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.
| 101. |
If `A, B, C` are the vertices of a triangle whose position vectros are `vec a,vec b, vec c and G` is the centroid of the `DeltaABC,` then `overline(GA)+overline(GB)+overline(GC) =` |
|
Answer» Correct Answer - A Position vectors of vertices A,B and C of the `DeltaABC=a,b and c`. We know that, position vector of centroid of the triangle, `G=(a+b+c)/(3)`. Therefore, GA+GB+GC `=(a-(a+b+c)/(3))+(b-(a+b+c)/(3))+(c-(a+b+c)/(3))` `=(1)/(3)(2a-b-c+2b-a-c+2c-a-b)=0`. |
|
| 102. |
For any three vectors `overset(to)(a), overset(to)(b) " and " overset(to)(C )` ` (overset(to)(a) - overset(to)(b)). {(overset(to)(b)-overset(to)(c))xx(overset(to)(c)-overset(to)(a))} = 2overset(to)(a).(overset(to)(b)xx overset(to)(c))` |
|
Answer» Correct Answer - 1 `(vec(a)-vec(b)).{(vec(b)-vec(c ))xx (vec(c )-vec(a)) }=(vec(a)-vec(b)).(vec(b)xx vec(c )-vec(b) xx vec(a) +vec(c )xx vec(a))` `=vec(a) ". "(vec(b) xx vec(c )) -vec(b) .(vec(c )xx vec(a)) = [vec(a)vec(b)vec(c )]+[vec(a)vec(b)vec(c )]` `=2 [vec(a)vec(b)vec(c )]=2vec(a).(vec(b)xx vec(c ))` Hence it is a true statement . |
|
| 103. |
Let `overset(to)(a) , overset(to)(b) " and " overset(to)(c )` be three vectors having magnitudes 1 , and 2 respectively . If `overset(to)(a) xx (overet(to)(a) xx overset(to)(c ) ) + overset(to)(b) = overset(to)(0)` then the actue angle between `overset(to)(a) " and " overset(to)(c )` is ...... |
|
Answer» Correct Answer - `(pi)/(6)` Given `vec(a) xx (vec(a) xx vec(c )) +vec(b) =vec(0)` `rArr (vec(a) ". " vec(c )) vec(a) - (vec(a) "." vec(a)) vec( c) + vec(b) =vec(0)` `rArr (2 cos 0) vec(a) -vec(c ) + vec(b) =vec(0)` `rArr (2 cos 0 vec(a) - vec( c))^(2) = (-vec(b))^(2)` `rArr 4 cos^(2) 0 . |vec(a)|^(2)+|vec( c)|^(2) 2 2 cos 0 vec(a) ". " vec(c ) = |vec(b)|^(2)` `rArr 4 cos^(2) 0-4 -8 cos^(2) 0=1` ` rArr 4 cos^(2) 0=3` `rArr cos 0 = +- (sqrt(3))/(2)` `rArr cos 0 = (sqrt(3))/(2) rArr 0= (pi)/(6)` For 0 to be acute |
|
| 104. |
If `overset(to)(a) , overset(to)(b) , overset(to)(c ) ` are non-coplanar unit vectors such that `overset(to)(a) xx (overset(to)(b) xx overset(to)(c )) = ((overset(to)(b) + overset(to)(c )))/(sqrt(2))`, then the angle between `overset(to)(a) " and " overset(to)(b)` isA. `(3pi)/(4)`B. ` (pi)/(4)`C. `(pi)/(2)`D. `pi` |
|
Answer» Correct Answer - A Since `vec(a) xx (vec(b) xx vec(c )) = (vec(b) + vec(c ))/(sqrt(2))` ` rArr (vec(a) " ." vec(c )) vec(b) - (vec(a) ". " vec(b)) vec(c ) = (1)/(sqrt(2)) vec(b) + (1)/(sqrt(2)) vec( c)` On equating the coefficient of `vec(c )` we get `rArr |vec(a)||vec(b)|| cos 0 = (1)/(sqrt(2))` ` :. cos 0 = (1)/(sqrt(2)) rArr 0= (3pi)/(4)` |
|
| 105. |
The number of vectors of unit length perpendicular to the vectors `veca=2hati+hatj+2hatkandvecb=hatj+hatk` isA. oneB. twoC. threeD. infinite |
|
Answer» The number of vectors of unit length perpendicular to the vectors `vecaandvecb" is "vecc` (say) i.e., `vecc=+-(vecaxxvecb)` So, there will be two vectors of unit length perpendicular to the vectors `vecaandvecb`. |
|
| 106. |
If `veca=hati+hatj+2hatkandvecb=2hati+hatj+2hatk`, then find the unit vector in the direction of (i) `6vecb` (ii) 2veca-vecb` |
|
Answer» Here, `veca=hatl+hatj+2hatkandvecb=2hati+hatj-2hatk` (i) since, `6vecb=12hatl+6hatj-12hatk` :. Unit vector in the direction of `6vecb=(6vecb)/(|6vecb|)` `=(12hati+6hatj-12hatk)/(sqrt12^(2)+6^(2)+12^(2))=(6(2hati+hatj-2hatk))/(sqrt324)` `=(6(2hati+hatj-2hatk))/(18)=(2hatj+hatj-2hatk)/(3)` (ii) Since, `2veca-vecb=2(hati+hatj+2hatk)-(2hati+hati-2hatk)` `=2hati+2hatj+4hatk-2hati-hatj+2hatk=hatj+6hatk` :. Unit vector in the direction of `2veca-vecb=(2veca-vecb)/(|2veca-vecb|)=(hatj+6hatk)/(sqrt1+36)=(1)/(sqrt37)(hatj+6hatk)` |
|
| 107. |
The position vectors of A and B are `hati-hatj+2hatk and 3hati-hatj+3hatk`. The position vector of the middle points of the line AB isA. `(1)/(2)hati-(1)/(2)hatj+hatk`B. `2hati-hatj+(5)/(2)hatk`C. `(3)/(2)hati-(1)/(2)hatj+(3)/(2)hatk`D. none of these |
|
Answer» Correct Answer - B `(a+b)/(2)=2hati-hatj+(5)/(2)hatk`. |
|
| 108. |
For any vector `veca` the value of `(vecaxxhati)^2+(vecaxxhatj)^2+(vecaxxhatk)^2` is equal toA. `vec(a^(2))`B. `3vec(a^(2))`C. `4vec(a^(2))`D. `2vec(a^(2))` |
|
Answer» Correct Answer - D Let `veca=xhati+yhatj+zhatk` `:.vec(a^(2))=x^(2)+y^(2)+z^(2)` `:.vecaxxhati=|[hati,hatj,hatk],[x,y,z],[1,0,0]|` `=hati[0]-hatj[-z]+hatk[-y]` `=zhati-yhatk` `:.(vecaxxhati)^(2)=(zhatj-yhatk)(zhatj-yhatk)` Similarly, `(vecaxxhatj)^(2)=x^(2)+z^(2)` and `(vecaxxhatk)^(2)=x^(2)+y^(2)` `:.(vecaxxhati)^(2)+(vecaxxhatj)^(2)+(vecaxxhatk)^(2)=y^(2)+z^(2)+x^(2)+z^(2)+x^(2)+y^(2)` `=2(x^(2)+y^(2)+z^(2))=2veca^(2)` |
|
| 109. |
The sides of a parallelogram are `2hati +4hatj -5hatk and hati + 2hatj +3hatk `. The unit vector parallel to one of the diagonals isA. `(1)/(sqrt(69))(hati+2hatj-8hatk)`B. `(1)/(69)(hati+2hatj-8hatk)`C. `(1)/(sqrt(69))(-hati-2hatj+8hatk)`D. `(1)/(69)(-hati-2hatj+8hatk)` |
|
Answer» Correct Answer - C Since, `AB+BD=AD` `BD=AD-AB` `implies =(hati+2hatj+3hatk)-(2hati+4hatj-5hatk)` `=-hati-2hatj+8hatk` Hence, unit vector in the direction of BD is `(-hati-2hatj+8hatk)/(|-hati-2hatj+8hatk|)=(-hati-2hatj+8hatk)/(sqrt(69))`. |
|
| 110. |
The vectors from origin to the points A and B are `veca=2hati-3hatj+2hatkandvecb=2hati+3hatj+hatk`A. 340B. `sqrt25`C. `sqrt229`D. `(1)/(2)sqrt229` |
|
Answer» :. Area of `DeltaOAB=(1)/(2)|vec(OA)xxvec(OB)|` `=(1)/(2)|(2hati-3hatj+2hatk)xx(2hatj+3hatj+hatk)|` `=(1)/(2)|[hati,hatj,hatl],[2,-3,2],[2,3,1]|` `=(1)/(2)|[hati(-3-6)-hatj(2-4)+hatk(6+6)]|` `=(1)/(2)|-9i+2hatj+12hatk|` :. Area of `DeltaOAB=(1)/(2)sqrt((81+4+144))=(1)/(2)sqrt229` |
|
| 111. |
Let the vectors `overset(to)(a), overset(to)(b), overset(to)( c) " and " overset(to)(d)` be such that `(overset(to)(a) xx overset(to)(b)) xx ( overset(to)(c ) xx overset(to)(d)) = overset(to)(0) . " If " P_(1) " and " P_(2)` are planes determined by the pairs of vectors `overset(to)(a) , overset(to)(b) " and " oerset(to)(c ) , overset(to)(d)` respectively then the angle between `P_(1) " and "P_(2)` is |
|
Answer» Correct Answer - A If 0 is the angle between `P_(1) " and " P_(2)` then normal to the plenes are `underset(N_(2) = vec(c ) xx vec(d))(`N_(1) = vec(a) xx vec(b)`)}` Then `|N_(1) | xx | N_(2) | sin 0 =0` `rArr sin 0 = rArr 0=0` |
|
| 112. |
Let `overset(to)(p) , overset(to)(q) , overset(to)(r )` be three mutually perpendicular vectors of the same magnitude. If a vectors `overset(to)(X)` satisfies the equation `overset(to)(p) xx [(overset(to)(x) -overset(to)(q)) xx overset(to)(p)] + overset(to)(q) xx [(overset(to)(x)-overset(to)(r ))xx overset(to)(q)]` `+ overset(to)(r ) xx [(overset(to)(x) - overset(to)(p)) xx overset(to)(r )]=overset(to)(0) " then " overset(to)(x) ` is given byA. `(1)/(2) (overset(to)(p) +overset(to)(q)-2overset(to)(r ))`B. `(1)/(2) (overset(to)(p) +overset(to)(q)+overset(to)(r ))`C. `(1)/(3) (overset(to)(p)+overset(to)(q)+overset(to)(r ))`D. `(1)/(2) (2overset(to)(p)+overset(to)(q) - overset(to)(r ))` |
|
Answer» Correct Answer - B Since `vec(p) ,vec(q) ,vec(r )` are mutually perpendicular vectors of same magnitude so let us consider `underset(vec(p) ". " vec(q)=vec(q) "." vec(r ) = vec(r ) ". " vec(p)=0)(|vec(p)|=|vec(q)|=|vec(r)|=lambda" and ")}` Given `vec(p) xx {(vec(x)-vec(q)) xx vec(p)}+ vec(q) xx {(vec(x) -vec(r)) xx vec(q)}` `+vec(r ) xx {(vec(x) - vec(p)) xx vec(r )}=vec(0)` `rArr (vec(p) ". " vec(p)) xx (vec(X) - vec(q)) -{vec(p).(vec(X)-vec(q))}vec(p)+(vec(q) ". " vec(q)) (vec(X) -vec(r))` `-{vec(q) ". " (vec(x) -vec(r )) } vec(q) + (vec(r).vec(r))(vec(x)-vec(p)) -{vec(r ).(vec(x) -vec(p))} vec(r )=0` `rArr vec(X) {(vec(p)"." vec(p)) + (vec(q) "." vec(q))+ (vec(r) ". " vec(r))} - (vec(p) ". " vec(p))vec(q)` `-(vec(q) ". " vec(q)) r- (vec(r ) ". " vec(r)) vec(p) = (vec(x)"." vec(q)) vec(p) + (vec(x ) ". " vec(q)) vec(q) + (vec(x) ". " vec(r)) vec(r)` `rArr 3 vec(X) |lambda|^(2) -(vec(p) +vec(q))|vec(r))|lambda|^(2) =(vec(X)"." vec(p))vec(p)` `+(vec(x ) ". " vec(q)) vec(q) + (vec(x) ". " vec(r)) vec(r)` Taking dot of Eq. (ii) with `vec(p)` we get `3(vec(X)"." vec(p)) |lambda|^(2) -|lambda|^(4) =(vec(x ) ". " vec(p)) |lambda|^(2) rArr vec(x) ". " vec(p) = (1)/(2) |lambda|^(2)` Similarly taking dot of Eq . (ii) with `vec(q) " and " vec(r )` respectively we get ` vec(x) ". " vec(q) = (|lambda|^(2))/(2) = vec(x) ". " vec(r )` `:. Eq ` (ii) becomes `3vec(x ) |lambda|^(2) -(vec(p) +vec(q) + vec(r )) |lambda|^(2) = (|lambda|^(2))/(2) (vec(p) +vec(q) + vec(r ))` `rArr 3vec(x) = (1)/(2) (vec(p) +vec(q) +vec(r )) + (vec(p) +vec(a) +vec(r ))` `rArr vec(X) = (1)/(2) (vec(p)+vec(q) +vec(r ))` |
|
| 113. |
Let `veca` and `vecb` are non collinear vectors. If vectors `vecalpha=(lambda-2)veca+vecb` and `vecbeta=(4lambda-2)veca+3vecb` are collinear, then `lambda` is equal to (a) `-4` (b) `4` (c) `2` (d) `-2`A. 4B. -3C. 3D. -4 |
|
Answer» Correct Answer - D Two vectors c and d are said to be collinear if we can write c`= lambda b` for some non-zero scalar `lambda` Let the vectors `a=(lambda -2) a+b` and `beta= (4lambda -2) a+3b ` are collinear where a and b are non -collinear `:. `We can write `alpha = k beta ` for some `k in R -{0}` `rArr (lambda-2 ) a+ b = k [(4 lambda -2)a+ 3 b]` `rArr [(lambda -2) -k (4lambda -2)] a + (1-3k) b=0` Now as a and b are non- collinear therefore they are linearly independent and hence `(lambda-2) -k (4lambda-2)=0` and `1-3k =0` `rArr lambda -2 =k (4lambda -2) " and " 3k=1` `rArr lambda -2 = (1)/(3) (4lambda -2)` `rArr 3lambda -6 =4 lambda -2` `rArr lambda =-4` |
|
| 114. |
If `a=(2,5) and b=(1,4)`, then vector parallel to (a+b) isA. (3,5)B. (1,1)C. (1,3)D. (8,5) |
|
Answer» Correct Answer - C `a+b=3hati+9hatjj=3(hati+3hatj)`. Hence, it is parallel to (1,3). |
|
| 115. |
If the position vectors of the points A and B are `hati+3hatj-hatk and 3hati-hatj-3hatk`, then what will be the position vector of the mid-point of ABA. `hati+2hatj-hatk`B. `2hati+hatj-2hatk`C. `2hati+hatj-hatk`D. `hati+hatj-2hatk` |
|
Answer» Correct Answer - B `(3hati-hatj-3hatk+hatj+3hatj-hatk)/(2)=2hati+hatj-2hatk`. |
|
| 116. |
The position vectors of thepoint `A ,B ,Ca n dDa r e3 hat i-2 hat j- hat k ,2 hat i+3 hat j-4 hat k ,- hat i+ hat j+2 hat k`and 4` hat i+5 hat j+lambda hat k ,`respectively. Ifthe points `A ,B ,Ca n dD`lie on a plane, find the value of `lambdadot` |
|
Answer» Correct Answer - `(146)/(17)` Here `vec(AB) =- hat(i) +5hat(j) - 3hat(k)` `vec(AC) =- 4hat(i) + 3hat(j) +3hat(k) " and " vec(AD) =hat(i) +7hat(j) + (lambda + 1) hat(k)` we know that A,B,C,D lie in a plane if `vec(AB) , vec(AC) , vec(AD)` are coplanar i.e, `[vec(AB) vec(AC) vec(AD)]=0 rArr |{:(-1,,5,,-3),(-4,,3,,3),(1,,7,,lambda+1):}|=0` `rArr -1 (3lambda +3 -21) -5 (-4lambda -4-3) -3 (-28 -3) =0` `rArr -17lambda + 146 =0` `:. lambda =(146)/(17)` |
|
| 117. |
If `overset(to)(a) = (hat(i) + hat(j) + hat(k)) , overset(to)(a) , overset(to)(b) , overset(to)(c ) =1 " and " overset(to)(a) xx overset(to)(b) = hat(j) - hat(k), " then " overset(to)(b)` is equal toA. `hat(i) - hat(j) + hat(k)`B. `2hat(j) - hat(k)`C. `hat(i)`D. `2hat(i)` |
|
Answer» Correct Answer - C We know that `vec(a) xx (vec(a) xx vec(b)) = (vec(a) ". " vec(b)) vec(a) - (vec(a) ". " vec(b)) vec(b) ` ` :. (hat(i) + hat(j) + hat(k)) xx (hat(j) - hat(k) ) = (hat(i) + hat(j) + hat(k)) - (sqrt(3))^(2) vec(b)` `rArr - 2hat(i) + hat(j) + hat(k) = hat(i) + hat(j) + hat(k) - 2vec(b) rArr 3 vec(b) = + 3hat(i)` `:. vec(b) = hat(i)` |
|
| 118. |
Let `veca = 2hati+(lambda)_1hatj+3hatk` , `vec(b)=4hati+(3-(labda)_2)hatj+6hatk` and `vec(c)=3hati+6hatj+((lambda)_3-1)hatk` be three vectors such that `vec(b)=2 vec(a)` and `vec(a)` is perpendicular to `vec(c)` then a possible value of `((lambda)_1,(lambda)_2,(lambda)_3)` is: (a) `(1,3,1)` (b) `((-1/2),4,0)` (c) `(1,5,1)` (d) `((1/2), 4, -2)`A. `(1,3,1)`B. `(1,5,1)`C. `(-(1)/(2), 4,0)`D. `((1)/(2) , 4, -2)` |
|
Answer» Correct Answer - C We have `a= 2hat(i) + lambda_(1) hat(j) + 3hat(k) , b=4hat(i) + (3-lambda_(2)) hat(j) + 6 hat(k)` and ` c= 3hat(i) + 6hat(j) + (lambda_(3)-1)hat(k)` such that `b=2a` Now ` b=2a` ` rArr 4hat(i) +(3 - lambda_(2)) hat(j)+ 6hat(k) =2 (2hat(i) + lambda_(1) hat(j) +3hat(k))` ` rArr 4hat(i) +(3 - lambda_(2)) hat(j)+6hat(k) =4 hat(i) +2lambda_(1) hat(j) +6hat(k)` `rArr (3-2lambda_(1)) -lambda_(2))hat(j)=0` `rArr 2lambda_(1) +lambda_(2) =3` Also as a is perpendicular to c , therefore a, c=0 `rArr(2hat(i) + lambda_(1) hat(j) +3hatk)) . (vec(3hat(i) +6hat(j) + (lambda_(3) -1)hat(k)) =0` `rArr 6 + 6lambda_(1) +3(lambda_(3) -1) =0` `rArr 6lambda_(1) +3lambda_(3) + 3 =0` `rArr 2lambda_(1) +lambda_(3) =-1` Now from Eq . (i) `lambda_(2) =3 -2lambda` and from Eq . (ii) `lambda_(3) = 2lambda_(1) -1` `:. (lambda_(1) , lambda_(2) , lambda_(3)) -= (lambda_(1) , 3- 2lambda_(1) -1)` If `lambda_(1) =- (1)/(2) ,` then `lambda_(2) = 4 " and " lambda_(3) =0` Thus a possible value of `(lambda_(1),lambda_(2) , lambda_(3)) = (-(1)/(2) , 4,0)` |
|
| 119. |
Find the value of `lamda` such that the vectors `veca=2hati+lamdahatj+hatkandvecb=hati+2hatj+3hatk` are orthogonal. |
|
Answer» Since, two non-zero vectors `vecaandvecb` are orthogonal i.e., `veca.vecb=0`. `:.(2i+lamdahati+hatk).(hati+2hatj+3hatk)=0` `implies2+2lamda+3=0` `:.lamda=(-5)/(2)` |
|
| 120. |
Prove that the ponts `A(1,2,3),B(3,4,7),C(-3 -2, -5)` are collinear and find the ratio in which B divides AC. |
|
Answer» Clearly, `AB=(3-1)hati+(4-2)hatj+(7-3)hatk` `=2hati+2hatj+4hatk` and `BC=(-3-3)hati+(-2-4)hatj+(-5-7)hatk` `=6hati-6hatj-12hatk` `=-3(2hati+2hatj+4hatk)=-3AB` `becauseBC=-3AB` `therefore` A,B and C are collinear, Now, let C diivide AB in the ratio k:1, then `OC=(kOB+1*OA)/(k+1)` `implies-3hati-2hatj-5hatk=(k(3hati+4hatj+7hatk)+(2hati+2hatj+3hatk))/(k+1)` `implies-3hati-2hatj-5hatk=((3k+1)/(k+1))hati+((4k+2)/(k+1))hatj+((7k+3)/(k+1))hatk` `implies(3k+1)/(k+1)=-3,(4k+2)/(k+1)=-2 and ((7k+3)/(k+1)=-5` From, all relations, we get `k=(-2)/(3)` Hence, C divides AB externally in the ratio 2:3. |
|
| 121. |
Let ` vec a= hat i+ hat j+ hat k , vec b= hat i- hat j+ hat ka n d vec c= hat i- hat j- hat k`be three vectors. A vector ` vec v`in the plane of ` vec aa n d vec b ,`whose projection on ` vec c`is `1/(sqrt(3))`is given by` hat i-3 hat j+3 hat k`b. `-3 hat i-3 hat j+3 hat k`c. `3 hat i- hat j+3 hat k`d. ` hat i+3 hat j-3 hat k`A. `hat(i)-3hat(j)+3hat(k)`B. `-3hat(i)-3hat(j)-hat(k)`C. `3hat(i)-hat(j)+3hat(k)`D. `hat(i)+3hat(j)-3hat(k)` |
|
Answer» Correct Answer - C Let `vec(v)= vec(a) +lambda vec(b)` `vec(b) =(1 + lambd) hat(i) +(1-lambda) hat(j) (1+lambda) hat(k)` Projection of `vec(v) " on " vec(c ) = (1)/(sqrt(3)) rArr (vec(v).vec(c ))/(|vec(c )|) = (1)/(sqrt(3))` `rArr ((1+lambda)-(1-lambda)-(1+lambda))/(sqrt(3)) =(1)/(sqrt(3))` `rArr 1+lambda -lambda -1 +lambda -1 -lambda =1` `rArr vec(v) =3 hat(i) - hat(j) + 3hat(k)` |
|
| 122. |
The unit vector which is orthogonal to the vector `3hati+2hatj+6hatk` and is coplanar with the vectors `2hati+hatj+hatk` and `hati -hatj+hatk` isA. `(2hat(i) -6hat(j)+hat(k))/(sqrt(41))`B. `(2hat(i) - 3hat(j))/(sqrt(13))`C. `(3hat(j) - hat(k))/(sqrt(10))`D. `(4hat(i) +3hat(j)-3hat(k))/(sqrt(34))` |
|
Answer» Correct Answer - C As we know that a vector coplanar to `vec(a) ". " vec(b)` and orthogonal to `vec(c ) " is " lambda {(vec(a) ". " vec(b)) xx vec(c )}` `:.` A vector coplanar to `(2hat(i) + hat(j) + hat(k)) , (hat(i) - hat(j) + hat(k)) ` and orthogonal to `3hat(i) + 2hat(j) +6hat(k)` `= lambda [{(2hat(i) + hat(j) + hat(k)) xx (hat(i) - hat(j)+ hat(k))} xx (3hat(i) + 2hat(j) + 6hat(k))]` `= lambda [(2hat(i) -hat(j) -3hat(k)) xx (3hat(i) + 2hat(j) +6hat(k))]` `=lambda (21 hat(j) = 7hat(k))` `:.` Unit vectors `=+ ((21 hat(j) -7hat(k)))/(sqrt((21)^(2) (7)^(2))) =+ ((3hat(j) -hat(k)))/( sqrt(10))` |
|
| 123. |
If `a=hati+2hatj+3hatk,b=-hati+2hatj+hatk and c=3hati+hatj`, then the unit vector along its resultant isA. `3hati+5hatj+4hatk`B. `(3hati+5hatj+4hatk)/(50)`C. `(3hati+5hatj+4hatk)/(5sqrt(2))`D. none of these |
|
Answer» Correct Answer - C `R=3hati+5hatj+4hatk` `impliesR=(3hati+5hatj+4hatk)/(5sqrt(2))`. |
|
| 124. |
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. `-8`B. `-6`C. 8D. 6 |
|
Answer» Correct Answer - B `AB=(hati-3hatj)-(2hati+hatj)=-hati-4hatj`. `CD=(hati+lamdahatj)-(3hati+2hatj)=-2hati+(lamda-2)hatj` `AB||CDimpliesAB=xCD` `-hati-4hatj=x{-2hati+(lamda-2)hatj}` `implies -1=-2x,-4=(lamda-2)x` `impliesx=(1)/(2) and lamda=-6`. |
|
| 125. |
If the points `a+b,a-b and a+kb` be collinear, then k is equal to |
|
Answer» Correct Answer - D `(a-b)-(a+b)=[(a+kb)-(a-b)]` `implies -2b=(k+1)b` Hence, `k in R` |
|
| 126. |
If the vectors `3hati+2hatj-hatk and 6hati-4xhatj+yhatk` are parallel, then the value of x and y will beA. `-1,-2`B. `1,-2`C. `-1,2`D. `1,2` |
|
Answer» Correct Answer - A Obviously, `(3)/(6)=(2)/(-4x)=-(1)/(y)` `implies x=-1 and y=-2`. |
|
| 127. |
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 Statement I and Statement II are 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 - C In `DeltaABC,AB+BC=AC=-CA` or `AB+BC+CA=0` OA+AB=OB is the triangle law of addition. Hence, statement 1 is true and statement 2 is false. |
|
| 128. |
If the position vectors of A,B,C and D are `2hati+hatj,hati-3hatj,3hati+2hatj and hati+lamda hatj` respectively and `vec(AB) || vec(CD)`. Then `lamda` will beA. `-8`B. `-6`C. 8D. 6 |
|
Answer» Correct Answer - B `AB=-hati-4hatj,CD=-2hati+(lamda-2)hatj` `because AB||CD` So, `(-1)/(-2)=(-4)/(lamda-2)implies lamda-2=-8` `implies lamda=-6` |
|
| 129. |
Statement I: `a=hati+phatj+2hatk and b=2hati+3hatj+qhatk` are parallel vectors, iff `p=(3)/(2) and q=4`. Statement II: `a=a_(1)hati+a_(2)hatj+a_(3)hatk and b=b_(1)hati+b_(2)hatj+b_(3)hatk` are parallel `(a_(1))/(b_(1))=(a_(2))/(b_(2))=(a_(3))/(b_(3))`.A. Both Statement I and Statement II are 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 - A `(1)/(2)=(p)/(3)=(2)/(q) implies p=(3)/(2) and q=4` |
|
| 130. |
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 Statement I and Statement II are 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 - A `2a+3b-5c=0` `3(b-a)=5(c-a)` `implies AB=(5)/(3)AC` Hence, AB and AC must be parallel since there is a common point A. the points A,B and C must be collinear. |
|
| 131. |
Three points A,B, and C have position vectors `-2veca+3vecb+5vecc, veca+2vecb+3vecc` and `7veca-vecc` with reference to an origin O. Answer the following questions? Which of the following is true?A. 2OA-3OB+OC=0B. 2OA+7OB+9OC=0C. OA+OB+OC=0D. none of these |
|
Answer» Correct Answer - A `2OA-3OB+OC` `=2(-2a+3b+5c)-3(a+3b+3c)+(7a-c)=0` |
|
| 132. |
Three points A,B, and C have position vectors `-2veca+3vecb+5vecc, veca+2vecb+3vecc` and `7veca-vecc` with reference to an origin O. Answer the following questions? Which of the following is true?A. AC=2ABB. AC=-3ABC. AC=3ABD. none of these |
|
Answer» Correct Answer - C `AB=OB-OA=3a-b-2c` `AC=OC-OA=9a-3b-6c=3AB` |
|
| 133. |
Prove that if `cos alpha ne 1, cos beta ne1 and cos gamma ne 1`, then the vectors `a=hati cos alpha+hatj+hatk,b=hati+hatj cos beta+hatk`. `c=hati+hatj+hatk cos gamma` can never be coplanar. |
|
Answer» Suppose that, a,b and c are coplanar. `implies|(cosalpha,1,1),(1,cosbeta,1),(1,1cos gamma)|=0` On applying `R_(2) to R_(2)-R_(1) and R_(3) to R_(3)-R_(1)` `implies|(cos alpha,1,1),(1-cosalpha,cosbeta-1,0),(1-cosalpha,0,cosgamma-1)|=0` `implies cosalpha(cosbeta-1)(cosgamma-1)-(1-cosalpha)(cosgamma-1)-(1-cosalpha)(cosbeta-1)=0` On dividing throughout by `(1-cosalpha)(1-cosbeta)(1-cosgamma),` we get `(cosalpha)/(1-cosalpha)+(1)/(1-cosbeta)+(1)/(1-cosgamma)=0` `implies(-(1-cosalpha)+1)/(1-cosalpha)+(1)/(1-cosbeta)+(1)/(1-cosgamma)=0` `implies-1+(1)/((1-cosalpha))+(1)/((1-cosbeta))+(1)/((1-cosgamma))=0` `(1)/(1-cosalpha)+(1)/(1-cosbeta)+(1)/(1-cosgamma)=1` `implies"cosec"^(2)(alpha)/(2)+"cosec"^(2)(beta)/(2)+"cosec"^(2)(gamma)/(2)=2,` which is not possible As, `"cosec"^(2)(alpha)/(2)ge1,"cosec"^(2)(beta)/(2)ge1` and `"cosec"^(2)(gamma)/(2)ge1` `because`They cannot be coplanar. |
|
| 134. |
If the vectors `xhati+hatj+hatk,hati+yhatj+hatk and hati+hatj+zhatk` are coplanar where, `x ne1,y ne1 and z ne1`, then prove that `(1)/(1-x)+(1)/(1-y)+(1)/(1-z)=1` |
|
Answer» The vectors are coplanar, if we can fiind two scalars `lamda and mu` such that `(xhati+hatj+hatk)=lamda(hati+yhatj+hatk)+mu(hati+hatj+zhatk)` `impliesx=lamda+mu,1=lamda y+mu,1=lamda+muz` `implies x=lamda+mu,y=(1-mu)/(lamda),z=(1-lamda)/(mu)` `implies 1-x=1-lamda-mu,1-y=(lamda-1+mu)/(lamda)` `1-z=(mu-1+lamda)/(mu)` `therefore(1)/(1-x)+(1)/(1-y)+(1)/(1-z)=(1)/(1-lamda-mu)+(lamda)/(lamda+mu-1)+(mu)/(lamda+mu-1)` `=(-1+lamda+mu)/(lamda+mu-1)=1` `implies(1)/(1-x)+(1)/(1-y)+(1)/(1-z)=1`. |
|
| 135. |
If the position vectors of the points A,B and C be `hati+hatj,hati-hatj` and `ahati+bhatj+chatk` respectively, then the points A,B and C are collinear, ifA. a=b=c=1B. a=1,b and c are arbitrary scalarsC. ab=c=0D. c=0,a=1 and b is arbitrary scalars |
|
Answer» Correct Answer - D Here, `AB=-2hatj,BC=(a-1)hati+(b+1)hatj+chatk` The points are collinear, then `AB=k(BC)` `-2hatj=k{(a-1)hati+(b+1)hatj+chatk}` On comparing, `k(a-1)=0,k(b+1)=-2,kc=0` Hence, c=0,a=1 and b is arbitrary scalar. |
|
| 136. |
If `ahati+hatj+hatk` `hati+bhatj+hatk` `hati+hatj+chatk` are coplaner then `1/(1-a)+1/(1-b)+1/(1-c)=` |
|
Answer» Correct Answer - 1 Since vectors are coplanar . `:. |{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|=0` Applying `R_(2) to R_(2) -R_(1) ,R_(3) to R_(3) -R_(1),` `|{:(a,,1,,1),(1-a,,b-1,,0),(1-a,,0,,c-1):}|=0` `|{:(a//(1-a),,1//(1-b),,1//(1-c)),(1,,-1,,0),(1,,0,,-1):}|=0` `rArr (a)/(1-a) (1)/(1-b) (-1) +(1)/(1-c)(1)=0` `rArr (a)/(1-a) +(1)/(1-b) +(1)/(1-c) =0` ` rArr -1 +(1)/(1-a) +(1)/(1-b)+(1)/(1-c) =0` `rArr (1)/(1-a) +(1)/(1-b) +(1)/(1-c)=1` |
|
| 137. |
If `A=(0,1)B=(1,0),C=(1,2),D=(2,1)`, prove that ` vec A B= vec C Ddot` |
|
Answer» Here, `AB=(1-0)hati+(0-1)hatj=hati-hatj` and `CD=(2-1)hati+(1-2)hatj=hati-hatj` clearly, AB=CD Hence proved. |
|
| 138. |
If the position vectors of A and B respectively `hati+3hatj-7hatk and 5 hati-2hatj+4hatk`, then find AB |
|
Answer» Let O be the origin, then we have `OA=hati+3hatj-7hatk` and `OB=5hati-2hatj+4hatk` Now, `AB=OB-OA=(5hati-2hatj+4hatk)-(hati+3hatj-7hatk)` `=4hati-5hatj+11hatk` |
|
| 139. |
Let `vec(alpha)=ai+bj+ck` and `vec(beta)=bi+cj+ak`, where `a`, `b`, `c` `in R` If `theta` be the angle between `alpha` and `beta` then, |
|
Answer» `vecalpha=ahati+bhatj+chatk` `vecbeta=bhati+chatj+ahatk` `vecalpha*vecbeta=ab+bc+ca` `|vecalpha||vecbeta|costheta=ab+bc+ca` `costheta=(ab+bc+ca)/(a^2+b^2+c^2)` `1/2((a+b+c)^2-(a^2+b^2+c^2))/(a^2+b^2+c^2)` `costheta> -1/2` `theta in (pi-pi/3,pi)` `theta in (2/3pi/pi)` option c is correct. |
|
| 140. |
Vectors drawn the origin `O`to the points `A , Ba n d C`are respectively ` vec a , vec ba n d vec4a- vec3bdot`find ` vec A Ca n d vec B Cdot` |
|
Answer» We have, `OA=a,OB=b and OC=4a-3b` Clearly, `AC=OC-OA=(4a-3b)-(a)` `=3a-3b` and `BC=OC-OB=(4a-3b)-(b)=4a-4b` |
|
| 141. |
If ` vec a , vec b`are the position vectors of the points `(1,-1),(-2,m),`find the value of `m`for which ` vec aa n d vec b`are collinear.A. 4B. 3C. 2D. 0 |
|
Answer» Correct Answer - C Condition for collinearly, `b=lamda=a` `implies(-2hati+mhatj)=lamda(hati-hatj)` Comparison of coefficient, we get `implies lamda=-2 and -lamda=m` so, m=2 |
|
| 142. |
The points with position vectors `10hati+3hatj,12hati-5hatj and ahati+11hatj` are collinear, if a is equal toA. `-8`B. 4C. 8D. 12 |
|
Answer» Correct Answer - C If given be A,B and C, then `AB=k(BC)` or `2hati-8hatj=k[(a-12)hati+16hatj]` `implies k=-(1)/(2)` also, `2=k(a-12)` `implies a=8` |
|
| 143. |
p=2a-3b,q=a-2b+c and r=-3a+b+2c, where a,b,c being non-coplanar vectors, then the vector -2a+3b-c is equal toA. `p-4q`B. `(-7q+r)/(5)`C. `2p-3q+r`D. `4p-2r` |
|
Answer» Correct Answer - B Let `-2a+3b-c=xp+yq+zr` `implies-2a+3b-c=(2x+y-3z)a+(-3x-2y+z)b+(y+2z)c` `therefore 2x+y-3z=-2,-3x-2y+z=3` and `y+2z=-1` On solving these we get `x=0,y=-(7)/(5),z=(1)/(5)` `therefore-2a+3b-c=((-7q+r))/(5)` trick check alternates one-by-ne ie.., (a) `p-4q=-2a+5b-4c` (b) `(-7q+r)/(5)=-2a+3b-c` |
|
| 144. |
Show that points with position vectors `2-2b+3c,-2a+3b-c and 4a-7b+7c` are collinear. It is given that vectors a,b and c and non-coplanar. |
|
Answer» The three points are collinear, if we can find `lamda_(1),lamda_(2) and lamda_(3),` such that `lamda_(1)(a-2b+3c)+lamda_(2)(-2a+3b-c)+lamda_(3)` `(4a-7b+7c)=0` with `lamda_(1)+lamda_(2)+lamda_(3)=0` On equating the coefficient a,b and c separately to zero, we get `lamda_(1)-2lamda_(2)+4lamda_(3)=0,-2lamda_(1)+3lamda_(2)-7lamda_(3)=0` and `3lamda_(1)-lamda_(2)+7lamda_(3)=0` On solving we get `lamda_(1)=-2,lamda_(2)=1,lamda_(3)=1` so that, `lamda_(1)+lamda_(2)+lamda_(3)=0` hence, the given vectors are collinear. |
|
| 145. |
If a,b,c are non-coplanar vectors and `lamda` is a real number, then the vectors `a+2b+3c,lamdab+4c and (2lamda-1)c` are non-coplanar forA. all value of `lamda`B. all except one value of `lamda`C. all except two value of `lamda`D. no value of `lamda` |
|
Answer» Correct Answer - C The three vectors (a+2b+3c),`(lamdab+4c) annd (2lamda-1)`c are non-coplanar, if `|(1,2,3),(0,lamda,4),(0,0,2lamda-1)|ne0` `implies (2lamda-1)(lamda) ne0` `implies lamda ne0,(1)/(2)` So, these three vectors are non-coplanar for all except two values of `lamda`. |
|