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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.
| 1. |
If ABCDE is a pentagon, then `vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) ` is equal toA. `4 vec(AC)`B. `2 vec(AC)`C. `3vec(AC)`D. `5 vec(AC)` |
| Answer» Correct Answer - C | |
| 2. |
Prove that 1,1,1 cannot be direction cosines of a straight line. |
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Answer» Sum of squares of direction cosines of a straight line is always `1`. Here, sum of squares of direction cosines ` = 1^2+1^2+1^2 = 3 ` As `3 !=1`, so `(1,1,1)` can not be direction cosines of a straight line. |
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| 3. |
If ABCD is a quadrilateral, then `vec(BA) + vec(BC)+vec(CD) + vec(DA)=`A. `2 vec(BA)`B. `2 vec(AB)`C. `2vec(AC)`D. `2(BC)` |
| Answer» Correct Answer - A | |
| 4. |
If `P`is a pointand `A B C D`is aquadrilateral and ` vec A P+ vec P B+ vec P D= vec P C`, show that `A B C D`is aparallelogram. |
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Answer» We can draw a diagram with the given details. Please refer to video to see the diagram. It is given that, `vec(AP) +vec(PB) +vec(PD) = vec(PC) ` `=>vec(AP) +vec(PB) = vec(PC) - vec(PD)->(1)` Now, in `Delta PDC`, `vec(PD)+vec(DC) = vec(PC)` `=>vec(DC) = vec(PC) - vec(PD)->(2)` Now, in `Delta PAB`, `vec(AP)+vec(PB) = vec(AB)->(3)` From (1),(2) and (3), `vec(AB) = vec(DC)` It implies that `ABCD` is a parallelogram. |
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| 5. |
A vector ` vec r`is inclinedat equal acute angles of `x-a xi s ,``y-a xi s`and `z-a xi sdot`if `| vec r|=6`units, find ` vec rdot` |
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Answer» As the given line is equally inclined with all three coordinate axis, let it create angle `alpha` with all three coordinate axis. Then, Direction cosines will be `(cos alpha,cos alpha,cos alpha)`. `:. cos^2alpha+cos^2alpha+cos^2alpha = 1` `=>3cos^2alpha = 1` `=>cos alpha = +-1/sqrt3` Since `vecr` is inclined at acute angle, therefore, `cos alpha = 1/sqrt3`. Direction cosines will be `(1/sqrt3,1/sqrt3,1/sqrt3)`. Now, `|vecr| = 6` Therefore, `vecr = |vecr|(lhati+mhatj+nhatk) ` `vecr = 6(1/sqrt3hati+1/sqrt3hatj+1/sqrt3hatk) ` `vecr = 2sqrt3(hati+hatj+hatk) .` |
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| 6. |
Prove that the sum of threevectors determined by the medians of a triangle directed from the vertices iszero. |
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Answer» Let `O` is the center of regular octagon `ABCDEFGH`. Please refer to video to see the regular octagon. Now, `vec(OA) = -vec(OE)` `vec(OB) = -vec(OF)` `vec(OC) = -vec(OG)` `vec(OD) = -vec(OH)` Now, sum of all vectors drawn from the center of this octagon `=vec(OA)+vec(OB)+vec(OC)+vec(OD)+vec(OE)+vec(OF)+vec(OG)+vec(OH)` `=(vec(OA)+vec(OE))+(vec(OB)+vec(OF))+(vec(OC)+vec(OG))+(vec(OD)+vec(OH))` `=0+0+0+0 = vec0`. Therefore, the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector. |
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| 7. |
A, B have vectors ` vec a , vec b ` relative to the origin O and X, Y divide ` vec(AB)` internally and externally respectively in the ratio `2:1` . Then , `vec (XY)=`A. `(3)/(2) (vec b - vec a) `B. `(4)/(3) (vec a - vec b) `C. `(5)/(6) (vec b - vec a) `D. `(4)/(3) (vec b - vec a) ` |
| Answer» Correct Answer - D | |
| 8. |
G is a point inside the plane of the triangle `ABC, vecGA + vecGB + vecGC=0`, then show that G is the centroid of triangle ABC.A. `vec(0)`B. `3vec(GA)`C. `3 G vec(B)`D. `3 vec(GC)` |
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Answer» Correct Answer - A We have, `G vec(B)+G vec(C ) =(1+1)vec(GD)` `rArr G vec(B) + G vec(C ) =2 vec(GD)`, where D is the mid point of BC. `rArr G vec(A) + G vec(B) + G vec(C ) =Gvec(A) +2 vec(GD) ` ` because ` G divides AD in the ratio `2 : 1 therefore 2 vec(GD) = -G vec(A)` `therefore G vec(A) +G vec(B) + G vec(C ) =G vec(A) - G vec(A) = vec(0)` |
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| 9. |
If A, B, C are vertices of a triangle whose position vectors are `vec a, vec b and vec c` respectively and G is the centroid of `Delta ABC, " then " vec(GA) + vec(GB) + vec(GC),` isA. `vec 0`B. `veca + vec b + vec c `C. `(veca + vec b + vec c )/(3)`D. `(veca - vec b - vec c )/(3)` |
| Answer» Correct Answer - A | |
| 10. |
If the position vectors of the vertices of a triangle of a triangle are `2 hati - hatj + hatk , hati - 3 hatj - 5 hatk and 3 hati -4 hatj - 4 hatk ,` then the triangle isA. equilateralB. isoscelesC. right angled but not isoscelesD. right angled |
| Answer» Correct Answer - D | |
| 11. |
A point O is the centre of a circle circumscribed about a triangle ABC. Then, `O vec A sin 2 A + O vec B sin 2 B + O vec C sin 2 C ` is equal toA. `(O vec a + O vec B + O vec C ) sin 2 A `B. `3 vec(OG), ` where G is the centroid of triangle ABCC. `vec 0`D. none of these |
| Answer» Correct Answer - C | |
| 12. |
The vectors `2 hati + 3 hatj , 5 hati + 6hatj and 8 hati + lambda hatj ` have their initial points at (1, 1). The value of `lambda` so that the vectors terminate on one straight line, is |
| Answer» Correct Answer - D | |
| 13. |
Let `vecAB = 3 hati + hatj - hatk and vecAC = hati -hatj + 3hatk` and a point P on the line segment BC is equidistant from AB and AC, then `vec (AP)` isA. `2 hati - hatk `B. `hati -2 hatk`C. `2hati +hatk`D. none of these |
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Answer» Correct Answer - C Clearly, a point equidistant from AB and AC is on the bisector of the `angle BAC.` A vector along the bisector of `angle BAC` is `(vec(AB))/(|vec(AB)|)=(vec(AC))/(|vec(AC)|)= (1)/(sqrt(11))(4hati +2hatk)=(2)/(sqrt(11))(2hati+hatk)` Let ` vec(AP) = lambda(2hati + hatk )` `therefore vec(BP) = vec(AP)-vec(AB) =lambda(2 hati+hat k)-(3hati + hatj -hatk)` `rArr vec(BP) = (2lambda -3) hati - hat j + (lambda + 1) hat k` Also, `vec(BC) = vec(AC) - vec(AB) = - 2 hati - 2 hatj + 4 hatk` Since ` vec(BP) ||vec(BC).` Therefore, `vec(BP) = t vec(BC)` `rArr (2lambda -3) hati -hatj + (lambda +1) hatk = t(-2hati - 2 hatj + 4hatk)` `rArr 2 lambda -3= -2t, -1= -2t, lambda +1= 4t` `rArr lambda =1, t=(1)/(2)` `therefore vec(AP) = 2 hati + hatk`. |
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| 14. |
If the points with position vectors ` 20 hati + p hatj , 5 hati - hatj and 10 hati - 13 hatj` are collinear, then p =A. 7B. -37C. -7D. 37 |
| Answer» Correct Answer - B | |
| 15. |
If `vec r = 3 hati + 2 hatj - 5 hatk , vec a= 2 hati - hatj + hatk, vec b = hati + 3 hatj - 2hatk` `and vec c=2 hati + hatj - 3 hatk " such that " hat r = x vec a +y vec b + z vec c` thenA. x, y, z are in APB. x, y, z are in GPC. x, y, z are in HPD. `y, (x)/(2), z ` are in AP |
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Answer» Correct Answer - D We have, `vec r = x vec a + y vec b + z vec c ` `rArr 3 hati + 2 hatj - 5 hatk =x (2 hati - hatj + hatk ) +y(hati + 3hatj -2 hatk ) + z( -2 hati + hatj - 3hatk)` `rArr 3= 2x+y-2z, 2= -x=3y + z, -5=x-2y-3z` `rArr x=3, y=1,z=2` Clearly, `x=y+z rArr y, (x)/(2),z` are in AP. |
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| 16. |
The position vectors of the points A, B, C are `2 hati + hatj - hatk , 3 hati - 2 hatj + hatk and hati + 4hatj - 3 hatk ` respectively . These pointsA. form an isosceles triangleB. form a right triangleC. are collinearD. form a scalene triangle |
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Answer» Correct Answer - C Let `veca = 2 hati + hatj - hatk , vec b = 3 hati - 2 hatj + hatk and vec c = hati + 4 hatj - 3 hat k . ` Then, ` A vec B = vec b - vec a = hati -3 hatj + 2 hatk and B vec C = 2 hati + 6 hat j - 4hatk ` Clearly , `B vec C = -2 A vec B = 2 B vec A ` So, points A, B, C are collinear. |
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| 17. |
D, E and F are the mid-points of the sides BC, CA and AB respectively of `Delta ABC` and G is the centroid of the triangle, then `vec(GD) + vec(GE) + vec(GF) = `A. `vec 0`B. `2 vec(AB)`C. `2 vec(GA)`D. `2 vec(GC)` |
| Answer» Correct Answer - A | |
| 18. |
If the vector `-hati +hatj-hatk` bisects the angle between the vector `vecc` and the vector `3hati +4hatj,` then the vector along `vecc` isA. `(1)/(15)(11 hati + 10hatj + 2 hatk)`B. `-(1)/(15)(11 hati - 10hatj + 2 hatk)`C. `-(1)/(15)(11 hati + 10hatj - 2 hatk)`D. `-(1)/(15)(11 hati + 10hatj + 2 hatk)` |
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Answer» Correct Answer - D Let `x hati + yhatj + z hatk` be the unit vector along `vec c`. Since `-hati + hatj - hatk` bisects the angle between `vec c and 3 hati + hatj.` `therefore lambda (-hati +hatj - hatk) = (xhati +yhatj + z hatk ) + (3hati +4hatj)/(5)` `rArr x +(3)/(5) = - lambda , y+(4)/(5)=lambda and z= -lambda ` Now, ` x^(2) +y^(2) +z^(2) =1 " " [ because x hati + y hatj + z hatk " is a unit vector" ]` ` rArr (-lambda - (3)/(5))^(2) + (lambda-(4)/(5))^(2) + lambda^(2) =1 rArr lambda =0 or, lambda = (2)/(15)` But, `lambda ne 0. " Because " lambda = 0` implies that the given vectors are parallel. ` therefore lambda=(2)/(15) rArr x = -(11)/(15) , y= - (10)/(15) and z = -(2)/(15)` Hence, `xhati +yhatj + z hatk =-(1)/(15) (11 hati + 10hatj + 2hatk)` . |
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| 19. |
The sides of a parallelogram are `2 hati + 4 hatj -5 hatk and hati + 2 hatj + 3 hatk `, then the unit vector parallel to one of the diagonals isA. `(1)/(7) (3 hati + 6 hatj - 2 hatk ) `B. `(1)/(7) (3 hati - 6 hatK - 2 hatk ) `C. `(1)/(7) (-3 hati + 6 hatj - 2 hatk ) `D. `(1)/(7) (3 hati + 6 hatj + 2 hatk ) ` |
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Answer» Correct Answer - A Let ` vec a = 2 hati + 4 hatj - 5 hat k , vec b = hati - 2 hatj + 3 hat k .` The diagonals of the parallelogram are `vec p = vec a + vec b , vec q = vec b -vec a ` `rArr vecp =3 hati + 6 hatj - 2 hat k , vecq= - hati - 2 hatj + 8 hatk ` So, unit vectors along the diagonals are `(1)/(7) (3 hati + 6 hat j - 2 hatk ) and (1)/ (sqrt(69)) (-hati - 2 hatj + 8 hatk ) ` Hence, option (a) is correct. |
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| 20. |
`A B C D`are fourpoints in a plane and `Q`is the pointof intersection of the lines joining the mid-points of `A B`and `C D ; B C`and `A Ddot`Show that ` vec P A+ vec P B+ vec P C+ vec P D=4 vec P Q ,`where `P`is any point. |
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Answer» Let`vec(PA)=veca,vec(PB)=vecb,vec(PC)=vecc,(PD)=vecd` `vec(PH)=(veca+vecd)/2` `vec(PF)=(vecb+vecc)/2` `vec(PE)=(veca+vecb)/2` `vec(PG)=(vecc+vecd)/2` Q is the mid point of HF anf EG `vec(PQ)=(vec(PH)+vec(PF))/2 or (vec(PE)+vec(PG))/2` `=(veca+vecb+vecc+vecd)/4 or (veca+vecb+vecc+vecb)/4` `vec(PA)+vec(PB)+vec(PC)+vec(PD)=4vec(PQ)` `veca+vecb+vecc+vecd=4[(veca+veccb+vecc+vecd)/4]` Hence proved. |
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| 21. |
Let `vec a=hati -2 hatj + 3 hatk, vec b = 3 hati + 3 hatj -hat k and vec c=d hati + hatj + (2d-1) hat k ." If " vec c` is parallel to the plane of the vectors `veca and vec b, " then " 11 d =`A. 2B. 1C. -1D. 0 |
| Answer» Correct Answer - C | |
| 22. |
In a `Delta ABC, " if " vec(AB) = hati - 7hatj + hatk and vec(BC) = 3 hatj + hatj + 2 hatk, " then " | vec(CA)|=`A. `sqrt(61)`B. `sqrt(52)`C. `sqrt(51)`D. `sqrt(41)` |
| Answer» Correct Answer - A | |
| 23. |
Show that the points (3,4), (-5, 16), (5,1) are collinear. |
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Answer» A(3,4) B(-5,16) and C(5,1) Their respective position vectors are=>`3hati+4hatj,-5hati+16hatj,5hati+hatj` `vec(BA)=8hati-12hatj` `vec(BC)=10hati-15hatj` `vec(BA)=4/5vec(BC)` thus these are parallel vectors. since B is common therefore A,B,C are collinear points. |
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| 24. |
Show that theline segments joining the mid-points of opposite sides of aquadrilateral bisects each other. |
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Answer» Let `ABCD` is the given quadrilateral. Please refer to video to see the diagram with their vector representation. Let `O` is the origin and `E,F,G and H` are the mid points of sides of quadrilateral. Then, `vec(OE) = (veca+vecb)/2` `vec(OF) = (vecb+vecc)/2` `vec(OG) = (vecc+vecd)/2` `vec(OH) = (veca+vecd)/2` Then, midpoint of `HF = (veca+vecb+vecc+vecd)/4` Midpoint of `EG = (veca+vecb+vecc+vecd)/4` As midpoints of opposite sides of the quadrilateral are same, it means they are bisecting each other. |
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| 25. |
`A B C D`is parallelogram.If the coordinates of `A , B , C`are `(-2,-1),(3,0)`and `(1,-2)`respectively,find the coordinates of `Ddot` |
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Answer» Let `(x,y)` are the coordinates of point `D`. As, `ABCD` is a parallelogram, `vec(AD) = vec(BC)` `:. (x+2)hati +(y+1) hatj = -2hati +(-2)hatj` `:. x+2 =- 2 and y+1 = -2` `=> x= -4 and y =-3` `:.` Coordinates of point `D` is `(-4,-3)`. |
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| 26. |
If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then ` O vecA + O vec B + O vec C + vec (OD) = `A. ` 2 vec(OG)`B. `4 vec(OG)`C. `5 vec (OG)`D. `3 vec (OG)` |
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Answer» Correct Answer - B Talking O as the origin, let the position vectors of A, B, C and D be ` vec a , vec b , vec c and vec d ` respectively. In `Delta` OAC, G is the mid-point of AC. ` O vec A + O vec C = 2 O vec G " " ` (i) ` In `Delta OBD,` G is the mid-point of BC. ` therefore O vec B + O vec D = 2 O vec G " " ` (ii) Adding (i) and (ii) , we get `O vec A + O vec B + O vec C + O vec D = 4 O vec G ` |
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| 27. |
The vector `cos alpha cos beta hati + cos alpha sin beta hatj + sin alpha hatk ` is aA. null vectorB. unit vectorC. constant vectorD. none of these |
| Answer» Correct Answer - B | |
| 28. |
In a `Delta` ABC, if `vec(AB) = 3 hati + 4 hatk, vec(AC) = 5 hati + 2 hatj + 4hatk ,` then the length of median through A , isA. `3sqrt(2)`B. `6sqrt(2)`C. `5sqrt(2)`D. `sqrt(33)` |
| Answer» Correct Answer - D | |
| 29. |
`A B C D`is parallelogram and `P`is the point of intersection of its diagonals. If `O`is the origin of reference, show that ` vec O A+ vec O B+ vec O C+ vec O D=4 vec O Pdot` |
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Answer» Please refer to video to see the vector diagram for the given details. From the diagram,`vec(OD) + vec(DP) = vec(OP)` `vec(OA) + vec(AP) = vec(OP)` `vec(OB) + vec(BP) = vec(OP)` `vec(OC) + vec(CP) = vec(OP)` `:. vec(OD) + vec(DP)+vec(OA) + vec(AP)+vec(OB) + vec(BP)+vec(OC) + vec(CP) = 4vec(OP)` `=>vec(OA)+vec(OB)+vec(OC)+vec(OD)+vec(AP)+vec(CP)+vec(BP)+vec(DP) = 4vec(OP)->(1)` Also, from the diagram, we can see that, `vec(PA)+vec(PC) = 0` `vec(PD)+vec(PB) = 0` So, equation (1) becomes, `=>vec(OA)+vec(OB)+vec(OC)+vec(OD) + 0 + 0 = 4vec(OP)` `=>vec(OA)+vec(OB)+vec(OC)+vec(OD) = 4vec(OP)` |
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| 30. |
If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then ` O vecA + O vec B + O vec C + vec (OD) = `A. `2vec(OG)`B. `4vec(OG)`C. `5vec(OG)`D. `3vec(OG)` |
| Answer» Correct Answer - B | |
| 31. |
Let `vec(A)D` be the angle bisector of `angle A" of " Delta ABC ` such that `vec(A)D=alpha vec(A)B+beta vec(A)C,` thenA. `alpha= (|vec(AB)|)/(|vec(AB)|+|vec(AC)|),beta=(|vec(AC)|)/(|vec(AB)|+|vec(AC)|)`B. `alpha= (|vec(AB)|+|vec(AC)|)/(|vec(AB)|),beta=(|vec(AB)|+|vec(AC)|)/(|vec(AC)|)`C. `alpha= (|vec(AC)|)/(|vec(AB)|+|vec(AC)|),beta=(|vec(AB)|)/(|vec(AB)|+|vec(AC)|)`D. `alpha= (|vec(AB)|)/(|vec(AC)|),beta=(|vec(AC)|)/(|vec(AB)|)` |
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Answer» Correct Answer - C Clearly, AD divides BC in the ratio AB : AC. `therefore vec(AD)=(|vec(AB)|vec(AC)+|vec(AC)|vec(AB))/(|vec(AB)|+|vec(AC)|)` `rArrvec(AD)=alpha vec(AB)+beta vec(AC),` where `alpha= (|vec(AC)|)/(|vec(AB)|+|vec(AC)|) and beta=(|vec(AB)|)/(|vec(AB)|+|vec(AC)|)` |
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| 32. |
The position vectors of P and Q are respectively `vec a and vec b `. If R is a point on `vec(PQ) ` such that `vec(PR) = 5 vec(PQ),` then the position vector of R, isA. `5 vec b - 4 vec a `B. `5 vec b + 4 vec a `C. `4 vec b - 5 vec a `D. `4 vec b + 5 vec a ` |
| Answer» Correct Answer - A | |
| 33. |
If `A B C D`is a rhombus whose diagonals cut at theorigin `O ,`then proved that ` vec O A+ vec O B+ vec O C+ vec O D+ vec Odot`A. `A vec(B) + A vec(C )`B. `vec(0)`C. `2(vec(AB)+vec(BC))`D. `A vec(C ) +vec(BD)` |
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Answer» Correct Answer - B Since the diagonals of a rhombus bisect each other. `therefore O vec(A) = -O vec(C ) and O vec(B)=-vec(OD)` `rArr O vec(A)+O vec(B)+O vec(C ) +vec(OD) = vec(0)` |
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| 34. |
If a and b are position vectors of A and B respectively the positionvector of a point C on AB produced such that `vec(AC)=3 vec(AB)` isA. `3 vec(a)-2 vec(b)`B. `3vec(b)-2vec(a)`C. `3vec(a)+2 vec(a)`D. `2vec(a)-3vec(b)` |
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Answer» Correct Answer - B Let the position vector for C be `vec(c )` Clearly, B divides AC internally in the ratio `1:2`. `therefore vec(b)=(2vec(a)+1*vec(c))/(2+1)rArr vec(c) = 3 vec(b) - 2 vec(a) ` |
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| 35. |
The non-zero vectors are `vec a,vec b and vec c` are related by `vec a= 8vec b and vec c = -7vec b`. Then the angle between `vec a and vec c` is |
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Answer» Correct Answer - D We have, `vec(a)=8vec(b) and vec(c ) = -7 vec(b)` `rArr vec(a) =(-8to)/(7) c` `rArr vec(a) and vec(c ) ` are unlike parallel vectors. `rArr " Angle between " vec(a) and vec(c ) " is " pi.` |
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| 36. |
Let `alpha, beta, gamma` be distinct real numbers. The points with position vectors `alpha hati + beta hatj +gamma hat k , beta hati + gamma hatj +alpha hat k , gamma hati +alpha hatj + beta hatk`A. are collinearB. form an equilateral triangleC. form a scalene triangleD. form a right angled triangle |
| Answer» Correct Answer - B | |
| 37. |
If A, B, C, D be any four points and E and F be the middle points of AC and BD respectively, then `A vec(B) + C vec(B) +C vec (D) + vec(AD)` is equal toA. `3 vec(EF)`B. `4vec(EF)`C. `4 vec(FE)`D. `3 vec(FE)` |
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Answer» Correct Answer - B Since F is the middle point of BD. Therefore, `A vec(B)+vec(AD)=2 vec(AF) " " ` …(i) Similarly, we have `vec(CB)+vec(AD)+2vec(CF) " " ` …(ii) Adding (i) and (ii), we get `A vec(B)+vec(AD)+C vec(B)+vec(CD)=2(vec(AF)+vec(CF))= -2(Fvec(A)+Fvec(C)) ` `rArr A vec(B) +vec(AD)+C vec(B)+vec(CD)= -2(2vec(FE))` `rArr A vec(B)+vec(AD) +C vec B +vec(CD)=4vec(EF) " " [ because E " is the mid-point of " AC]` |
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| 38. |
If ABCDEF is a regualr hexagon, then `vec(AC) + vec(AD) + vec(EA) + vec(FA)=`A. `2 vec(AB)`B. `3 vec (AB)`C. `vec(AB)`D. `vec0` |
| Answer» Correct Answer - B | |
| 39. |
Thevector ` vec a=""alpha hat i+2 hat j+""beta hat k`lies in the plane of the vectors ` vec b="" hat"i"+ hat j`and` vec c= hat j+ hat k`and bisects the angle between ` vec b`and ` vec c`.Then which one of the following gives possible values of `alpha"and"beta`?(1) `alpha=""2,beta=""2`(2) `alpha=""1,beta=""2`(3) `alpha=""2,beta=""1`(4)`alpha=""1,beta=""1`A. `alpha = 2, beta = 2 `B. `alpha = 1, beta = 2 `C. `alpha = 2, beta = 1 `D. `alpha = 1, beta = 1 ` |
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Answer» Correct Answer - D It is given that `veca` besects the angle between `vecb and vec c`. Therefore , `vec a` is parallel `(vec b + vec c )/(2)` So ` vec a = lambda (hat b + hat c ) ` `rArr alpha hati + 2 hatj + beta hatk = lambda ((hati + hatj ) /( sqrt(2))+(hatj + hatk )/(sqrt(2)))` `rArr alpha hati + 2 hatj + beta hatk = (lambda) /( sqrt(2)) (hati + 2 hat j + hatk ) ` `rArr alpha = (lambda )/(sqrt(2)) , 2 =sqrt(2) lambda and beta = (lambda )/(sqrt(2)) rArr alpha 1, beta =1` |
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| 40. |
`veca, vecb ,vecc` are three non zero vectors no two of which are collonear and the vectors `veca + vecb` be collinear with `vecc,vecb+vecc` to collinear with `veca` then `veca+vecb+vecc` the equal to ?(A) `veca` (B) `vecb` (C) `vecc` (D) None of theseA. `vec a `B. `vec b`C. `vec c`D. `vec 0` |
| Answer» Correct Answer - D | |
| 41. |
`veca,vecb` and `vecc` are three non-zero vectors, no two of which are collinear and the vectors `veca+vecb` is collinear with `vecb`, `vecb+vecc` is collinear with `veca`, then `veca+vecb+vecc=`A. `vec a`B. `vec b`C. `vec c`D. none of these |
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Answer» Correct Answer - D It is given that `vec a + vec b ` is collinear with `vec c and vec b + vec c ` is collinear with `vec a.` `therefore vec a + vec b = lambda vec c and vec b + vec c = mu vec a ` `rArr vec a + vec b = lambda (mu vec a - vec b) " " [" On eliminating " vec c ]` `rArr (lambda mu -1) vec a - (lambda +1) vec b = vec 0` `rArr lambda mu -1=0 and lambda +1=0 " " [ vec a , vec b " are non- collinear "] ` ` rArr lambda = -1 and mu = -1` Substituting `lambda = -1 " in " vec a + vec b = lambda vec c , " we get " vec a + vec b + vec c = vec 0 ` |
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| 42. |
Given that the vectors `vec(a) and vec(b)` are non- collinear, the values of x and y for which the vector equality ` 2 vec(u) -vec(v)= vec(w)` holds true if `vec(u) = x vec(a) + 2y vec(b), vec(v)= - 2 y vec (a) + 3 x vec(b), vec(w) = 4 vec(a)-2 vec(b) ` areA. `x=(4)/(7), y= (6)/(7)`B. `x=(10)/(7), y= (4)/(7)`C. `x=(8)/(7), y= (2)/(7)`D. `x=2, y= 3` |
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Answer» Correct Answer - B We have, `2 vec(u) - vec(v) = vec(w)` `rArr (2x+2y-4)vec(a)+(-3x+4y+2)vec(b) =0` `rArr 2x+2y-4=0 and -3x+4y+2=0[ because vec(a) vec(b) " are non- collinear "]` `rArr x=10//7, y=4//7.` |
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| 43. |
In a regualr hexagon ABCDEF, `A vecB = vec a, B vec C = vecb and C vec D = vec c. " Then " A vec E = `A. `veca + vec b + vec c`B. `2 veca + vec b + vec c `C. `vec a + vec c `D. `vec a + 2 vec b + 2 vec c ` |
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Answer» Correct Answer - C We have, `A vec B = vec a , B vec C = vec b and C vec D = vec c ` `therefore A vec C = vec a + vec b ` In ` Delta ACD`, we have `A vec C + C vec D = A vec D ` `rArr vec a + vec b + vec c = A vec D ` In ` Delta AED`, we have `A vec D + D vec E = A vec E ` `rArr vec a + vec b + vec c - vec a = A vec E " " [ because D vec E = B vec A = - vec a ]` ` rArr A vec E = vec b + vec c ` |
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| 44. |
Let ` vec a , vec b, vec c`be three non-zero vectors such that any two ofthem are non-collinear. If ` vec a+2 vec b`is collinear with ` vec ca n d vec b+3 vec c`is collinear with ` vec a`then prove that ` vec a+2 vec b+6 vec c= vec0`A. `lambda vec(a)`B. `lambda vec(b)`C. `lambda vec(c)`D. `vec(0)` |
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Answer» Correct Answer - D It is given that `vec(a) + 2 vec(b) ` is collinear with ` vec(c ) and vec(b) + 3 vec( c) ` is collinear with `vec(a).` ` therefore vec(a) + 2 vec(b) = x vec(c ), and vec(b) + 3 vec( c) =y vec(a) " for some " x, y in R` `therefore vec(a) + 2 vec(b) +6 vec(c)=(x+6) vec(c ) ` Also, `vec(a)+2 vec(b)+6 vec(c ) = (1+2y) vec(a)` `therefore (x+6) vec(c) = (1+2y) vec(a) ` `rArr x+6=0 and 1+2y =0 " " [ because vec(a), vec(c ) " are non- collinear ]` `rArr x= -6 and y= -1//2` `rArr vec(a) + 2 vec(b) + 6 vec(c ) = vec(0)` |
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| 45. |
If ` vec a , vec b , vec c and vec d ` are the position vectors of points A, B, C, D such that no three of them are collinear and ` vec a + vec c = vec b + vec d ,` then ABCD is aA. rhombusB. rectangleC. squareD. parallelogram |
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Answer» Correct Answer - D We have, `vec a + vec c = vec b + vec d ` `rArr vec b - vec a = vec c - vec d and vec d - vec a = vec c - vec b ` `rArr A vec B = D vec C and A vec D = B vec C ` `rArr ABCD ` is a parallelogram. |
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| 46. |
Statement -1 : If `veca and vecb` are non- collinear vectors, then points having position vectors `x_(1) vec(a) + y_(1) vec(b) , x_(2)vec(a)+ y_(2) vec(b) and x_(3) veca + y_(3) vecb` are collinear if `|(x_(1),x_(2),x_(3)),(y_(1),y_(2),y_(3)),(1,1,1)|=0` Statement -2: Three points with position vectors `veca, vecb , vec c` are collinear iff there exist scalars x, y, z not all zero such that `x vec a + y vec b + z vec c = vec 0, " where " x+y+z=0.`A. Statement - 1 is True, Statement - 2 is True , Statement - 2 is a correct explanation for Statement - 1.B. Statement -1 is True, Statement - 2 is True, Statement -2 is not a correct explanation for Statement - 1.C. Statement - 1 is True, Statement - 2 is False.D. Statement - 1 is False, Statement - 2 is True. |
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Answer» Correct Answer - A Statement -2 si true Using statement -2, points `x_(1) veca + y_(1) vec b , x_(2) veca + y_(2) vec b and x_(3) vec a + y_(3) vecb` will be collinear iff there exist scalars l, m, n such that `l (x_(1)veca +y_(1)vecb) + m(x_(2) veca + y_(2)vec b)+ n( x_(3) vec a + y_(3) vec b) = vec 0,` where `l + m+ n =0` `rArr (lx_(1)+mx_(2)+nx_(3))vec a + (ly_(1)+my_(2)+ny_(3)) vec b = vec 0` `rArr lx_(1)+mx_(2)+nx_(3) =0 and ly_(1)+my_(2) + ny_(3) =0` ` " " [ because vec a , vec b " are non-collinear " ]` Thus, we have, `lx_(1)+mx_(2)+nx_(3)=0` `ly_(1)+my_(2) + ny_(3) =0` `l+m+n=0` This is a homogeneous system of equations having non-trivial solutions (as l, m, n are not all zero). `therefore |(x_(1),x_(2),x_(3)),(y_(1),y_(2),y_(3)),(1,1,1)|=0` So statement -1 is true and statement -2 is a correct explanation for statement -1. |
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| 47. |
Vectors ` vec aa n d vec b`are non-collinear.Find for what value of `x`vectors ` vec c=(x-2) vec a+ vec ba n d vec d=(2x+1) vec a- vec b`are collinear?A. `1//3`B. `1//2`C. 1D. 0 |
| Answer» Correct Answer - A | |
| 48. |
If `vec a , vec b , vec c ` are three non-zero vectors (no two of which are collinear), such that the pairs of vectors `(vec a + vec b, vec c) and (vec b + vec c , vec a) ` are collinear, then `vec a + vec b + vec c =`A. `vec a`B. `vec b`C. ` vec c `D. `vec 0` |
| Answer» Correct Answer - D | |
| 49. |
If ` vec a , vec b , vec c`and ` vec d`are the position vectors of points `A , B , C , D`such that no three of them are collinear and ` vec a+ vec c= vec b+ vec d , t h e n A B C D`is aa. rhombus b.rectanglec. square d.parallelogramA. rhombusB. rectangleC. squareD. parallelogram |
| Answer» Correct Answer - D | |
| 50. |
If the vectors `vec a =2hati + 3hatj +6hatk and vec b` are collinear and ` |vec b |=21, " then " vec b=`(A) `pm 3(2hati + 3 hatj + 6 hatk)`(B) `pm (2hati + 3hatj - 6 hatk)`(C)`pm 21(2hati + 3 hatj + 6 hatk)`(D)`pm 21(hati + hatj + hatk)`A. `pm 3(2hati + 3 hatj + 6 hatk)`B. `pm (2hati + 3hatj - 6 hatk)`C. `pm 21(2hati + 3 hatj + 6 hatk)`D. `pm 21(hati + hatj + hatk)` |
| Answer» Correct Answer - A | |