatj+hatk` The required unit vector parallel to `vecR`. `vecR/\|vecR\|=(6hati+0hatj+hatk)/sqrt(6^(2)+0^(2)+1^(2)) = (6hati+0hatj+hatk)/sqrt(37)` `=1/sqrt(37)(hati+hatk)`

Discussion

427.	The position of the particle is given by `vec(r )=(vec(i)+2vec(j)-vec(k))` momentum `vec(P)= (3vec(i)+4vec(j)-2vec(k))`. The angular momentum is perpendicular toA. x-axisB. y-axisC. z-axisD. Line at equal angles to all the three axes
Answer» Correct Answer - A `vec(L)= vec(r )xxvec(P)=\| (hat(i),hat(j),hat(k)), (1,2,-1), (3,4,-2)\|=-hat(j)-2hat(k)` i.e., the angular momentum is perpendicular to x-axis.

Discussion

428.	If the position vectors of P and Q are `hati+2hatj-7hatk` and `5hati-3hatj+4hatk` respectively, the cosine of the angle between `vec(PQ)` and z-axis isA. `4/sqrt(162)`B. `11/sqrt(162)`C. `5/sqrt(162)`D. `-5/sqrt(162)`
Answer» Correct Answer - B `vec(PQ)=vec(OQ) - vec(OP) = 4hati-5hatj+11hatk` `therefore (vec(PQ))/(\|vec(PQ)\|)= 4/sqrt(162) hati - 5/sqrt(162)hatj + 11/sqrt(162)hatk` `therefore coslambda=11/sqrt(162)`, where `lambda` is the angle of `vec(PQ)` with z-axis.

Discussion

429.	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 B and C areA. `veci+2hatk, hati+hatj+3hatk`B. `hatj+2hatk, hati+hatj+3hatk`C. `-hatj+2hatk, hati-hatj+3hatk`D. `-hatj+2hatk,hati+hatj+3hatk`
Answer» Correct Answer - D `vec(OA) =hati+2hatj-hatk, vec(OB)= hati+hatj+hatk, vec(OC) = 2hati+3hatj+2hatk` Position vector of B w.r.t. origin at A is `vec(AB) = vec(OB)-vec(OA) =-hatj+2hatk` Position vector of C w.r.t. origin at A is `vec(AC) = vec(OC) -vec(OA) = hati+hatj+3hatk`

Discussion

430.	i. Prove that the points `veca - 2vecb + 3 vecc, 2 veca + 3vecb- 4 vecc and -7 vecb + 10 vecc` are collinear, where `veca, vec b and vecc` are non-coplanar. ii. Prove that the points `A(1, 2, 3), B(3,4, 7) and C(-3, -2, -5)` are collinear. Find the ratio in which point C divides AB.
Answer» Let the given points be A, B and C. Therefore, `" " vec(AB)` = P.V. of B- P.V. of A `" " = ( 2veca + 3 vecb - 4vecc) - (vec a - 2 vecb + 3vecc)` `" " = veca + 5 vecb - 7 vecc ` `" " vec(AC) `= P.V of C - P.V of A `" "=(-7vecb + 10 vec c) - (veca - 2vecb + 3 vecc)` `" " = - veca - 5 vecb + 7vecc = - vec(AB)` Since `vec(AC) =- vec(AB)`, it follows that the points A, B and C are collinear. ii. Let C divide AB in the ratio `k:1`, then `C(-3, -2, -5) -= ((3k+1)/(k+1), (4k+2)/(k+1), (7k+3)/(k+1))` `rArr " " (3k+1)/(k+1) = -3, (4k+2)/(k+1) = -2 and (7k+3)/(k+1) =-5` `rArr " " k = -(2)/(3) ` from all relations Hence, C divides AB externally in the ratio `2 : 3`.

Discussion

431.	Orthocenter of an equilateral triangle ABC is the origin O. If `vec(OA)=veca, vec(OB)=vecb, vec(OC)=vecc`, then `vec(AB)+2vec(BC)+3vec(CA)=`A. `3vecc`B. `3veca`C. `vec0`D. `3vecb`
Answer» Correct Answer - B For an equilateral triangle, centroid is the same as orthocenter `therefore (vec(OA) + vec(OB) + vec(OC))/(3) =vec0` `therefore vec(OA)+vec(OB)+vec(OC) =vec0` Now, `vec(AB) + 2vec(BC) + 3vec(CA)` `=vec(OB) - vec(OA) +2vec(OC) - 2vec(OB) + 3vec(OA)-3vec(OC)` `=-vec(OB) + 2vec(OA) -vec(OC)` `=-(vec(OB) +vec(OA) +vec(OC))+3vec(OA)` `=3vec(OA)` `=3veca`

Discussion

432.	If `veca, vecb` and `vecc` are position vectors of A,B, and C respectively of `DeltaABC` and `if\|veca-vecb\|=4,\|vecb-vec(c)\|=2, \|vecc-veca\|=3`, then the distance between the centroid and incenter of `triangleABC` isA. 1B. `1/2`C. `1/3`D. `2/3`
Answer» Correct Answer - C Let G be centroid and I be incenter. `\|vec(GI)\|=\|vec(OI)-vec(OG)\|=\|(2veca+3vecb+4vecc)/(9)-(veca+vecb+vecc)/(3)\|` `=\|(-veca+vecc)/(9)\|=3/9=1/3`

Discussion

433.	The momentum of a particle is given by `vecP= (2 sin t hati- 2cos t hatj)kgm//s`. Select the correct option:A. Momentum `vecp` of the particle is always perpendicular to `vecF`B. Momentum `vecp` of the particle is always parallel t `vecF`C. Magnitude of momentum remains constantD. None of the above
Answer» Correct Answer - A::C `vecp =2 sin t hati - 2 cos t hatj` `vecF = (d vecp)/(dt)= 2 cos t hati + 2 sin that j` `therefore" "vecF. vecp = 0 rArr vecF botvecp` Also, `\|vecp\|= sqrt((2sin t)^(2)+ (2 cos t)^(2) )= 2=` constant.

Discussion

434.	Given ` vecP=3 hati+2hatj+5hatk, veca=hati+hatj, vecb=hatj+hatk, vecc=hati+hatk and vecP=xveca+yvecb+zvecc`, then x,y,z are respectivelyA. `(3)/(2), (1)/(2), (5)/(2)`B. `(1)/(2),(3)/(2),(5)/(2)`C. `(5)/(2),(3)/(2),(1)/(2)`D. `(1)/(2),(5)/(2),(3)/(2)`
Answer» Correct Answer - B `p=xa+yb+zc` `implies3hati+2hatj+4hatk=x(hati+hatj)+y(hatj+hatk)+z(hati+hatk)` `implies 3hati+2hatj+4hatk=(x+z)hati+(x+y)hatj+(y+z)hatk` On comparing both sides, the coefficients of `hati, hatj, hatk`, we get `x+z=3" ... (i)"` `x+y=2" ... (ii)"` and `y+z=4" ... (iii)"` On solving Eqs. (i), (ii) and (iii), we get `x=(1)/(2), y=(3)/(2), z=(5)/(2)`

Discussion

435.	The vector area of triangle position vectors of whose vertices are `veca,vecb,vecc` isA. `(vecA.(vecBxx vecC) + vecB. (vecC xx vecA)+ vecC. (vecAxxvecB))/(3)`B. `vecA.(vecB xx vecC)`C. `(vecA xx vecB + vecBxx vecC +vecC xx vecA)/(6)`D. `vecA xx (vecB xx vecC)`
Answer» Correct Answer - C Area `=(1)/(2) [vecA xx vecB]= (1)/(2) [vecB xx vecC]= (1)/(2)[vecC xx vecA]`

Discussion

436.	Let `veca=hatj-hatk` and `vecc=hati-hatj-hatk`. Then the vector `vecb` satisfying `vecaxxvecb+vecc=vec0` and `veca.vecb=3` isA. `-hati+hatj-2hatk`B. `2hati-hatj+2hatk`C. `hati-hatj-2hatk`D. `hati+hatj-2hati`
Answer» Correct Answer - A Given , `a xx b + c = 0` `rArr a xx (a xx b) + a xx c = 0` `rArr (a.b)a - (a.a) + a xx c = 0` `rArr 3a - 2b +a xx c = 0` `rArr 2b = 3a+ a xx c` `rArr 2b = 3hatj - 3hatk -2hati - hatj - hatk` `rArr 2b = - hati + 2hatj - 4hatk` `:. b = - hati + hatj - 2hatk`

Discussion

437.	Let `veca=hati + 2hatj +hatk, vecb=hati - hatj +hatk andvecc= hathatj-hatk` A vector in the plane of `veca and vecb` whose projections on `vecc is 1//sqrt3` isA. `hati+hatj-2hatk`B. `3hati+hatj-3hatk`C. `4hati-hatj+4hatk`D. `4hati+hatj-4hatk`
Answer» Correct Answer - C Any vector r in the plane containing a and b is `a + lambda b` `therefore r = ( hati + 2 hatj + hatk ) + lambda(hati-hatj+hatk)` `=(1+lambda)hati + (2 -lambda)hatj + (1+lambda)hatk` Projection of r on `c=(rc)/(\|c\|)` `rArr ([(1+lambda)hati+(2-lambda)hatj+(1+lambda)hatk](hati+hatj-hatk))/(sqrt((1)^(2)+(1)^(2)+(-1)^(2)))=\|(1)/(sqrt(3))\|` `rArr (1+lambda + 2 -lambda - 1-lambda)/(sqrt(3))=\|(1)/(sqrt(3))\|` `rArr 2-lambda = pm 1 rArr lambda 1,3 ` `therefore r = 2 hati +hatj + 2 hatk , 4 hati -hatj + 4 hatk`

Discussion

438.	Let `veca=hati+hatj+hatk, vecb=hati-hatj+2hatk and vecc=xhati+(x-2)hatj-hatk`. If the vector `vecc` lies in the plane of `veca and vecb` then x equals (A) 0 (B) 1 (C) -4 (D) -2
Answer» Correct Answer - D Since, given vectors a, b and c are coplanar. `:. \|{:(1,1,1),(1,-1,2),(x,x-2,-1):}\| = 0` `rArr 1{1-2(x-2)} -1(-1-2x)+ 1(x-2 +x) = 0` `rArr 1 -2x + 4 + 1 + 2x + 2x- 2 = 0` `rArr 2x = - 4` `x = - 2`

Discussion

439.	The position of a particle is given by `vecr=(hati+2hatj-hatk)` and momentum `vecp=(3hati+4hatj-2hatk)`. The angular momentum is perpendicular to theA. `x` -axisB. `y`-axisC. `z`-axisD. Line at equal angles to all the three axes
Answer» Correct Answer - A

Discussion

440.	The angle between `(vecA+vecB)&(vecAxxvecB)`
Answer» Correct Answer - C

Discussion

441.	Linear momentum of an object can be expressed as `vecP= (4 cos t)hati+( 4 sin t)hatj`. Angle between the forces acting on the object and its linear momentum is :A. `(pi)//(2)`B. `(pi)//(4)`C. `(3pi)//(4)`D. `pi`
Answer» Correct Answer - A `" "vecP=(4 cos t)hati + (4 sin t)hatj` `therefore" "vecF= (dvecP)/(dt)=(-4 sin t)hati +(4 cos t)hatj` `thereforevecF. vecP= 0 rArr theta = pi//2`

Discussion

442.	The vector area of triangle whose sides are `veca,vecb,vecc` isA. `(1)/(2)\|vecbxxvecc+veccxxveca+vecaxxvecb\|`B. `(1)/(2)\|vecbxxvecc+veccxxveca+vecb\|`C. `(1)/(3)\|vecbxxvecc+veccxxveca+vecb\|`D. `(1)/(2)\|-vecbxxvecc+veccxxveca+vecb\|`
Answer» Correct Answer - A

Discussion

443.	If the constant forces `2hati-5hatj+6hatk and -hati+2hatj-hatk` act on a particle due to which it is displaced from a point A (4,-3,-2) to a point B (6,1,-3), then the work done by the forces isA. 10 unitsB. `-10` unitsC. `-15` unitsD. `-9` units
Answer» Correct Answer - A Resultant force, `F=(2hati-5hatj+6hatk)+(-hati+2hatj-hatk)=hati-3hatj+5hatk` and displacement `d=(6hati+hatj-3hatk)-(4hati-3hatj-2hatk)=2hati+4hatj-hatk` `:." "` Work done W `= F* d=( hati-3hatj+5hatk)*(2hati+4hatj-hatk)=-15` `=-15` units [ neglecting - ve sign]

Discussion

444.	The position vector `vecr` and linear momentum `vecp` and `vecr=hati` and `vecp=4hatj` the angular momentum vector is perpendicular toA. x-axisB. y-axisC. z-axisD. xy-plane
Answer» Correct Answer - D

Discussion

445.	If vector A is perpendicular to vectors B and `\|vecA + vecB\| =n\|vecA +vecB\|` then the value of n.
Answer» Correct Answer - 1 `\|vecA + vecB\|=n\|vecA - vecB\|` `sqrt(A^(2) +B^(2) )=nsqrt(A^(2) + B^(2))" "(therefore theta= 90^(@))` n=1

Discussion

446.	If `0.8 hati + 0.2chatj` represents direction, then the value of c will be:
Answer» Correct Answer - 3 `sqrt((0.8)^(2) + (0.2c)^(2))=1` `0.64 + 0.04c^(2) = 1` `0.04c^(2) = 0.36` `c^(2)= 9` ` therefore" " c= 3`

Discussion

447.	Show that the vectors `2veca-vecb+3vecc, veca+vecb-2vecc and veca+vecb-3vecc` are non-coplanar vectors (where `veca, vecb, vecc` are non-coplanar vectors).
Answer» Consider `2veca-vecb+3vecc=x (veca+vecb-2vecc) + y(veca+vecb-3vecc)` or `" "2veca-vecb+3vecc= (x+y)veca+ (x+y)vecb+ (-2x-3y)vecc` `" "x+y=2" "` (i) `" "x+y=-1" "` (ii) `" "-2x-3y=3" "` (iii) Multiplying (i) by 3 and adding it to (iii), we get `x=9` From (i), `9+y=2 or y =-7` Now putting `x=9 and y=-7` in (ii), we get `" "9-7=-1` or `2=-1`, which is not true. Therefore, the given vectors are not coplanar. Alternate method : We have determinant of co-efficients as `" "\|{:(2,,-1,,3),(1,,1,,-2),(1,,1,,-3):}\| = -3 ne 0` Hence, vectors are non-coplanar.

Discussion

448.	If `veca, vecb and vecc` are three non-zero, non-coplanar vectors,then find the linear relation between the following four vectors : `veca-2vecb+3vecc, 2veca-3vecb+4vecc, 3veca-4vecb+ 5vecc, 7veca-11vecb+15vecc`.
Answer» Any vector `vecr` can be uniquely expressed as a linear combinatioin of three non-coplanar vectors. Let us choose that `" "7veca -11 vecb + 15 vecc= x(veca-2vecb+ 3vecc) + y (2veca-3vecb+4vecc)+ z(3veca-4vecb+5vecc)` Comparing the coefficients of `veca, vecb and vecc` on both sides, we get `x+2y+3z=7, -2x-3y-4z=-11, 3x+4y+ 5z=15` Eliminating x and then solving for `y and z`, we get x=1, y=3, z=0`

Discussion

449.	In a four-dimensional spacewhere unit vectors along the axes are ` hat i , hat j , hat ka n d hat l ,a n d vec a_1, vec a_2, vec a_3, vec a_4`are four non-zero vectors such that novector can be expressed as a linear combination of others and `(lambda-1)( vec a_1- vec a_2)+mu( vec a_2+ vec a_3)+gamma( vec a_3+ vec a_4-2 vec a_2)+ vec a_3+delta vec a_4=0,`thena. `lambda=1`b. `mu=-2//3`c. `gamma=2//3`d. `delta=1//3`A. `lamda =1`B. `mu = -2//3`C. `gamma = 2//3`D. `delta = 1//3`
Answer» Correct Answer - A::B::D `(lamda -1) (veca_1 - veca_2) + mu(veca_2 + veca_3)+ gamma (veca_3 + veca_4- 2veca_2) + veca_3 + deltaveca_4 = vec0` i.e., `(lamda -1) veca_1 + (1-lamda +mu- 2gamma)veca_2 + (mu + gamma +1)veca_3 + ( gamma + delta) veca_4=vec0` Since `veca_1, veca_2, veca_3 and veca_4` are linearly inependent, we have `lamda-1 =0, 1-lamda + mu - 2gamma =0, mu+gamma +1=0 and gamma +delta =0` i.e., `lamda =1, mu=2gamma, mu +gamma+1 =0, gamma +delta =0` Hence, `lamda =1, mu = - (2)/(3), gamma = - (1)/(3), delta = (1)/(3)`

Discussion

450.	Five vectors `vecA, vecB, vecC,vecD` and `vecE` have magnitude `10, 12 sqrt(2), 20, 20` and 10 unit respectively, they are direacted as shown in the fig. 3.58 Answer the following questions :
Answer» Correct Answer - A `vecA = 10hati," "vecB= 12 hati +12 hatj` `vecC = - sqrt(3) 10hati + 10 hatj" ", " " vecp= 10hati-10sqrt(3) hatj` `vecD = -10hatj` `(vecA + vecB +vecC ) = (22-10sqrt(3)) hati + 22hatj` x component = `22-10sqrt(3)` y component = `22`

Discussion

Explore topic-wise InterviewSolutions in .

Which of the following relation are worng ?A. `vecA + vecB= C`B. `vecA + vecB= C`C. `vecA + vecB= vecC`D. `A + vecB= vecC`

Area of a parallelogram formed by vectors `(3hati-2hatj+hatk)m` and `(hati+2hatj+3hatk)` m as adjacent sides isA. 8B. `8sqrt(3)`C. `3sqrt(8)`D. 192

Area of a parallelogram formed by vectors `(3hati-2hatj+hatk)m` and `(hati+2hatj+3hatk)` m as adjacent sides isA. `3sqrt(8)m^(2)`B. `24m^(2)`C. `8sqrt(3)m^(2)`D. `4sqrt(3)m^(2)`

The magnitude of scalar and vector products of two vectors are 144 and `48sqrt(3)` respectivley. What is the angle between the two vectors?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`

The magnitude of scalar and vector products of two vectors are `48sqrt(3)` and 144 respectively. What is the angle between the two vectors?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`

The modulus of the vector product of two vector is `(1)/(sqrt(3)` times their scalar product . The angle between vectors isA. `(pi)/(6)`B. `(pi)/(2)`C. `(pi)/(4)`D. `(pi)/(3)`

The value of the sum of two vectors `vecA` and `vecB` with `theta` as the angle between them isA. `sqrt(A^(2)+B^(2)+2ABcos theta)`B. `sqrt(A^(@)-B^(2)+2ABcostheta)`C. `sqrt(A^(@)+B^(2)-2ABsintheta)`D. `sqrt(A^(2)+B^(2)+2ABsintheta)`

Column-I gives operation of vectors `vecA and vecB` and column- II gives the angle `(theta)` between `vecA` and `vecB` (`vecA` and `vecB` are not zero vectors ).

Find the torque of a force `vecF=3hati+2hatj+hatk` acting at the point `vecr=8hati+2hatj+3hatk` about originA. `14hati-38hatj+3hatk`B. `4hati+4hatj+6hatk`C. `-14hati+38hatj-16hatk`D. `-4hati-17hatj+22hatk`

If `theta` is the angle between unit vectors `vecA` and `vecB`, then `((1-vecA.vecB))/(1+vecA.vecB))` is equal toA. `tan^(2)(theta//2)`B. `sin^(2)(theta//2)`C. `cot^(2)(theta//2)`D. `cos^(2)(theta//2)`

A force of `2i+3j+2k N` acts on a body for 4 and produces a displacement of `3hati+4hatj+5hatk` m calculate the power?A. 5WB. 6WC. 7WD. 9W

What is the unit vector perpendicular to the following Vector `2hat(i)+2hat(j)-hat(k)` and `6hat(i)-3hat(j)+2hat(k)?`A. `(hat(i)+10hat(j)-18hat(k))/(5sqrt(17))`B. `(hat(i)-10hat(j)+18hat(k))/(5sqrt(17))`C. `(hat(i)-10hat(j)-18hat(k))/(5sqrt(17))`D. `(hat(i)+10hat(j)+18hat(k))/(5sqrt(17))`

What is the angle between the following pair of vectors? `vec(A)=hat(i)+hat(j)+hat(k)` and `vec(B)=-2hat(i)-2hat(j)-2hat(k)`.

Find the value of m so that the vector `3hat(i)-2hat(j)+hat(k)` may be perpendicular to the vector `2hat(i)+6hat(j)+mhat(k)`.

A boat crosses a river with a velocity of `8(km)/(h)`. If the resulting velocity of boat is `10(km)/(h)` then the velocity of river water isA. `4km//h`B. `6km//h`C. `8km//h`D. `10km//h`

A man walk 20 m at an angle` 60^(@)` north of east . How far towards east has he he travellled ?A. 10 mB. 20 mC. `20sqrt(3)` mD. `10sqrt(3)`m

The diagonals of a parallelogram are `vecA=2hati-3hatj+hatk` and `vecB=-2hati+4hatj-hatk` what is the area of the paralleogram?A. 2 unitsB. 4 unitsC. `sqrt(5)` unitsD. `sqrt(5)/(2)` units

The area of the parallelogram represented by the vectors `vecA=2hati+3hatj` and `vecB=hati+4hatj` isA. 14 unitB. 7.5 unitsC. 10 unitD. 5 units

A unit vector of modulus 2is equally inclined to `x`- and `y`-axes angle at an angle `pi//3`. Find the length ofprojection of the vector on the `z`-axis.

The unit vector parallel to the resultant of the vectors `2hati+3hatj-hatk` and `4hati-3hatj+2hatk` isA. `1/sqrt(37)(6hati+hatk)`B. `1/sqrt(37)(6hati+hatj)`C. `1/sqrt(37)(6hati+hatk)`D. none of these

The position of the particle is given by `vec(r )=(vec(i)+2vec(j)-vec(k))` momentum `vec(P)= (3vec(i)+4vec(j)-2vec(k))`. The angular momentum is perpendicular toA. x-axisB. y-axisC. z-axisD. Line at equal angles to all the three axes

If the position vectors of P and Q are `hati+2hatj-7hatk` and `5hati-3hatj+4hatk` respectively, the cosine of the angle between `vec(PQ)` and z-axis isA. `4/sqrt(162)`B. `11/sqrt(162)`C. `5/sqrt(162)`D. `-5/sqrt(162)`

i. Prove that the points `veca - 2vecb + 3 vecc, 2 veca + 3vecb- 4 vecc and -7 vecb + 10 vecc` are collinear, where `veca, vec b and vecc` are non-coplanar. ii. Prove that the points `A(1, 2, 3), B(3,4, 7) and C(-3, -2, -5)` are collinear. Find the ratio in which point C divides AB.

Orthocenter of an equilateral triangle ABC is the origin O. If `vec(OA)=veca, vec(OB)=vecb, vec(OC)=vecc`, then `vec(AB)+2vec(BC)+3vec(CA)=`A. `3vecc`B. `3veca`C. `vec0`D. `3vecb`

If `veca, vecb` and `vecc` are position vectors of A,B, and C respectively of `DeltaABC` and `if|veca-vecb|=4,|vecb-vec(c)|=2, |vecc-veca|=3`, then the distance between the centroid and incenter of `triangleABC` isA. 1B. `1/2`C. `1/3`D. `2/3`

The momentum of a particle is given by `vecP= (2 sin t hati- 2cos t hatj)kgm//s`. Select the correct option:A. Momentum `vecp` of the particle is always perpendicular to `vecF`B. Momentum `vecp` of the particle is always parallel t `vecF`C. Magnitude of momentum remains constantD. None of the above

Given ` vecP=3 hati+2hatj+5hatk, veca=hati+hatj, vecb=hatj+hatk, vecc=hati+hatk and vecP=xveca+yvecb+zvecc`, then x,y,z are respectivelyA. `(3)/(2), (1)/(2), (5)/(2)`B. `(1)/(2),(3)/(2),(5)/(2)`C. `(5)/(2),(3)/(2),(1)/(2)`D. `(1)/(2),(5)/(2),(3)/(2)`

The vector area of triangle position vectors of whose vertices are `veca,vecb,vecc` isA. `(vecA.(vecBxx vecC) + vecB. (vecC xx vecA)+ vecC. (vecAxxvecB))/(3)`B. `vecA.(vecB xx vecC)`C. `(vecA xx vecB + vecBxx vecC +vecC xx vecA)/(6)`D. `vecA xx (vecB xx vecC)`

Let `veca=hatj-hatk` and `vecc=hati-hatj-hatk`. Then the vector `vecb` satisfying `vecaxxvecb+vecc=vec0` and `veca.vecb=3` isA. `-hati+hatj-2hatk`B. `2hati-hatj+2hatk`C. `hati-hatj-2hatk`D. `hati+hatj-2hati`

Let `veca=hati + 2hatj +hatk, vecb=hati - hatj +hatk andvecc= hathatj-hatk` A vector in the plane of `veca and vecb` whose projections on `vecc is 1//sqrt3` isA. `hati+hatj-2hatk`B. `3hati+hatj-3hatk`C. `4hati-hatj+4hatk`D. `4hati+hatj-4hatk`

Let `veca=hati+hatj+hatk, vecb=hati-hatj+2hatk and vecc=xhati+(x-2)hatj-hatk`. If the vector `vecc` lies in the plane of `veca and vecb` then x equals (A) 0 (B) 1 (C) -4 (D) -2

The position of a particle is given by `vecr=(hati+2hatj-hatk)` and momentum `vecp=(3hati+4hatj-2hatk)`. The angular momentum is perpendicular to theA. `x` -axisB. `y`-axisC. `z`-axisD. Line at equal angles to all the three axes

The angle between `(vecA+vecB)&(vecAxxvecB)`

Linear momentum of an object can be expressed as `vecP= (4 cos t)hati+( 4 sin t)hatj`. Angle between the forces acting on the object and its linear momentum is :A. `(pi)//(2)`B. `(pi)//(4)`C. `(3pi)//(4)`D. `pi`

The vector area of triangle whose sides are `veca,vecb,vecc` isA. `(1)/(2)|vecbxxvecc+veccxxveca+vecaxxvecb|`B. `(1)/(2)|vecbxxvecc+veccxxveca+vecb|`C. `(1)/(3)|vecbxxvecc+veccxxveca+vecb|`D. `(1)/(2)|-vecbxxvecc+veccxxveca+vecb|`

If the constant forces `2hati-5hatj+6hatk and -hati+2hatj-hatk` act on a particle due to which it is displaced from a point A (4,-3,-2) to a point B (6,1,-3), then the work done by the forces isA. 10 unitsB. `-10` unitsC. `-15` unitsD. `-9` units

The position vector `vecr` and linear momentum `vecp` and `vecr=hati` and `vecp=4hatj` the angular momentum vector is perpendicular toA. x-axisB. y-axisC. z-axisD. xy-plane

If vector A is perpendicular to vectors B and `|vecA + vecB| =n|vecA +vecB|` then the value of n.

If `0.8 hati + 0.2chatj` represents direction, then the value of c will be:

Show that the vectors `2veca-vecb+3vecc, veca+vecb-2vecc and veca+vecb-3vecc` are non-coplanar vectors (where `veca, vecb, vecc` are non-coplanar vectors).

If `veca, vecb and vecc` are three non-zero, non-coplanar vectors,then find the linear relation between the following four vectors : `veca-2vecb+3vecc, 2veca-3vecb+4vecc, 3veca-4vecb+ 5vecc, 7veca-11vecb+15vecc`.

Five vectors `vecA, vecB, vecC,vecD` and `vecE` have magnitude `10, 12 sqrt(2), 20, 20` and 10 unit respectively, they are direacted as shown in the fig. 3.58 Answer the following questions :