InterviewSolution
Saved Bookmarks
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. |
Given a parallelogram `OACB`. The lengths of the vectors `vec (OA)`, `vec (OB)` & `vec (AB)` are `a`, `b` & `c` respectively. The scalar product of the vectors `vec (OC)` & `vec (OB)` is(A) `(a^2-3b^2+c^2)/2`(B) `(3a^2+b^2-c^2)/2`(C) `(3a^2-b^2+c^2)/2`(D) `(a^2+3b^2-c^2)/2` |
|
Answer» With the given details, we can create a diagram. Please refer to video for the diagram. Here,` vec(OC).vec(OB) =|vec(OB)||vec(OC)|costheta` `=> vec(OC).vec(OB) =b|vec(OC)|costheta ->(1)` In `Delta AOB`, using cosine law, `c^2 = a^2+b^2-2abcos(theta+alpha)->(2)` In `Delta AOC`, using cosine law, `OC^2 = a^2+b^2-2abcos(pi-(theta+alpha))` `=>OC^2 = a^2+b^2+2abcos(theta+alpha)->(3)` Adding (2) and (3), `c^2+OC^2 = 2(a^2+b^2)` `=> OC^2 = 2(a^2+b^2)-c^2` Now, in `Delta BOC`, using cosine law, `a^2 = b^2+OC^2-2b|vecOC|costheta` `=>a^2 = b^2+2a^2+2b^2-c^2-2b|vecOC|costheta` `=>2b|vecOC|costheta = 3b^2+a^2-c^2` `=>b|vecOC|costheta = (3b^2+a^2-c^2)/2` From (1), `vec(OC).vec(OB) = (a^2+3b^2-c^2)/2` |
|
| 2. |
ABCD is a parallelogram. L is a point on BC which divides BC in the ratio `1:2`. AL intersects BD at P.M is a point on DC which divides DC in the ratio `1 : 2` and AM intersects BD in Q. Point Q divides DB in the ratioA. `1:2`B. `1:3`C. `3:1`D. `2:1` |
|
Answer» Correct Answer - B Hence, P divides AL in the ratio 3:1 and P divides DB in the ratio 1:3 similarly Q divides DB in the ratio 1:3, thus, `DQ=(1)/(4)DB` and `PB=(1)/(4)DB` |
|
| 3. |
ABCD is a parallelogram. L is a point on BC which divides BC in the ratio `1:2`. AL intersects BD at P.M is a point on DC which divides DC in the ratio `1 : 2` and AM intersects BD in Q. `PQ : DB` is equal toA. `(2)/(3)`B. `(1)/(3)`C. `(1)/(2)`D. `(3)/(4)` |
|
Answer» Correct Answer - B `thereforePQ=(1)/(2)DB`, i.e., `PQ:DB=1:2` |
|
| 4. |
Let `overset(to)(a),overset(to)(b),overset(to)(c )` be unit vectors such that `overset(to)(a)+overset(to)(b)+overset(to)(c ) = overset(to)(0).` Which one of the following is correct ?A. `overset(to)(a)xxoverset(to)(b)=overset(to)(b)xxoverset(to)(c)=overset(to)(c)xxoverset(to)(a)=overset(to)(0)`B. `overset(to)(a)xxoverset(to)(b)=overset(to)(b)xxoverset(to)(c)=overset(to)(c)xxoverset(to)(a)neoverset(to)(0)`C. `overset(to)(b)xxoverset(to)(b)=overset(to)(b) xx overset(to)(c) =overset(to)(a)xxoverset(to)(c)=overset(to)(0)`D. `overset(to)(a)xxoverset(to)(b),overset(to)(b)xxoverset(to)(c),overset(to)(c)xxoverset(to)(a)` are mutually perpendicular |
|
Answer» Correct Answer - B Since `vec(a) ,vec(b) , vec( c) ` are unit vectors and `vec(a) + vec(b) + vec( c) =vec(0)` then `vec(a) , vec(b) , vec( c) ` represent and aquilateral triangle . `:. , vec(a) xx vec(b) = vec(b) xx vec(c ) = vec(c ) xx vec(a) be vec(0) ` |
|
| 5. |
If `veca=hati+hatj+hatk, vecb=4hati+3hatj+4hatk and vecc=hati+alphahati+betahatk` are linearly dependent vectors and `|vecc|=sqrt3.` thenA. `alpha=1, beta =-1`B. `alpha= 1,beta =+-1`C. `alpha=- 1,beta = +-1`D. ` alpha= +- 1,beta =1` |
|
Answer» Correct Answer - D Since `vec(a) , vec(b) ,vec(c ) ` are linearly dependent vectors . `rArr [vec(a) ,vec(b) vec(c )]=0` ` rArr |{:(1,,1,,1),(4,,3,,4),(1,,alpha,,beta ):}|=0` applying `C_(2) to C_(2)- C_(1), C_(3) to C_(3) - C_(1)` `|{:(1,,0,,0),(4,,-1,,0),(1,,alpha-1,,beta-1):}|=0 rArr - (beta - 1) =0 rArr beta =1` Also `|vec(c )| = sqrt(3)` `rArr 1+alpha + beta^(2) =3" ""[given " c = hat(i) + alpha hat(j) + beta hat(k)"]"` `rArr 1+ alpha^(2) +1 =3 rArr alpha^(2) =1rArr alpha = +-1` |
|
| 6. |
Let `a= 2hat(i)+ hat(j) -2hat(k) , b=hat(i) +hat(j) ` and c be a vectors such that `|c-a| =3, |(axxb)xx c|=3` and the angle between c and `axx b" is "30^(@)` . Then a. c is equal toA. `(25)/(8)`B. `2`C. `5`D. `(1)/(8)` |
|
Answer» Correct Answer - B we have `a= 2hat(i)+ hat(j) -2hat(k)` `rArr |a|= sqrt(4+1+4)=3` `" and " b=hat (i) + hat(j)` `rArr |b| = sqrt(1 +1)= sqrt(2)` Now `|c-a|=3 rArr |c-a|^(2) =9` ` rArr (c-a) .(c-a) =9` `rArr |c|^(2) +|a|^(2)- 2 c.a =9` Again , `|(a xx b) xx c|=3` `rArr | a xx b| | c| " sin " 30^(@) = 3` `|c| = (6)/(|axxB|)` But `axx b = |{:(hat(i),,hat(j),,hat(k)),(2,,1,,-2),(1,,1,,0):}| = 2hat(i) - 2hat(j) + hat(k)` ltbgt `:. |c| = (6)/(sqrt(4+4+1)) = 2` From Eqs . (i) and (ii) we get `(2)^(2) +(3)^(2) - 2c. a=9` `rArr 4+9 -3c. a =9` `rArr c.a =2` |
|
| 7. |
Four non zero vectors willalways bea. linearly dependent b. linearly independentc. either a or b d. none of theseA. linearly dependentB. linearly independentC. either (a) or (b)D. none of these |
|
Answer» Correct Answer - A Four or more than four non-zero vectors are always linearly dependent. |
|
| 8. |
Let `a=3hat(i) +2hat(j)+xhat(k) " and " b=hat(i)-hat(j) +hat(k) ` for some real x Then `|axx b| =r` is possible ifA. `0 lt r le sqrt((3)/(2))`B. `sqrt((3)/(2)) lt r le 3 sqrt((3)/(2))`C. `3sqrt((3)/(2)) lt r lt 5sqrt((3)/(2))`D. `r ge 5 sqrt((3)/(2))` |
|
Answer» Correct Answer - D Given vectors are ` a = 3hat(i) + 2hat(j) + xhat(k)` ` " and " b= hat(i) - hat(j) + hat(k)` `:. , a xx b = |{:(hat(i) ,,hat(j) ,,hat(k) ),(3,,2,,1),(1,,-1,,1):}|= hat(i) (2 +x) - hat(j) (3-x) + hat(k) (-3-2)` `= (x+2)hat(i) + (x-3) hat(j) = 5hat(k)` ` rArr | axx b | = sqrt( (x+2)^(2) + (x-3)^(2) + 25)` `=sqrt(2x^(2) -2x+ 4+9 +25)` `=sqrt(2(x^(2)-x +(1)/(4)) -(1)/(2) + 38) = sqrt(2(x-(1)/(2))^(2) + (75)/(2))` `= sqrt(2(x^(2) -x+ (1)/(4)) - (1)/(2) + 38 )= sqrt(2(x-(1)/(2))^(2) + (75)/(2))` `So | axx b| ge sqrt(75)/(2)) [ "at " x=(1)/(2) ,| a xx b| " is minimum "]` `rArr r ge 5 sqrt((3)/(2))` |
|
| 9. |
If the vectors `vec a` and `vec b`are linearly independent and satisfying `(sqrt3tantheta-1)vec a + (sqrt3sectheta-2)vec b=vec 0`,then the most general values of `theta` are:A. `npi-(pi)/(6),n inZ`B. `2npi+-(11pi)/(6)n in Z`C. `n pi+-(pi)/(6),n in Z`D. `2npi+(11pi)/(6), n in Z` |
|
Answer» Correct Answer - D `sqrt(3) tan theta+1 =0 and sqrt(3) sec theta-2=0` `implies theta=(11pi)/(6)` `implies theta=2n pi+(11pi)/(6), n in Z`. |
|
| 10. |
If `overset(to)(a),overset(to)(b),overset(to)(c ),overset(to)(d)` are four distinct vectors satisfying the conditions `overset(to)(a)xxoverset(to)(b)=overset(to)(c )xx overset(to)(d) " and " overset(to)(a)xxoverset(to)(c ) = overset(to)(b)xx overset(to)(d)` then prove that `, overset(to)(a).overset(to)(b)+overset(to)(c ). overset(to)(d) ne overset(to)(a). overset(to)(c)+overset(to)(b).overset(to)(d)`. |
|
Answer» Answer - Given `vec(a) xx vec(b) xx vec(d) " and " vec(a) xx vec( c) = vec(b) xx vec(d)` `rArr vec(a) xx vec(b) - vec(a) xx vec( c) = vec( c) xx vec(d) - vec(b) xx vec(d)` `rArr vec(a) xx (vec(b) - vec( c)) = (vec(c )- vec(b)) xx vec(d)` `rArr vec(a) xx (vec(b) -vec(c )) -(vec(c )-vec(b)) xx vec(d) =0` `rArr vec(a) xx (vec(b) -vec(c )) - vec(d) xx (vec(b)-vec(c )) =0` `rArr (vec(a) - vec(d)) xx (vec(b)-vec(c )) =0 rArr (vec(a) -vec(d))||(vec(b)-vec(c ))` `:. (vec(a) xx vec(d)) .(vec(b) - vec(c )) ne 0` `rArr vec(a) "." vec(b) = vec(d) ". " vec(c ) ne vec(d) ". " vec(b) + vec(a) ". " vec(c )` |
|
| 11. |
If ` vec xa n d vec y`are two non-collinear vectors and a, b, andc represent the sides of a ` A B C`satisfying `(a-b) vec x+(b-c) vec y+(c-a)( vec xxx vec y)=0,`then ` A B C`is (where ` vec xxx vec y`is perpendicular to the plane of `xa n dy`)a. an acute-angled triangle b.an obtuse-angled trianglec. a right-angled triangle d. a scalene triangleA. an acute angled triangleB. ann obtuse angled triangleC. a right angled triangleD. a scalene triangle |
|
Answer» Correct Answer - A As x,y and `x xxy` are non-collinear vectors, vectors are linearly independent. hence, `a-b=0=b-c=c-a` or a=b=c therefore, the triangle is equilateral. |
|
| 12. |
The vector `hat(i)+xhat(j)+3hat(k)` is rotated through an angle `theta` and doubled in magnitude then it becomes `4hat(i)+(4x-2)hat(j)+2hat(k)`. The value of x isA. `{-(2)/(3),2}`B. `((1)/(3),2)`C. `{(2)/(3),0}`D. `{2,7}` |
|
Answer» Correct Answer - A Since, the vector `hati+xhatj+3hatk`k is doubled in magnitude, then it becomes `4hati+(4x-2)hatj+2hatk` `therefore 2|hati+xhatj+3hatk|=4hati+(4x-2)hatj+2hatk|` `implies 2sqrt(2+x^(2)+9)=sqrt(16+(4x-2)^(2)+4)` `implies 40+4x^(2)=20+(4x-2)^(2)` `implies 3x^(2)-4x-4=0` `implies (x-2)(3x+2)=0` `implies x=2,-(2)/(3)`. |
|
| 13. |
Let `O`be the origin, and ` O X , O Y , O Z `be three unit vectors in the direction of the sides ` Q R `, ` R P `, ` P Q `, respectively of a triangle PQR.`| O X xx O Y |=``s in(P+R)`(b) `sin2R``(c)sin(Q+R)`(d) `sin(P+Q)dot`A. sin(P+Q)B. sin(P+R)C. sin(Q+R)D. sin2R |
|
Answer» Correct Answer - A sin R = sin (P+Q) |
|
| 14. |
if `overset(to)(b) " and " overset(to)(c )` are any two non- collinear unit vectors and `overset(to)(a)` is any vector then `(overset(to)(a).overset(to)(b))overset(to)(b).(overset(to)(a).overset(to)(c )) overset(to)(c ) + .(overset(to)(a).(overset(to)(b)xxoverset(to)(c)))/(|overset(to)(b)xxoverset(to)(c)|^(2)).(overset(to)(b)xxoverset(to)(c))=.........` |
|
Answer» Correct Answer - `overset(to)(a)` Let `hat(i) ` be a unit vector in the direction of `vec(b) , hat(j)` in the direction of `vec(c )` . Note that `vec( c) = hat(j)` and `(vec(b) xx vec(c )) = |vec(b)||vec(c )| " sin " alpha hat(k) = " sin " alpha hat(k)` where `hat(k) ` is a vector perpendicular to `vec(b) " and " vec(c )` `rArr |vec(b) xx vec(c )| " = sin " alpha rArr hat(k) = (vec(b) xx vec( c))/(|vec(b) xx vec(c )|)` Let `vec( a) = a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k)` Now `vec(a) ". " vec(b) = vec(a) ". " hat(i) = hat(i) "." (a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k)) = a_(1)` and `vec(a) ". " vec(c ) = vec(a) " ." hat(j) = hat(j) ". " (a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k)) = a_(2)` and `vec(a) ". " (vec(b) xx vec( c))/( |vec(b) xx vec( c)|) = vec(a) ". "vec(k) = a_(3) ` `:. (vec(a) ". " vec(b)) vec(b) + (vec(a) ". " vec(c )) vec( c) + (vec(a) (vec(b)xx vec(c )))/(|vec(b) xx vec(c )|^(2)) (vec(b) xx vec(c ))` `=a_(1) vec(b) + a_(2) vec(c ) + a_(3) .((vec(b) xx vec(c )))/(|vec(b) xx vec(c )|) = a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k) = hat(a)` |
|
| 15. |
The points with position vectors `overset(to)(a)+overset(to)(b),overset(to)(a)-overset(to)(b)` and `overset(to)(a)+koverset(to)(b)` are collinear for all real values of k. |
|
Answer» Let position vectors of points `vec(A) ,vec(B) " and " vec(C )` be `vec(a)+ vec(b), vec(a) -vec(b) " and " vec(a)+ kvec(b)` respectively . `:. (vec(a)-vec(b)) -(vec(a)+vec(b)) =(vec(a)+kvec(b)) - (vec(a)- vec(b))` `rArr -2vec(b) =(k +1) vec(b)` `rArr k+1 =-2` `rArr k=-3` Hence it is false statement . |
|
| 16. |
If `|{:(a,,a^(2),,1+a^(3)),(b,,b^(2),,1+b^(3)),(c,,c^(2),,1+c^(3)):}|=0` and the vectors `overset(to)(A) =(1, a, a^(2)) , overset(to)(B) = (1, b, b^(2)) , overset(to)(C )(1,c,c^(2))` are non-coplanar then the product abc = ….A. 2B. `-1`C. `1`D. 0 |
|
Answer» Correct Answer - B Since, `|(a,a^(2),1+a^(3)),(b,b^(2),1+b^(3)),(c,c^(2),a+c^(3))|=|(a,a^(2),1),(b,b^(2),1),(c,c^(2),1)|+|(a,a^(2),a^(3)),(b,b^(2),b^(3)),(c,c^(2),c^(3))|=0` `implies |(a,a^(2),1),(b,b^(2),1),(c,c^(2),1)|=abc|(a,a^(2),1),(b,b^(2),1),(c,c^(2),1)|=0` `implies (1+abc)|(a,a^(2),1),(b,b^(2),1),(c,c^(2),1)|=0" "[because|(a,a^(2),1),(b,b^(2),1),(c,c^(2),1)| ne 0]` `implies 1+abc=0` `implies abc=-1` |
|
| 17. |
` vec a , vec b , vec c`are three coplanar unit vectors such that ` vec a+ vec b+ vec c=0.`If three vectors ` vec p , vec q ,a n d vec r`are parallel to ` vec a , vec b ,a n d vec c ,`respectively, and have integral butdifferent magnitudes, then among the following options, `| vec p+ vec q+ vec r|`can take a value equal toa. `1`b. `0`c. `sqrt(3)`d. `2`A. 1B. 0C. `sqrt(3)`D. 2 |
|
Answer» Correct Answer - C::D Let a,b and c lie in the XY-plane Let `a=hati,b=-(1)/(2)hati+(sqrt(3))/(2)hatj and c=-(1)/(2)hati-(sqrt(3))/(2)hatj` Therefore, `|p+q+r|=|lamda a+mub+vc|` `=|lamdahati+mu(-(1)/(2)hati+(sqrt(3))/(2)hatj)+v(-(1)/(2)hati-(sqrt(3))/(2)hatj)|` `|(lamda+(mu)/(2)-(v)/(2))hati+(sqrt(3))/(2)(mu-v)hatj|` `=sqrt((lamda-(mu)/(2)-(v)/(2))^(2)+(3)/(4)(mu-v)^(2))` `=sqrt(lamda^(2)+mu^(2)+v^(2)-lamdamu-lamdav-muv)` `=(1)/(sqrt(2))sqrt((lamda-mu)^(2)+(mu-v)^(2)+(v-lamda)^(2))` `=(1)/(sqrt(2))sqrt(1+1+4)=sqrt(3)`. |
|
| 18. |
Suppose that `vec p, vec q and vec r` are three non-coplanar vectors in `R^3`. Let the components of a vector `vec s` along `vec p, vec q and vec r` be 4, 3 and 5, respectively. If the components of this vector `vec s` along `(-vec p+vec q +vec r), (vec p-vec q+vec r) and (-vec p-vec q+vec r)` are x, y and z, respectively, then the value of `2vec x+vec y+vec z` is |
|
Answer» Correct Answer - `(9)` Here `s= 4P + 3q +5r` and `s= (-P+q+r )x+ (p-q+r) y+(-P-q+r) z ……(1)` ` :. 4p+3q+5r =p (-x+y-z) +q(x-y+z)+ r(x+y+z)` On comparing both sides we get `-x+y -z= 4, x -y =3 " and " x+y+z=6` On solving above equations we get `x= 4 , y= (9)/(2), =(-7)/(2)` `:. 2x +y+ z=8 + (9)/(2)- (7)/(2) = 9` |
|
| 19. |
The components of a vectors `overset(to)(a)` along and perpendicular to a non-zero vectors `overset(to)(b) ` are …….and …….. Respectively. |
|
Answer» Correct Answer - `((overset(to)(a)"."overset(to)(b))/(|overset(to)(b)|^(2))) overset(to)(b) " and " overset(to)(a) - ((overset(to)(a)"."overset(to)(b))/(|overset(to)(b)|^(2)))overset(to)(b)` Vector component of `vec(a) ` along and perpendicular to `vec(b)` are `((vec(a)"."vec(b))/(|vec(b)|^(2)))vec(v) " and " vec(a) = ((vec(a)"." vec(b))/(|vec(b)|^(2)))vec(b)` |
|
| 20. |
Let `vec(a)` and `vec(b)` be two unit vectors such that `vec(a).vec(b)=0` For some `x,y in R`, let `vec(c)=xvec(a)+yvec(b)+(vec(a)xxvec(b)` If `(|vec(c)|=2` and the vector `vec(c)` is inclined at same angle `alpha` to both `vec(a)` and `vec(b)` then the value of `8cos^2alpha` is |
|
Answer» Correct Answer - 3 We have `vec(c ) = x vec(a) +y vec(b) + vec(a) xx vec(b) " and " vec(a) "." vec(b) =0` `|vec(a)|=|vec(b)|=1 " and " |vec(c ) |=2` Also given `vec( c) ` is inclined on `vec(a) " and " vec(b)` with same angle `alpha`. `:. ,vec(a)"." vec(c ) = x|vec(a)|^(2) +y(vec(a)"."vec(b))+ vec(a)"."(vec(a) xx vec(b))` `|vec(a)||vec(c )| cos alpha =x +0+0` Similarly `|vec(b)||vec(c)| cos alpha=0 +y+0` `rArr y=2 cos alpha` `|vec(c )|^(2) =x^(2) +y^(2) +|vec(a)xx vec(b)|^(2)` `4=8 cos^(2) alpha+ |alpha|^(2) |vec(b)|^(2) sin^(2) 90^(@)` `4=8 cos^(2) alpha+1 rArr 8 cos^(2) alpha=3` |
|
| 21. |
Write the direction ratios of the vector ` -> a= hat i+ hat j-2 hat k`and hence calculate its direction cosines. |
|
Answer» Here, `vec(a) = hati+hatj-2hatk` So, direction ratios will be `(1,1,-2)`. Let its direction cosines are (k,k,-2k). Then,`k^2+k^2+(-2k)^2 = 1` `6k^2 = 1` `k=+-1/sqrt6` We will take `k=1/sqrt6` for direction cosines in the same direction of vector. So, direction cosines will be `(1/sqrt6,1/sqrt6,-2/sqrt6)` |
|
| 22. |
If `a +b+c = alphad, b+c+d=beta a and a, b, c` are non-coplanar, then the sum of `a +b+c+d =` |
|
Answer» Correct Answer - A We have, `a+b+c =alphad` and b+c+d=`beta a` `therefore a+b+c+d=(alpha+1)d` and `a+b+c+d=(beta+1)a` `implies (alpha+1)d=(beta+1)a` if `alphane-1`, then `(alpha+1)d=(beta+1)a` `implies d=(beta+1)/(alpha+1)a` `implies a+b+c=alphad` `implies a+b+c=alpha((beta+1)/(alpha+1))a` `implies[1-(alpha(beta+1))/(alpha+1)]a+b+c=0`. a,b and c are coplanar which is contradiction to the given condition. `therefore alpha=-1` and so a+b+c+d=0. |
|
| 23. |
If the vectors `hati-hatj, hatj+hatk and veca` form a triangle then `veca` may be (A) `-hati-hatk` (B) `hati-2hatj-hatk` (C) `2hati+hatj+hatjk` (D) hati+hatk`A. `-hati-hatk`B. `hati-2hatj-hatk`C. `2hatj+hatj+hatk`D. `hati+hatk` |
|
Answer» Correct Answer - A::B::D `a=[+-(hati-hatj)+-(hatj+hatk)]` `=+-(hati+hatk),+-(hati-2hatj-hatk)` |
|
| 24. |
The value of the` lambda` so that P, Q, R, S on the sides OA, OB, OC and AB of a regular tetrahedron are coplanar. When `(OP)/(OA)=1/3 ;(OQ)/(OB)=1/2` and `(OS)/(AB)=lambda` is (A)`lamda=1/2` (B) `lamda=-1` (C) `lamda=0` (D) `lamda=2`A. `lamda=1/2`B. `lamda=-1`C. `lamda=0`D. fo no value of `lamda` |
|
Answer» Correct Answer - B Let `OA=a,OB=b and OC=c`. then `AB=b-a and OP=(1)/(3)a`. `OQ=(1)/(2)b,OR=(1)/(3)c`. Since, P,Q,R and S are coplanar, then `PS=alphaPQ+betaPR` (PS can be written as a linear combination of PQ and PR) `=alpha(OQ-OP)+beta(OR-OP)` i.e., `OS-OP=-(alpha+beta)(a)/(3)+(alpha)/(2)b+(beta)/(3)c` `implies OS=(1-alpha-beta)(a)/(3)+(alpha)/(2)b+(beta)/(3)c` . . (i) Given `OS=lamdaAB=lamda(b-a)` . . . (ii) From Eq. (i) and Eq. (ii), `beta=0,(1-alpha)/(3)=-lamda and (alpha)/(2)=lamda`. `implies 2 lamda=1+3lamda` or `lamda=-1`. |
|
| 25. |
ABCDE is a pentagon. Forces AB,AE,DC and ED act at a point. Which force should be added to this systemm to make the resultant 2AC?A. ACB. ADC. BCD. BD |
|
Answer» Correct Answer - C `AE+ED+DC+AB` `=AD+DC+AB=AC+AB` Obviously, if BC is added to this system, then it will be `AC+AB+BC=AC+AC=2AC`. |
|
| 26. |
If `hati+hatj+muhatk, muhati+hatj+hatk, hati+muhatj+hatk` are coplanar vectors, then sum of all distinct values of `mu` is (A) `0` (B) `2` (C) `-1` (D) `1`A. 2B. 0C. 1D. -1 |
|
Answer» Correct Answer - D Given vectors `mu hat(i) + hat(j) +hat(k) , hat(i) muhat(j) + hat(k) , hat(j) + mu hat(k) ` will be coplanar if `|{:(mu,,1,,1),(1,,mu,,1),(1,,1,,mu):}|=0` `rArr mu(mu^(2) -1) -1(mu -1) + 1(1-mu) =0` `rArr (mu -1) [mu , (mu+1) -1-1]=0` `rArr (mu-1) [mu^(2) + mu-2]=0` `rArr (mu-1) [(mu +2) (mu-1)]=0` So sum of tthe distinct real values of `mu =1 -2=-1` |
|
| 27. |
If `A_(1),A_(2)….., A_(n)` are the vertices of a regular plane polygon with n sides and o is its centre. Then show that `Sigma_(n-1)^(i-1) (overset(to)(OA)_(i) xx overset(to)(OA)_(i+1) ) =(1-n)(overset(to)(OA)_(2)xxoverset(to)(OA)_(1))` |
|
Answer» Answer - Since `vec(OA)_(1) , vec(OA)_(2) ,…….vec(OA)_(n)` are ll vectors of same magnitude and angle between any two consective vectors is same i.e., `(2pi//n)` `:. , vec(OA)_(1) xx vec(OA)_(2) = a^(2) sin .(2pi)/(n). P` where `hat(p) ` is perpendicular to plane of polygon. Now `overset(n-1)underset(i=1)(Sigma) (vec(OA)_(i) xx vec(OA)_(i+1) ) = overset(n-1)underset(i=1)(Sigma) a^(2). sin.(2pi)/(n).p` ` =(n-1) .a^(2) sin.(2pi)/(n).hat(p)` `=(n-1) [vec(OA)_(1) xx vec(OA)_(2)]` `=(1-n)[vec(OA)_(2) xx vec(OA)_(1)] =RHS` |
|
| 28. |
Five points given by A,B,C,D and E are in a plane. Three forces AC,AD and AE act at A annd three forces CB,DB and EB act B. then, their resultant isA. 2ACB. 3ABC. 3DBD. 2BC |
|
Answer» Correct Answer - B Points `A,B,C,D and E` are in a plane. Resultant`=(AC+AD+AE)+(CB+BD+EB)` `=(AC+CB)+(AD+DB)+(AE+EB)` `=AB+AB+AB=3AB` |
|
| 29. |
A line segment has length 63 and direction ratios are `3, -2, 6.` The components of the line vector areA. `-27,18,54`B. `27,-18,54`C. `27,-18,-54`D. `-27,-18,-54` |
|
Answer» Correct Answer - B Let the components of line segment on axes are x,y and z. So, `x^(2)+y^(2)+z^(2)=63^(2)` Now, `(x)/(3)=(y)/(-2)=(z)/(6)=k` `because(3k)^(2)+(-2k)^(2)+(6k)^(2)=63^(2)` `k=+-(63)/(7)=+-9` `therefore` Components are `(27,-18,54)` or `(-27,18,-54)`. |
|
| 30. |
Points X and Y are taken on the sides QR and RS, respectively of a parallelogram PQRS, so that `QX=4XR` and `RY=4YS` The line XY cuts the line PR at Z Find the ratio PZ: ZR |
|
Answer» `Yvecq+YvecS=lambda/(5(lambda+1))vecq+4/5*1/(lambda+1)vecs` `Y=lambda/(5(lambda+1))` `Y=4/(5(lambda+1))` `lambda=4` `Y=4/25` `Z=4/25vec(PR)` ratio=21:4. |
|
| 31. |
If a and b are position vector of two points A,B and C divides AB in ratio 2:1, then position vector of C isA. `(a+2b)/(3)`B. `(2a+b)/(3)`C. `(a+2)/(3)`D. `(a+b)/(2)` |
| Answer» Correct Answer - A | |
| 32. |
Let `veca=(1,1,-1), vecb=(5,-3,-3)` and `vecc=(3,-1,2)`. If `vecr` is collinear with `vecc` and has length `(|veca+vecb|)/(2)`, then `vecr` equalsA. `+-3c`B. `+-(3)/(2)c`C. `+-c`D. `+-(2)/(3)c` |
|
Answer» Correct Answer - C let `r=lamdac` Given `|r|=|lamda||c|` `therefore(|a+c|)/(2)=|lamda||c|` `therefore|6hati-2hatj-4hatk|=2|lamda|3hati-hatj+2hatk|` `therefore sqrt(56)=2|lamda|sqrt(14)` `therefore lamda=+-1` `therefore r=+-c`. |
|
| 33. |
The volume of a tetrahedronformed by the coterminous edges ` vec a , vec b ,a n d vec c`is 3. Then the volume of the parallelepipedformed by the coterminous edges ` vec a+ vec b , vec b+ vec ca n d vec c+ vec a`is`6`b. `18`c. `36`d. 9 |
|
Answer» Volume of tetrahedron= `1/6*[veca vecb vecc]=3` Hence, `[vec(a) vec(b) vec(c)]=18` Volume of parallelopiped=` [vec(a+b) vec(b+c) vec(c+a)]` = `2[veca vecb vecc]` =2*18=36. |
|
| 34. |
The vector in the direction of the vector `hati-2hatj+2hatk` that has magnitude 9 isA. `hati-2hatj+2hatk`B. `(hati-2hatj+2hatk)/(3)`C. `3(hati-2hatj+2hatk)`D. `9(hati-2hatj+2hatk)` |
|
Answer» Let `veca=hati-2hatj+2hatk` Any vector in the dirction of a vector `veca` is given by `veca/(|veca|)`, `(hati-2hatj+2hatk)/(sqrt(1^(2)+2^(2)+2^(2)))=(hati-2hatj+2hatk)/(3)` :. Vector in the direction of `veca` with magnitude `9=9.(hati-2hatj+2hatk)/(3)` `=3(hati-2hatj+2hatk)` |
|
| 35. |
The vector having initial and terminal points as (2, 5, 0) and (-3,7,4), respectively isA. `-hati+12hatj+4hatk`B. `5hati+2hatj+4hatk`C. `-5hati+2hatj+4hatk`D. `hati+hatj+hatk` |
|
Answer» Required vector `=-(-3-2)hati+(7-5)hatj+(4-0)hatk` `=-5hati+2hatj+4hatk` Similarly, we can say that for having initial and terminal points as (i) (4, 1, 1) and (3, 13, 5), respectively. (ii) (1, 1, 9) and (6, 3, 5), respectively (iii) (1, 2,3) and (2, 3, 4) respectively, we shall get (a), (b) and (d) as its correct otions. |
|
| 36. |
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are `2( -> a+ -> b)`and `( -> a-3 -> b)`externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ |
|
Answer» `(RP)/(RQ)-1/2` `(RP)/(RP+PQ)=1/2` RP=PQ So, P is mid point of RQ `vecP-vecR=vecQ-vecP``vecr=2vecp-vecq``vecr=2(2veca+vecb)-(veca-3vecb)``vecr=3veca+5vecb` |
|
| 37. |
The position vector of the point which divides the join of points `2veca-3vecbandveca+vecb` in the ratio 3:1, isA. `(3veca-2vecb)/(2)`B. `(7veca-8vecb)/(7)`C. `(3veca)/(4)`D. `(5veca)/(4)` |
|
Answer» Let the position vector of the point R divides the join of points `2veca-3vecbandveca+vecb`. :.Position vector `R=(3(veca+vecb)+1(2veca-3vecb))/(3+1)` Since, the position vector of a point R dividing the line segment joining the points P and Q, whose position vectors are `vecpandvecq` in the ratio m:n internally, is given by `(mvecq+nvec p)/(m+n)` `:.R=(5veca)/(4)` |
|
| 38. |
Find the position vector of the point which divides the join of the points `(2veca - 3vecb)` and `(3veca - 2vecb)` (i) internally and (ii) externally in the ratio `2 : 3` .A. `(12)/(5)a+(13)/(5)b`B. `(12)/(5)a-(13)/(5)b`C. `(3)/(5)a-(2)/(5)b`D. none of these |
|
Answer» Correct Answer - B Position vectors of the points which divides internally is `(3(2a-3b)+2(3a-2b))/(5)=(12a-13b)/(5)` |
|
| 39. |
If the position vector of a point A is `vec a + 2 vec b and vec a ` divides AB in the ratio `2:3`, then the position vector of B, isA. `2a-b`B. `b-2a`C. `a-3b`D. `b` |
|
Answer» If x be the position vector of B, then a divides AB in the ratio 2:3. `thereforea=(2x+3(a+2b))/(2+3)` `implies5a-3a-6b=2x` `impliesx=a-3b`. |
|
| 40. |
Let `A` and `B` be points with position vectors `veca` and `vecb` with respect to origin `O`. If the point `C` on `OA` is such that `2vec(AC)=vec(CO), vec(CD) ` is parallel to `vec(OB)` and `|vec(CD)|=3|vec(OB)|` then `vec(AD)` is (A) `vecb-veca/9` (B) `3vecb-veca/3` (C) `vecb-veca/3` (D) `vecb+veca/3`A. `3b-(a)/(2)`B. `3b+(a)/(2)`C. `3b-(a)/(3)`D. `3b+(a)/(3)` |
|
Answer» Correct Answer - C Since, `OA=a,OB=b and 2AC=CO` by section formula, `OC=(2)/(3)a` Therefore, `|CD|=3|OB|` `impliesCD=3b` `impliesOD=OC+CD=(2)/(3)a+3b` Hence, `AD=OD-OA=(2)/(2)a+3b-a` `=3a-(1)/(3)a`. |
|
| 41. |
If ` vec a`,` vec b`,` vec c`,` vec d`are the position vector of point `A , B , C`and `D`, respectively referred tothe same origin `O`such that no three of these point arecollinear and ` vec a`+ ` vec c`= ` vec b`+ ` vec d`, than prove that quadrilateral `A B C D`is a parallelogram.A. squareB. rhombusC. rectangleD. parallelogram |
|
Answer» Correct Answer - D Given, `a+c=b+d` `implies(1)/(2)(a+c)=(1)/(2)(b+d)` Here, mid-points of AC and BD coincide, where AC and BD are diagonals. In addition, we know that, diagonals of a parallelogram bisect each other. Hence, quadrilateral is parallelogram. |
|
| 42. |
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 angled triangleC. are collinearD. form a scalene triangle |
|
Answer» Correct Answer - C `AB=(3-2)hati+(-2-1)hatj+(1+1)hatk` `=hati-3hatj+2hatk` `BC=(1-3)hati+(4+2)hatj+(-3-1)hatk` `=-2hati+6hatj-4hatk` `CA=(2-1)hati+(1-4)hatj+(-1+3)hatk` `=hati-3hatj-2hatk` `|AB|=sqrt(1+9+4)=sqrt(24)` `|BC|=sqrt(4+36+16)=sqrt(56)=2sqrt(14)` `|CA|=sqrt(1+9+4)=sqrt(24)` So, `|AB|+|AC|=|BC|` and angle between AB and BC is `180^(@)` so, points A,B and C cannot form an isosceles triangle. Hence, A,B and C are collinear. |
|
| 43. |
In a `DeltaABC`, if 2AC=3CB, then 2OA+3OB is equal toA. 5OCB. `-OC`C. `OC`D. none of these |
|
Answer» Correct Answer - A `2OA+3OB=2(OC+CA)+3(OC+CB)` `=5OC+2CA+3CB=5OC(because2CA=-3CB)`. |
|
| 44. |
If position vector of points A,B and C are respectively `hati,hatj, and hatk` and `AB=CX`, then position vector of point X isA. `-hati+hatj+hatk`B. `hati-hatj+hatk`C. `hati+hatj-hatk`D. `hati+hatj+hatk` |
|
Answer» Correct Answer - A `AB=Cx implies hatj-hati`= position vecto of point `X-hatk` `therefore`Position vector of point `X=-hati+hatj+hatk`. |
|
| 45. |
The position vector of a point C with respect to B is `hat i +hat j` and that of B with respect to A is `hati-hatj`. The position vector of C with respect to A isA. `2hati`B. `2hatj`C. `-2hatj`D. `-2hati` |
|
Answer» Correct Answer - A Since, position vectors of a point C with respect to B is `BC=hati+hatj` . . (i) similarly, `AB=hati-hatj` . . . (ii) Now, by Eqs. (i) and (ii), `AC=AB+BC=2hati`. |
|
| 46. |
Let `ABC and PQR` be any two triangles in the same plane. Assume that the perpendiculars from the points `A, B, C` to the sides `QR, RP, PQ` respectively are concurrent. Using vector methods or otherwise,prove that the perpendiculars from `P, Q, R to BC, CA, AB` respectively are also concurrent. |
|
Answer» Answer - Let the position vectors of A,B,C be `vec(a) ,vec(b) vec( c)` and `vec( c)` respectively and that of P,Q,R be `vec(p) , vec(q)` and `vec(r )` respectively . Let `vec(h)` be the the position vectors of the orthocentre H of the `Delta PQR` . We have `HP bot QR.` Equation of straignt line passing throught A and perpendicular to QR i.e., parallel to HP=`vec(P)-h` is `vec( r) =vec(a) + l_(1) (vec(p)- vec(h))` where `t_(1)` is parameter. Similarly equations of straight line through B and perpendicular to PR is `vec(r ) = vec( b) + t_(2) (vec(q) - vec(h))` Again equation of straight line through C and perpendicular to `PQ " is " vec(r ) = vec( c) + t_(3) (vec(r ) - hat(h))` If the lines (i) , (ii) and (iii) are concurrent then there exists a point D with position vector `vec(d)` which lies on all of them that is for some values of `t_(1) , t_(2)` and `t_(3)` which implies that `(1)/(t_(1)) vec(d) =(1)/(t_(1)) vec(a) + vec(p) - vec(h)` `(1)/(t_(2))vec(d) =(1)/(t_(2)) vec(b) +vec(q) - vec(h) ......(v)` `(1)/(t_(3)) vec(d) =(1)/(t_(3)) vec( c) +vec( r) - vec(h)......(vi)` From Eqs (iv) and (v) `((1)/(t_(1)) -(1)/(t_(3))) vec(d) =(1)/(t_(1)) vec(a) -(1)/(t_(2)) vec(b) + vec(p) -vec(q)....(vii)` and from Eqs (v) and (vi) `((1)/(t_(2)) -(1)/(t_(3))) vec(d) =(1)/(t_(2)) vec(b)- (1)/(t_(3)) vec( c) + vec(q) + vec(r )` Eliminating `vec(d)` from Eqs (vii) and (viii) we get `=((1)/(t_(1)) -(1)/(t_(2))) [(1)/(t_(2)) .vec(b) -(1)/(t_(3)) vec(c )+ vec(q) -vec(r )]` ` rArr (t_(3) -t_(2)) [t_(2) vec(a) -t_(1) vec(b) + t_(1)t_(2) (vec(p)-vec(q))]` ` =(t_(2)-t_(1)) [t_(3) vec(b) -l_(2) vec(c) +l_(2)l_(3) (vec(q)-vec(r ))]` [multiplying both sides by `t_(1) t_(2) t_(3)`] `rArr t_(2)(t_(3)-t_(2))vec(a) +t_(2)(t_(1)-t_(3))vec(b)` `+t_(2) (t_(2)-t_(1))vec(c) +t_(1)t_(2)(t_(3)-t_(2))vec(p)` `+t_(2)^(2) (t_(1) -t_(3))vec(q) +t_(2)t_(3)(t_(2) -t_(1))vec(r) =vec(0)` Thus lines (i) , (ii) and (iii) are concurrent is equivalent to say that there exist scalarst `t_(1),t_(2) " and " t_(3)` such that `(t_(2)-t_(3))vec(a) +(t_(3)-t_(1))vec(b) +(t_(1)-t_(2))vec(c)+t_(1)(t_(2)-t_(3))vec(p)` `+t_(2)(t_(3)-t_(1))vec(q)+t_(3)(t_(1)-t_(2))vec(r ) - vec(0)` On dividing by `t_(1) t_(2) t_(3)` we get `(lambda_(2)-lambda_(3))vec(p)+(lambda_(3)-lambda_(1)) vec(q) +(lambda_(1)-lambda_(2))vec(r)` `+lambda_(1)(lambda_(2)-lambda_(3))vec(a) +lambda_(2)(lambda_(3)-lambda_(1))vec(b) +lambda_(3)(lambda_(1) -lambda_(2)) vec(c )=0` where` lambda_(1) =(1)/(t_(1)) " for " i = 1 ,2,3` So this is the condition that the lines from p perpendicular to BC from Q perpendicular to CA and from R perpendicular to AB are concurrent (by changing ABC and PQR simultaneously) . |
|
| 47. |
Statement I: If `a=2hati+hatk,b=3hatj+4hatk and c=lamda a+mub` are coplanar, then `c=4a-b`. Statement II: A set vector `a_(1),a_(2),a_(3), . . ,a_(n)` is said to be linearly independent, if every relation of the form `l_(1)a_(1)+l_(2)a_(2)+l_(3)a_(3)+ . . .+l_(n)a_(n)=0` implies that `l_(1)=l_(2)=l_(3)= . . .=l_(n)=0` (scalar).A. Statement-I and statement II ar correct and Statement II is the correct explanation of statement IB. Both statement I and statement II are correct but statement II is not the correct explanation of statement IC. Statement I is correct but statement II is incorrectD. Statement II is correct but statement I is incorrect |
|
Answer» Correct Answer - B a,b and c are coplanar `c=lamda a+mubimplieslamda=4 and mu=-1`. |
|
| 48. |
In a four-dimensional space where unit vectors along the axes are `hati,hatj,hatk and hatl, and a_(1),a_(2),a_(3),a_(4) ` are four non-zero vectors such that no vector can be expressed as a linear combination of other `(lamda-1) (a_(1)-a_(2))+mu(a_(2)+a_(3))+gamma(a_(3)+a_(4)-2a_(2))+a_(3)+deltaa_(4)=0`, thenA. `lamda=1`B. `mu=-(2)/(3)`C. `gamma=(2)/(3)`D. `delta=(1)/(3)` |
|
Answer» Correct Answer - A::B::D `(lamda-1)(a_(1)-a_(2))+mu(a_(2)+a_(3))+gamma(a_(3)+a_(4)-2a_(2))+a_(3)+delta a_(4)=0` `i.e., (lamda-1)a_(1)+(1-lamda+mu-2gamma)a_(2)+(mu+gamma+1)a_(3)+(gamma+delta)a_(4)=0` Since, `a_(1),a_(2),a_(3) and a_(4)` are linearly independent, 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)`. |
|
| 49. |
Statement 1: If ` vec ua n d vec v`are unit vectors inclined at an angle `alphaa n d vec x`is a unit vector bisecting the angle betweenthem, then ` vec x=( vec u+ vec v)//(2sin(alpha//2)dot`Statement 2: If `"Delta"A B C`is an isosceles triangle with `A B=A C=1,`then the vectorrepresenting the bisector of angel `A`is given by ` vec A D=( vec A B+ vec A C)//2.`A. Statement-II and statement II ar correct and Statement III is the correct explanation of statement IB. Both statement I and statement II are correct but statement II is not the correct explanation of statement IC. Statement I is correct but statement II is incorrectD. Statement II is correct but statement I is incorrect |
|
Answer» Correct Answer - D We know that the unit vector along bisector of unit vectors u and v is `(u+v)/(2"cos"(theta)/(2))`, where `theta` is the angle between vector u and v. Also, in an isosceles `DeltaABC` in which AB=AC, the median and bisector from A must be same line. |
|
| 50. |
Statement 1: `| vec a|=3,| vec b|=a n d| vec a+ vec b|=5,t h e n| vec a- vec b|=5.`Statement 2: The length ofthe diagonals of a rectangle is the same.A. Statement-II and statement II ar correct and Statement III is the correct explanation of statement IB. Both statement I and statement II are correct but statement II is not the correct explanation of statement IC. Statement I is correct but statement II is incorrectD. Statement II is correct but statement I is incorrect |
|
Answer» Correct Answer - A We have, adjacent sides of triangle `|a|=3,|b|=4` the length of the diagonal is `|a+b|=5` Since, it satisfies the Pythagoras theorem, `a bot b` So, the parallelogram is a rectangle. Hence, the length of the other diagonal is `|a-b|=5`. |
|