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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. |
Find the length and the equations of the line of shortest distance between the lines `(x-8)/(3)=(y+9)/(-16) =(z-10)/(7) " and " (x-15)/(3)=(y-29)/(8)=(z-5)/(-5)` |
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Answer» The equations of the given lines are `(x-8)/(3)=(y+9)/(-16) =(z-10)/(7)=lambda`(say) `(x-15)/(3)=(y-29)/(8) =(z-5)/(-5)=mu `(say) Any point on the line (i) is `P(3lambda +8,-16lambda -9,7lambda +10)` Any point on the line(ii) is `Q( 3mu+ 15,8mu +29 ,- 5mu +5)` The direction rations of PQ are `(3mu -3lambda +7, 8mu+ 16lambda +38 , -5mu -7lambda -5)` Now PQ will be the shortest distance between (i) and (ii) only when PQ is perpendicular to each one of (i) and (ii) `:.{underset(3(3mu -3lambda+7) +8(8mu+16lambda+38)-5(-5mu-7lambda-5)=0)underset(" and ")(3(3mu -3lambda +7) -16(8mu +16lambda+38)+7(-5mu-7lambda-5)=0)` `rArr {underset(154lambda +98mu +350=0 )(314lambda +154mu+622 =0)rArr {underset(77lambda +49mu +175 =0)(157lambda +77mu +311=0)` On multiplying (iii) by 7 and (iv) by 11 and subtracting , we get `(1099 lambda -847 lambda) +(2177- 1925) =0 rArr 252 lambda=-252 rArr lambda =-1` Putting `lambda=-1 ` in (iv) we get `49 mu +(175 -77) =0 rArr 49 mu +98 =0 rArr 49mu =-98 rArr mu =-2` Now `lambda =-1` gives P (5,7,3) and `mu =-2 ` gives Q(9,13,15) `:.SD =|PQ| =sqrt((9-5)^(2)+(13-7)^(2) +(15 -3)^(2))` `=sqrt(16+36 +144)=sqrt(196) =14` units Equations of the line of shortest distance is the equation of PQ given by `(x-5)/(9-5) =(y-7)/(13-7) =(z-3)/(15-3)` `rArr (x-5)/(4)=(y-7)/(6)=(z-3)/(12)rArr (x-5)/(2)=(y-7)/(3)=(z-3)/(6)` Hence the equations of the line of shortest distance is `(x-5)/(2)=(y-7)/(3)=(z-3)/(6)` |
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| 2. |
रेखाएँ, जिनकी सदिश समीकरण निम्नलिखित है, के बीच की न्यूनतम ज्ञात कीजिए : ` vecr = (1- t ) hati + ( t - 2) hatj + ( 3- 2t ) hatk ` और ` vec r = ( s + 1 ) hati + ( 2s - 1 ) hatj - ( 2s + 1 ) hatk ` |
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Answer» The given equations can be written as `vec (r ) =(hat(i) -2hat(j) + 3hat(k)) + t (-hat(i) +hat(j) -2hat(k)) ` and ` vec(r ) =(hat(i) -hat(j) -hat(k)) + s (hat(i) +2hat(j) -2hat(k))` Comparing the given equations with the standard equations `vec( r) = vec(a)_(1) + t vec(b)_(1) " and " vec(a)_(2) + s vec(b)_(2) ` we get `vec( a)_(1) =(hat(i) -2hat(j) +2hat(k)), vec(b)_(1) =(-hat(i) +hat(j) -2hat(k))` `vec(a)_(2) =(hat(i)- hat(j)-hat(k)) " and " vec(b)_(2) =(hat(i) +2hat(j) -2hat(k))` `:. (vec(a)_(2) -vec(a)_(1)) =(hat(i) -hat(j) -hat(k)) =(hat(i) -2hat(j) +3hat(k)) =(hat(j) -4hat(k))` and `(vec(b)_(1)xx vec(b)_(2)) =|{:(hat(i),,hat(j),,hat(k)),(-1,,1,,-2),(1,,2,,-2):}|` ` =(-2 +4) hat(i) -(2+2) hat(j) + (-2 -1) hat(k)` `=(2hat(i)- 4hat(j)-3hat(k))` `:. |vec(b)_(1)xx vec(b)_(2)| =sqrt(2^(2)+(-4)^(2) +(-3)^(2))=sqrt(29)` `:. SD =|((vec(a)_(2)-vec(a)_(1)).(vec(b)_(1)xxvec(b)_(2)))/(|vec(b)_(1)xx vec(b)_(2)|)|` `=(|(hat(j) -4hat(k)).(2hat(i)-4hat(j)-3hat(k))|)/(sqrt(29)) =(|0-4+12|)/(sqrt(29))` ` =(8sqrt(29))/(29) ` units |
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| 3. |
Find the Cartesian and vector equations of a line which passes through the point (1,2,3) and is parallel to the line . `(-x-2)/(1) =(y+3)/(7) =(2z-6)/(3)` |
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Answer» Correct Answer - `(x-1)/(-2) =(y-2)/(14)=(z-3)/(3) ,vec( r) =(hat(i)+2hat(j) +3hat(k)) +lambda (-2hat(i) +14hat(j) +4hat(k))` The given equations may be written as `(x+2)/(-1) =(y+3)/(7) +(z-3)/(3) rArr (x+2)/(-2) =(y+3)/(14)=(z-3)/(3)` |
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| 4. |
Show that the lines `(x-1)/(2)=(y-2)/(3)=(z-3)/(4) " and " (x-4)/(5)=(y-1)/(2)=z` intersect each other . Find their point of intersection. |
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Answer» The equations of the given lines are `(x-1)/(2)=(y-2)/(3)=(z-3)/(4)=lambda`(say) `(x-4)/(5)=(y-1)/(2)=(z-0)/(1)=,mu` Any point on the line (i) is `P(2lambda+1,3lambda +2,4lambda +3)` Any point on the line (ii) is `Q(5mu+4,2 mu+1,mu)` If the lines (i) and (ii) intesect then P and Q must coincide for some particular values of `lambda " and " mu` this gives `2lambda +1= 5mu+ 4,3lambda+ 2=2mu +1 " and " 4lambda +3=mu` `rArr {underset(4lambda -mu=-3)underset(3lambda -2mu =-1)(2lambda-5mu =3)` On solving (i) and (ii) we get `lambda =-1 " and " mu =-1` These values of `lambda " and " mu` also satify (iii) Hence the given lines intersect . putting `lambda=-1 " we get " P(-1,-1,-1)` Note that putting `mu =-1 `we get Q(-1,-1,-1) Hence thhe point of intersection of the given lines is (-1,-1,-1) |
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| 5. |
Show that the lines `vec( r) =(hat(i) +hat(j) -hat(k)) +lambda (3hat(i) -hat(j)) " and " vec( r) =(4hat(i) -hat(k)) + mu (2hat(i) +3hat(k))` intersect . Find the point of the intersection. |
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Answer» Comparing the given equations with the standard equations `vec(r ) =vec(a)_(1) =lambda vec(b)_(1) " and " vec(r) =vec(a)_(2) +mu vec(b)_(2)` we get `vec(a)_(1) =(hat(i) +hat(j)-hat(k)) ,vec(b)_(1) =(3hat(i) -hat(j))` `vec(a)_(2) =(4hat(i) -hat(k)) " and " vec(b)_(2) =(2hat(i) +3hat(k))` `:. (vec(a)_(2) -vec(a)_(1)) =(4hat(i)-hat(k)) -(hat(i) +hat(j)-hat(k)) =(3hat(i) -hat(j))` And `(vec(b)_(1) xx vec(b)_(2)) =|{:(hat(i),,hat(j),,hat(k)),(3,,-1,,0),(2,,0,,3):}|=(-3-0)hat(i)-(9-0)hat(j) +(0+2) hat(k)` `=(-3hat(i) -9hat(j) +2hat(k))` `:. |vec(b)_(1)xx vec(b)_(2)| =sqrt((-3)^(2)+(-9)^(2) +2^(2))=sqrt(94)` `:. SD =|((vec(a)_(2)-vec(a)_(1)).(vec(b)_(1)xxvec(b)_(2)))/(|vec(b)_(1)xx vec(b)_(2)|)|` `=(|(3hat(i)-hat(j)).(-3hat(i)-9hat(j)+2hat(k))|)/(sqrt(94))` `=(|-9+9+0|)/(sqrt(94))=0` Thus the shorest distance between the given lines is 0 Hence the given lines intersect. Thus for some particular values of `lambda " and " mu` we have `(hat(i)+hat(j)-hat(k)) +lambda (3hat(i)-hat(j))=(4hat(i)-hat(k)) +mu (2hat(i) +2hat(k))` `rArr (1+3lambda)hat(i) +(1-lambda)hat(j) -hat(k) =(4+2mu)hat(i) +(3mu-1)hat(k)` , `rArr 1+3lambda =4 + 2 mu , 1 - lambda =0 " and " 3 mu -1 =-1` `rArr lambda=1 " and " mu=0` Thus the position vector of the point of intersection of the given lines is given by `vec(r ) =(hat(i) +hat(j) -hat(k)) +(3hat(i) -hat(j))` [putting `lambda=1`] ,i.e., `vec(r ) =(4hat(i)-hat(k))` Hence the point of intersection of the given lines is P(4,0,-1) |
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| 6. |
Show that the lines `(x+1)/(3)=(y+3)/(5) =(z+5)/(7) " and " (x-2)/(1)=(y-4)/(3)=(z-6)/(5)` intersect. Also find their point of intersection. |
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Answer» The given lines are `(x-1)/(3)=(y+3)/(5)=(z+5)/(7)=lambda `(say) and `(x-2)/(1)= (y-4)/(3)=(z-6)/(5)=mu` (say) The general point on (1) is P `(3lambda -1 ,5lambda -3,7lambda -5)` The general point on (2) is Q `(mu +2 ,3mu +4, 5mu +6)` The given lines will intersect only when they have a common point. This happens when P and Q coincide for some particular values of `lambda " and " mu` So , the given lines will intersect only when `3lambda-1=mu +2,5lambda-3=3mu +4 "and " 7lambda -5=5mu+6` `rArr 3lambda -mu =3....(i) ,5lambda -3mu,=7 .....(ii) "and " 7lambda -5mu =11` On solving (i) and (ii) we get `lambda =(1)/(2) " and " mu =(-3)/(2)` Clearly these values of `lambda " and " mu ` also satisfy (iii) Hence the given lines intersect . Putting `lambda=(1)/(2) " in P or " mu =(-3)/(2) ` in Q we get the point of intersection of the given lines as `((1)/(2),(-1)/(2),(-3)/(2))` |
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| 7. |
Find the angle between the lines ` vec r=2 hat i-5 hat j+ hat k+lambda(3 hat i+2 hat j+6 hat k)`and ` vec r=7 hat i-6 hat k+mu( hat i+2 hat j+2 hat k)dot` |
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Answer» Correct Answer - `cos^(-1) ((19)/(21))` the angle between the lines and `vec(r ) = a_(1) + lambda b_(1) " and " vec(r )=vec(a)_(2) + lambda b_(2) ` is given by `" cos " theta =(vec(b)_(1).vec(b)_(2))/(|vec(b)_(1)||vec(b)_(2)|)` `:. " cos " theta =(|(3hat(i) +2hat(j) +6hat(k)).(hat(i) +2hat(j) +2hat(k))|)/{{sqrt(sqrt(3^(2)+2^(2)+6^(2)))}{sqrt(1^(2)+2^(2)+2^(2))}} =((3 +4 +12))/((7xx3)) =(19)/(21)` `rArr theta = " cos "^(-1) ((19)/(21))` |
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| 8. |
Show that the lines `(x-1)/(3)=(y+1)/(2) =(z-1)/(5) " and " (x+2)/(4)=(y-1)/(3)=(z+1)/(-2)` do not intersect . |
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Answer» The given lines are `(x-1)/(3)=(y+1)/(2)=(z-1)/(5)= ` (say) `(x+2)/(4) =(y-1)/(3)=(z+1)/(-2) = mu` (say) The general point on (1) is `P(3lambda+ +1,2lambda-1,5lambda+1)` The general point on (2) is Q `(4mu -2, 3mu +1, -2mu -1)` If possible let the given lines intersect . Then P and Q coincide for some particular values of `lambda` and `mu` In that case we have `3lambda +1 =4mu -2, 2lambda -1 , =3mu +1 " and " 5lambda +1 =-2mu -1` `hArr 3lambda -4mu =-3 ...(i) ,2lambda -3mu =2 ....(ii) ,5lambda +2mu =-2 .....(iii)` solving (i) and (ii) we get `lambda =-17 " and " mu=-12` However these values of `lambda " and " mu ` do not satisfy (iii) Hence the given lines do not intersect . |
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| 9. |
Show that the lines `vec(r) =(hat(i) +2hat(j) +hat(k)) +lambda (hat(i)-hat(j)+hat(k)) " and " vec(r ) =(hat(i) +hat(j) -hat(k)) + mu (hat(i)- hat(j) + 2hat(k))` Do not intersect . |
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Answer» The given lines are `vec(r ) =(hat(i) +2hat(j) +hat(k)) +lambda (hat(i) -hat(j) +hat(k))` ` vec(r) =(hat(i) +hat(j)+hat(k)) +mu (hat(i)-hat(j)+hat(k))` These lines will intesect if for some particular values of `lambda " and " mu ` the values of `vec(r ) ` given by (1) and (2) are the same. This happens when `(hat(i) +2hat(j) + hat(k)) +lambda (hat(i)-hat(j) +hat(k)) =(hat(i) +hat(j) +hat(k))+mu (hat(i)-ha(j) +2hat(k))` `rArr (1+lambda)hat(i) +(2-lambda)hat(j)+(1+lambda)hat(k)=(1+mu)hat(i) +(1-mu)hat(j)+(1+2mu)hat(k)` `rArr 1+lambda =1+mu , 2-lambda =1 -mu " and " 1+lambda =1+2mu` `rArr lambda-mu=0 ....(i) , lambda-mu=1 .....(ii) , lambda-2mu=0 .....(iii)` From (ii) and (iii) we get `lambda=2 " and " mu =1 ` And these values of `lambda "and " mu` do not satisfy (i) Hence the given lines do not intersect . |
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| 10. |
Show that the lines ` vec r=( hat i+ hat j- hat k)+lambda(3 hat i- hat j)a n d vec r=(4 hat i- hat k)+mu(2 hat i+3 hat k)`are coplanar. Also, find the plane containing these two lines. |
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Answer» The given lines are `vec(r ) =(hat(i) +hat(j) -hat(k)) +lambda (3hat(i) -hat(j))` `vec( r) =(4hat(i) -hat(k)) +mu (2hat(i) +3hat(k))` These lines will intersect if for some particular values of `lambda " and " mu` the values `vec( r) ` given by (1) and (2) are the same So, we must have `(hat(i) +hat(j)-hat(k))+lambda(3hat(i)-hat(j))=(4hat(i)-hat(k))+mu(2hat(i)+3hat(k))` `rArr (1+3lambda)hat(i) +(1-lambda)hat(j) -hat(k)=(4+2mu) hat(i) +(3mu-1)hat(k)` `rArr 1+3lambda =4+2mu,1 -lambda =0 " and " 3mu-1 =-1` `rArr 3lambda -2mu =3 .....(i),lambda , =1 ...(i) " and " mu =0 .....(iii)` Clearly `lambda=1 " and " mu =0` also satisfy (i) Hence the given lines intersect . Putting `lambda =1 " in (1) we get " vec(r ) =(4hat(i) +0hat(j) -hat(k))` Hence the point of intersection of the given lines is (4,0 ,-1 ) |
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| 11. |
Show that the lines `vec( r)=(2hat(i) -3hat(k)) + lambda(hat(i) +2hat(j)+3hat(k)) " and " vec(r )=(3hat(i) +6hat(j)+3hat(k)) + mu(2hat(i) +3hat(j) +4hat(k))` intersect. Also find their point of intersection. |
| Answer» Correct Answer - `(2,6,3)` | |
| 12. |
Show that the lines `vec(r )=(3hat(i) -15hat(j) + 9hat(k)) + lambda(2hat(i) -7hat(j) +5hat(k))` and `vec(r )=(-hat(i) + hat(j) + 9hat(k)) + mu (2hat(i) + hat(j) -3hat(k))` do not intersect. |
| Answer» Correct Answer - N//A | |
| 13. |
`vec(r ) =(6hat(i) +3hat(k) ) + lambda(2hat(i) -hat(j) +4hat(k))` `vec(r )=(-9hat(i) +hat(j) -10hat(k)) + mu (4hat(i) +hat(j) +6hat(k))` |
| Answer» Correct Answer - `sqrt(38)` units | |
| 14. |
If A(1,2,3),B(4,5,7) ,C(-4,3,-6) and D(2,9,2) are four given points then find the angle between the lines AB and CD. |
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Answer» Correct Answer - `0^(@)` `vec(AB) =(p.v. of B) -(p.v.of A) =(4hat(i)+5hat(j)+7hat(k)) -(hat(i) +2hat(j)+3hat(k)) =(3hat(i)+3hat(j)+4hat(k))` `vec(CD) =(p.v.of D) -(p.v.of C) =(2hat(i) +9hat(j) +2hat(k))-(-4hat(i) +3hat(j)-6hat(k)) =(6hat(i) +6hat(j)+8hat(k))` `" cos " theta =(|vec(AB).vec(CD)|)/(|vec(AB).vec(CD)|) =((3xx6 +3 xx 6 + 4xx8))/((sqrt(9+9+16))(sqrt(36+36+64)))` `=(68)/(sqrt(34)xxsqrt(136))=(68)/(34xx2)=1` |
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| 15. |
A line passes through the points A(2,-1,1) and B(1,2-2) .The equations of the lines AB areA. `(x-2)/(-1) =(y+1)/(2)=(z-4)/(-6)`B. `(x+2)/(-1) =(y+1)/(2)=(z-4)/(6)`C. `(x-2)/(1) =(y+1)/(2)=(z-4)/(6)`D. none of these |
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Answer» Correct Answer - A The given line passes through the point A(2,-1,4) and its direction rations are `(1-2),(1+1),(-2-4), i.e., -1,2-6 ` `:.` required equations of the line are `(x-2)/(-1) =(y+1)/(2) =(x-4)/(-6)` |
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| 16. |
Show that the lines `(x-3)/(2)=(y+1)/(-3) =(z-2)/(4) " and " (x+2)/(2) =(y-2)/(4) =(z+5)/(2)` are perpendicular to each other . |
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Answer» Given lines in standard `(x)/(2) =(y-0)/(-2) =(z-5)/(-1) " and " (x+2)/(2)=(y-(1)/(2))/(1)=(z-1)/(-2)` |
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| 17. |
Find the angle between the lines `(5-x)/(3)=(y+3)/(-4) ,z=7 " and " (x)/(1) =(1-y)/(2)=(z-6)/(2)` |
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Answer» The given lines in standard form are `(x-5)/(-3) =(y+3)/(-4) =(z-7)/(0) " and " (x)/(1) =(y-1)/(-2) =(z-6)/(2)` Here `a_(1)=-3,b_(1) =-4 ,c_(1) =0 "and " a_(2) =1, b_(2) =-2 ,c_(2) =2` Let `theta ` be the angle between the given lines. Then `" cos " theta =(|a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)|)/({sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2)}}{sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2))}}` `=(|(-3)xx1+(-4)xx(-2)+0xx2|)/({sqrt(9+16+0}}{sqrt(1+4+4}}}=(5)/(5xx3) =(1)/(3)` `:. theta = cos ^(-1) .((1)/(3))` Hence the angle between the given lines is `cos^(-1) .((1)/(3))` |
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| 18. |
Find the shortest distance between lines ` -> r=6 hat i+2 hat j+ hat k+lambda( hat i-2 hat j+2 hat k)`and ` -> r=-4 hat i- hat k+mu(3 hat i-2 hat j-2 hat k)`. |
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Answer» Comparing the given equations with the standard equations `vec( r)=vec( a)_(1) =lambda vec( b)_(1) " and " vec( r) =vec(a)_(2) +lambda vec( b)_(2) ` we have `vec(a)_(1) =(6hat(i) +2hat(j) +2hat(k)) ,vec(b)_(1) =(hat(i) -2hat(j) +2hat(k))` `vec(a)_(2) =(-4hat(i) -hat(k)) " and " vec( b)_(2) =(3hat(i) -2hat(j) -2hat(k))` `:. (vec(a)_(2) -vec(a)_(1))=(-4hat(i) -hat(k)) -(6hat(i) +2hat(j)+2hat(k))=(-10hat(i) -2hat(j) -2hat(k))` and `(vec(b)_(1) xx vec(b)_(2)) =|{:(hat(i),,hat(j),,hat(k)),(1,,-2,,2),(3,,-2,,-2):}|=(4+4)hat(i)-(-2-6) hat(j) +(-2+6)hat(k)` ` =(8hat(i)+8hat(j) +4hat(k))` `:. |vec(b)_(1) xx vec(b)_(2)|=sqrt(8^(2) +8^(2)+4^(2))=sqrt(64+64 +16)` `:. SD =|((vec(a_(2)-vec(a)_(1))).(vec(b)_(1)xxvec(b)_(2)))/(|vec(b)_(1)xxvec(b)_(2)|)|` `=|((-10hat(i)-2hat(j)-3hat(k)).(8hat(i)+8hat(j)+4hat(k)))/(12)| =|(-80 -16 -12)/(12)|` `=|(-108)/(12)| =|-9| =9` units |
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| 19. |
Find the value of k so that the lines `(1-x)/(3)=(7y-14)/(2k) =(z-3)/(2) " and " (7-7x)/(3k) =(5-y)/(1)=(6-z)/(5)` are at right angles. |
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Answer» The given equation in standard form are `(x-1)/(-3) =(y-2)/((2k)/(7)) =(z-3)/(2) rArr (x-1)/(-21) =(y-2)/(2k) =(z-3)/(14)` `(x-1)/((-3k)/(7)) =(y-5)/(-1) =(z-6)/(-5) rArr (x-1)/(-3k) =(y-5)/(-7) =(z-6)/(-35)` Here `a_(1)=-21 , b_(1) =2k ,c_(1) =14 " and " a_(2) =-3k , b_(2) =-7 ,c_(2) =-35` Since the given lines are at right angles we have , `a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0` `rArr (-21) xx (-3k) +(2k) xx(-7)+14 xx(-35) =0` `rArr 63k -14k-490 =0 rArr 49k =490 rArr k=10` Hence k=10 |
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| 20. |
Assertion: If the lines `(x-1)/(-3)=(y-2)/(2k)=(z-3)/2 and (x-1)/(3k)=(y-1)/1=(z-6)/(-5) ` are perpendicular to each other , then `k=10/7`, Reason: Two lines having diection ratios `l_1,m_1,n_1 and l_2,m_2,n_2` are perpendiculr to each other if and only if `l_1l_2+m_1m_2+n_1n_2=0` (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.A. `(-5)/(7)`B. `(5)/(7)`C. `(10)/(7)`D. `(-10)/(7)` |
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Answer» Correct Answer - D Since the given lines are are perpendicular to each other we have , `(-3) (3k) +(2kxx1) +2xx (-5) =0 rArr 7k =-10 rArr k = (-10)/(7)` |
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| 21. |
Show that the lines `(x-5)/(7) =(y+2)/(-5)=(z)/(1) " and " (x)/(1) =(y)/(2)=(z)/(3)` are at right angles . |
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Answer» Here `(a_(1) =7 ,b_(1)=5,c_(1) =1) " and " (a_(2) =1,b_(2) =2 ,c_(2)=3)` `(a_(1)a_(2) +b_(1)b_(2) +c_(1)c_(2))=(7xx1) +(-5) xx2 +(1xx3) =0` Hence the given lines are at right angles. |
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| 22. |
Write the cartesianequation of the following line given in vector form :` vec r=2 hat i+ hat j-4 hat k+lambda ( hat i- hat j- hat k)` |
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Answer» Correct Answer - `(x-2)/(1)=(y-1)/(-1) =(z+4)/(-1)` The given line passes through the point (2,1,-4) and it is parallel to a line whose direction ratios are 1,-1,-1. So, its Cartesian equation is `(x-2)/(1) =(y-1)/(-1) =(z+4)/(-1)` |
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| 23. |
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3 hat i+2 hat j-2 hat k`. |
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Answer» Correct Answer - `vec(r )=(hat(i)+2hat(j) +3hat(k)) + lambda (3hat(i) +2hat(j) -2hat(k))` Clearly the required equation is `vec(r )=(hat(i) +2hat(j) +3hat(k)) + lambda(3hat(i) +2hat(j)-2hat(k))` |
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| 24. |
Find the vector equation of a line which is parallel to the vector `2 hat i- hat j+3 hat k`and which passes through the point (5, -2,4). Also reduce it toCartesian form. |
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Answer» Vector equation of the given line: The given line passes through the point `A(5,-2,4) " and " `it is parallel to the vector `vec(m) =(2hat(i) -hat(j) +3hat(k))` the position vector of A is given by `vec(r )_(1) =(5hat(i) -2hat(j) +4hat(k))` Hence the vector equations of the given line is `vec(r ) =vec(r )_(1) +lambdavec(m ) rArr vec( r) =(5hat(i) -2hat(j) +4hat(k)) +lambda (2hat(i) -hat(j) +3hat(k)) " ".....(i)` Cartesisan equation of the given line : Taking `vec(r ) =(xhat(i) +yhat(j) +zhat(k))` equation (i) becomes : `(xhat(i) +yhat(j) +zhat(k)) =(5hat(i) -2hat(j)+4hat(k)) +lambda(2hat(i)-hat(j)+3hat(k))` `rArr (xhat(i) +yhat(j) +zhat(k)) =(5+2lambda) hat(i) -(2+lambda ) hat(j) +(4 +3lambda) hat(k)` `rArr x=5 +2lambda ,y=-(2+lambda) " and " z=4 +3lambda` `rArr (x-5)/(2)=(y+2)/(-1) =(z-4)/(3)= lambda` Hence `(x-5)/(2)=(y+2)/(-1) =(z-4)/(3)` are the required equations of the given line in Cartesian form. |
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| 25. |
Find the vector and Cartesian equations of the line passing through the points A(2,-3,0) and B(-2,4,3). |
| Answer» Correct Answer - `vec(r )=(hat(i) -2hat(j) +hat(k)) +lambda( -4hat(i) +7hat(j) +3hat(k)) ,(x-2)/(-4) =(y+3)/(7) =(z)/(3)` | |
| 26. |
Find the image of the point `(0,2,3)` in the line `(x+3)/(5)= (y-1)/(2) = (z+4)/(3)`. |
| Answer» Correct Answer - (4,4,-5) | |
| 27. |
The vector equation of the x-axis is given byA. `vec(r ) =hat(i)`B. `vec(r )=hat(j) +hat(k)`C. `vec(r ) =lambdahat(i)`D. none of these |
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Answer» Correct Answer - C Note that the Cartesian equation of x-axis is `(x-0)/(1)=(y-0)/(0)=(z-0)/(0)` and its vector equations is `vec( r) = lambda hat(i)` |
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| 28. |
Find the angle between the lines `(x+1)/(1)=(2y-3)/(3) =(z-6)/(2) " and "(x-4)/(3)=(y+3)/(-2) ,z=5` |
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Answer» The given equations may be written as `(x+1)/(1) =(y-(3)/(2))/((3)/(2)) =(z-6)/(2) rArr (x+1)/(2)=(y-(3)/(2))/(3)=(z-6)/(4)....(i)` and `(x-4)/(3) =(y+3)/(-2) =(z-5)/(0)` Here `a_(1) =2,b_(1) =3 , c_(1)=4 " and " a_(2) =3 ,b_(2) =-2 ,c_(2) =0` Let `theta` be the angle between the given lines. Then `" cos " theta = (|a_(1)a_(2) +b_(1)b_(2)+c_(1)c_(2)|)/((sqrt(a_(1)^(2) +b_(1)^(2) +c_(1)^(2)))(sqrt(a_(2)^(2) +b_(2)^(2)+c_(2)^(2))))` `=(|(2xx3)+3xx(-2)+4xx0|)/((sqrt(4+9+16))(sqrt(19+4+0)))=(0)/((sqrt(29)xxsqrt(13)))=0` `:. theta =(pi)/(2)` Hence the angle between the given lines is `(pi)/(2)` |
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| 29. |
The cartesian equations of a line are 3x -3 =2y +1=5-6z (a) Write these equations in standard form and find the direction ratios of the given line . (b) write the equations for the given line in vector form. |
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Answer» (a) The given equations may be written as `3(x-1) =2(y+(1)/(2)) =-6 (z-(5)/(6))` `rArr ((x-1))/(((1)/(3)))= ((y+(1)/(2)))/(((1)/(2)))= ((z-(5)/(6)))/(((-1)/(6)))` `rArr ((x-1))/(2)= ((y+(1)/(2)))/(3)=((z+(5)/(6)))/(-1)` This is the standard form of the given equations in cartesian form. Clearly its direction rations are 2,3,-1 (b) The given line passes through the point `A(1,(-1)/(2),(5)/(6))` and it is parallel to the vector `vec( m) =(2hat(i) +3hat(j) -hat(k))` The position vector of the point A is `vec(r )_(1) =( hat(i) -(1)/(2) hat(j)+ (5)/(6) hat(k))` So the vector equation of the given line is `vec(r ) =vec( r) _(1) +lambdavec(m ) rArr vec( r) =(hat(i) -(1)/(2) hat(j) +(5)/(6) hat(k)) + lambda (2hat(i) +3hat(j) -hat(k))` |
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| 30. |
Find the angle between the following pair oflines: `(-x+2)/(-2)=(y-1)/7=(z+3)/(-3)`and `(x+2)/(-1)=(2y-8)/4=(z-5)/4`and check whether the lines areparallel or perpendicular. |
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Answer» The given equations in standard from are `(x-2)/(2)= (y-1)/(7)= (z+3)/(-3) " and " (x+2)/(-1) =(2y-8)/(y-4)/(2)=(z-5)/(4)` Here `a_(1)= 2 ,b_(1)=7 , c_(1)= -3 " and " a_(2)=-1 ,b_(2) =2 ,c_(2) =4` Let `theta ` be the angle between the given lines . Then `" cos " theta = (|a_(1)a_(2)+b_(1)b_(2) +c_(1)c_(2)|)/((sqrt((a_(1)^(2)+b_(1)^(2)+c_(1)^(2))))(sqrt((a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))))` `=(|2xx(-1)+7xx2 +(-3)xx4|)/({sqrt(2^(2)+7^(2)+(-3)^(2)}}{sqrt((-1)^(2)+2^(2)+4^(2))}}` `=(|(-2)+14+(-12)|)/((sqrt(62))(sqrt(21)))=0` `:. theta =(pi)/(2)` Hence the angle between the given lines is `(pi)/(2)` |
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| 31. |
`(x+1)/(1)=(y-4)/(1)=(z-5)/(2) " and " (x+3)/(3) =(y-2)/(5)=(z+5)/(4)` Find angle between the lines . |
| Answer» Correct Answer - `cos^(-1).((8sqrt(3))/(15))` | |
| 32. |
Find the vector and Cartesian equations of the line passing through the point (1,2-4) and perpendicular to each of the lines `(x-8)/(3)=(y+19)/(-16) =(z-10)/(7) " and " (x-15)/(3)=(y+29)/(8) =(z-5)/(-5)` |
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Answer» The given lines are `(x-8)/(3)=(y+19)/(-16) =(z-10)/(7)` `" and " (x-15)/(3)=(y+29)/(8) =(z-5)/(-5)` Let a,b,c be the direction rations of the required line . Then it being perpendicular to each of the lines (i)and (ii) we have 3a-16b+7c=0 and 3a+8b-5c =0 On solving these equations by cross multiplications we get `(a)/((80-56)) =(b)/((21+15))=(c)/((24+48))` `rArr (a)/(24)=(b)/(36) =(c )/(72) rArr (a)/(2)=(b)/(3)=(c)/(6)` Thus the desired line has direction rations 2,3,6 So we have to find the equations of a line passing through te point A(1,2-4) and having 2,3,6 as its directions rations So, the required equations in Cartesian form are `((x-1))/(2)=((y-2))/(3) =((z+4))/(6)` The position vector of point A is `vec(r ) =(hat(i) +2hat(j) -4hat(k))` Also the required line has direction rations 2,3,6 and so it is parallel to the vector `vec( m) =(hat(i) +3hat(j) +6hat(k))` So , its equation in vector form is `vec(r ) =vec(r )_(1) +lambda vec( m) rArr vec(r ) =(hat(i)+ 2hat(j) -4hat(k)) + lambda (2hat(i) +3hat(j) +6hat(k))` |
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| 33. |
Find the equations of the line passing through the point (1,-2,3) and parallel to the line `(x-6)/(3) =(y-2)/(-4) =(z+7)/(5)` Also find the vector form of this equations so obtained . |
| Answer» Correct Answer - `(x-1)/(3)=(y+2)/(-4) =(z-3)/(5) , vec( r) =(hat(i) -2hat(j) +3hat(k)) +lambda (3hat(i) -4hat(j) +5hat(k))` | |
| 34. |
The Cartesian equations of a line are `(x-3)/(2)=(y+2)/(-5) =(z-6)/(4)` Find the vector equations of the line . |
| Answer» Correct Answer - `vec(r ) =(3hat(i) -2hat(j) +6hat(k)) +lambda(2hat(i)-5hat(j)+4hat(k))` | |
| 35. |
The Cartesian equations of a line are `(x-2)/(2)=(y+1)/(3) =(z-3)/(-2)` .What is its vector equations ?A. `vec(r ) =(2hat(i) +3hat(j) -2hat(k)) +lambda (2hat(i) -hat(j) +3hat(k))`B. `vec(r )=(2hat(i) -hat(j) +3hat(k)) +lambda (2hat(i) +3hat(j) -2hat(k))`C. `vec(r ) =(2hat(i) +3hat(j) -2hat(k))`D. none of these |
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Answer» Correct Answer - B The line passes through the point `(2hat(i) - hat(j) +3hat(k))` and it is parallel to the vector `(2hat(i) + 3hat(j) -2hat(k))` So, its vector equations `vec(r ) = (2hat(i) - hat(j) +3hat(k)) + lambda (2hat(i) +3hat(j) -2hat(k))` |
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| 36. |
The straight line `(x-2)/(3)=(y-3)/(1)=(z+1)/(0)` isA. parallel to the x-axisB. parallel to the y-axisC. parallel to the z-axisD. perpendicular to the z-axis |
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Answer» Correct Answer - D If a line is perpendicular to the z-axis then cos `gamma = " cos " (pi)/(2)=0` So, the given line is perpendicular to the z-axis . |
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| 37. |
`vec(r )=(hat(i) +hat(j)) +lambda(2hat(i) -hat(j) +hat(k))` `vec(r )=(2hat(i) +hat(j) -hat(k)) + mu (3hat(i) -5hat(j) +2hat(k))` |
| Answer» Correct Answer - `(10)/(sqrt(59))` units | |
| 38. |
`(5-x)/(3) =(y+3)/(-2) ,z=5 " and " (x-1)/(1)=(1-y)/(3)=(z-5)/(2)` Find the angle between the lines. |
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Answer» Correct Answer - `cos^(-1).((3)/(sqrt(182)))` Given lines in standard form are : `(x-5)/(-3) =(y+3)/(-2) +(z-5)/(0) " and " (x-1)/(1)=(y-1)/(-3) =(z-5)/(2)` |
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| 39. |
`vec(r ) =(3hat(i) +hat(j)-2hat(k)) +lambda (hat(i)-hat(j)-2hat(k)) " and " vec(r ) =(2hat(i) -hat(j) -5hat(k)) + mu (3hat(i) -5hat(j) -4hat(k))` FInd angle between the lines |
| Answer» Correct Answer - `cos^(-1) .((8sqrt(3))/(15))` | |
| 40. |
`vec(r ) =(3hat(i) -4hat(j) +2hat(k)) + lambda (hat(i)+3hat(k)) " and " vec(r ) = 5hat(i) + mu (-hat(i) +hat(j) + hat(k))` Find angle between the lines . |
| Answer» Correct Answer - `cos^(-1).((sqrt(30))/(15))` | |
| 41. |
`vec(r )=(hat(i)-2hat(j)) + lambda (2hat(i) -2hat(j)) " and " vec( r) =3hat(k) + mu (hat(i) + 2hat(j) -2hat(k))` Find angle |
| Answer» Correct Answer - `cos^(-1).((-4)/(9))` | |
| 42. |
`(x-12)/(-9) =(y-1)/(4)=(z-5)/(2) " and " (x-23)/(-6) =(y-19)/(-4) =(z-25)/(3)` |
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Answer» Correct Answer - `26` units` Change the given equations in vector form. |
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| 43. |
Find the angle between the following pairs of lines:(i) ` -> r=2 hat i-5 hat j+ hat k+lambda(3 hat i+2 hat j+6 hat k)`and ` -> r=7 hat i-6 hat k+mu( hat i+2 hat j+2 hat k)`(ii) ` -> r=3 hat i+ hat j-2 hat k+lambda( hat i- hat j-2 hat k)`and ` -A. `cos^(-1)((8sqrt(3))/(15))`B. `cos^(-1) ((6sqrt(2))/(5))`C. `cos^(-1)((5sqrt(3))/(8))`D. `cos^(-1)(5sqrt(2))/(6))` |
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Answer» Correct Answer - A The given lines are `vec(r ) =vec(a)_(1) + lambda vec(b)_(1) " and " vec( r) vec(a)_(2) +lambda vec(b)_(2)` `:. " cos "theta =(|vec(b)_(1).vec(b)_(2)|)/(|vec(b)_(1)||vec(b)_(2)|)=(|(hat(i)-hat(j)-2hat(k)).(3hat(i)-5hat(j) -4hat(k))|)/{{sqrt(1^(2)+(-1)^(2))+(-2)^(2)}{sqrt(3^(2)+(-5)^(2)+(-4)^(2))}}` ``=((3+5 +8))/((sqrt(6)xx sqrt(50))) =(16)/(10sqrt(3)) =(8)/(5)xx(sqrt(3))/(3)=(8)/(15) sqrt(3)` `rArr theta = "cos"^(-1) ((8sqrt(3))/(15))` |
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| 44. |
A line passes through the point (3,4,5) and is parallel to the vector `(2hat(i) +2hat(j) -3hat(k))` . Find the equations of the line in the vector as well as Cartesian forms. |
| Answer» Correct Answer - `vec(r )=(2hat(i)+4hat(j)+5hat(k)) +lambda(2hat(i) +2hat(j)-3hat(k)) ,(x-3)/(2)=(y-4)/(2)=(z-5)/(-3)` | |
| 45. |
The Cartesian equations of a line are `(x-1)/(2)=(y+2)/(3) =(z-5)/(-1)` . Its vector equations isA. `vec(r ) =(-hat(i) +2hat(j) -5hat(k)) + lambda (2hat(i) +3hat(j) -hat(k))`B. `vec(r )=(2hat(i) +3hat(j) -hat(k)) +lambda (hat(i)-2hat(j) +5hat(k))`C. `vec(r )=(hat(i) -2hat(j) +5hat(k)) +lambda (2hat(i) +3hat(j)-4hat(k))`D. none of these |
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Answer» Correct Answer - C The line passes through the point `(1-2,5)`and it has direction rations 2,3,-1 So, it vector equations is `vec(r ) =(hat(i) -2hat(j) +5hat(k)) +lambda (2hat(i) +3hat(j) -hat(k))` |
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| 46. |
`vec(r ) =(lambda-1) hat(i) + (lambda+1) hat(j) -(lambda+1) hat(k)` vec(r ) =(1-mu) hat(i) + (2mu -1) hat(j) + (mu +2) hat(k).` |
| Answer» Correct Answer - `(5sqrt(2))/(2)` units | |
| 47. |
Prove that the points `A(2,0,-3), B (1,-2,-5)` and `C(3,2,-1)` are collinear. |
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Answer» The equations of the lines AB are `(x-2)/(3-2) =(y-0)/(2-0) =(z-3)/(-1-3) rArr (x-2)/(1)=(y)/(2) =(z-3)/(-4)` putting x=1 y =-2 and z=-5 in (i) we get `(1-2)/(1)=(-2)/(2)=(-5-3)/(-4)` which is clearly true . Thus the point C(1,-2,-5) satifies the equations of line AB. Hence the given points A ,B and C are collinear . |
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| 48. |
The shortest distance between line `(x-3)/(3)=(y-8)/(-1)=(z-3)/(1) and` `(x+3)/(-3)=(y+7)/(2)=(z-6)/(4) is` |
| Answer» Correct Answer - `3sqrt(30) " units "(x-3)/(2)=(y-8)/(5)=(z-3)/(-1)` | |
| 49. |
If `(a_(1) ,b_(1),c_(1)) " and "(a_(2),b_(2),c_(2))` be the direction rations of two parallel lines thenA. `a_(1) =a_(2) ,b_(1) =b_(2) ,c_(1)=c_(2)`B. `(a_(1))/(a_(2)) =(b_(1))/(b_(2)) =(c_(1))/(c_(2))`C. `a_(1)^(2) +b_(1)^(2) +c_(1)^(2) =a_(2)^(2) +b_(2)^(2) +c_(2)^(2)`D. `a_(1)a_(2) +b_(1)b_(2) +c_(1)c_(2)=0` |
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Answer» Correct Answer - B If the given line are parallel then `(a_(1))/(a_(2)) =(b_(1))/(b_(2)) =(c_(1))/(c_(2))` |
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| 50. |
`(x-3)/(-1) =(y-4)/(2)=(z+2)/(1) "and " (x-1)/(1)=(y+7)/(3) =(z+2)/(2)` |
| Answer» Correct Answer - `sqrt(35) " units "(x-4)/(-1) =(y-2)/(-3) =(z+3)/(5)` | |