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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 equation of a line passing through the point `(2hati-3hatj-5hatk)` and perpendicular to the plane`vecr.(6hati-3hatj+5hatk) +2=0`. Also find the point of intersection of this line and the plane. |
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Answer» Correct Answer - `vecr.(2hati-3hatj-5hatk)+lambda(6hati-3hatj+5hatk), (76/35,-108/35,-34/4)` Clearly, the required line passes through the point `(2hati-3hatj-5hatk)` and is parallel to the normal of the given plane, which is `(6ahti-3hatj+5hatk)`. So, the rquired vector equation is `vecr=(2hati-3hatj-5hatk)+lambda(6hati-3hatj+5hatk)`. The Cartesian equation of the line is `(x-2)/6=(y+3)/-3=(z+5)/5=k`. A general point on this line cut the plane `6x-3y+5z+2=0`. ltrbgt Then, `6(6k+2)-3(3k-3)+5(5k-2)=0`. `rArr (36k+9k+25k)=2 rArr 70k=2` `rArr k-1/35`. `therefore` required point of intersection of the line and plane is `P(6/35+2,-3/35-3,1/7-5)`, i.e., `P(76/35,-108/35,-34/7)`. |
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| 2. |
If the points `(1,1,p)` and `(-3,0,1)` be equidistant from the plane `vecr.(3hati+4hatj-12hatk)+13`, find the values of p. |
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Answer» The eqation of the given plane is Cartesian form is `xhati+yhatj+zhatk. (3hati+4hatj-12hatk)+13=0`. `rArr 3x+4y-12z+13=0`. …………(i) It is being given that the distances of the plane (i) from each of the points A(1,1,p) and B(-3,0,1) are equal. `therefore (|(3xx1)+(4xx1)-(12xxp)+13|)/sqrt(3^(2)+4^(2)+(-12)^(2))=|(3xx(-3)+(4xx0)-(12xx1))|/sqrt(3^(2)+4^(2)+(-12)^(2))` `|20-12p|=|-8| rArr |20-12p|=8` `rArr (20-12p)=8` or `-(20-12p)=8` `rArr 12p=2` or `12p=28` `rArr p=1` or `p=7/3`. Hence, `p=1` or `p=7/3`. |
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| 3. |
Find the distance of the point (1,2,5) from the plane `vecr.(hati+hatj+hatk)+17=0` |
| Answer» Correct Answer - `(25sqrt(3))/3` units | |
| 4. |
Find the co-ordinates of the foot of perpendicular and the length of perpendicular drawn from the point `(2,3,7)` to the plane `3x-y-z=7`. |
| Answer» Correct Answer - `sqrt(11)`units, (5,2,6) | |
| 5. |
Find the length of the foot of the perpendicular from the point (1,1,2) to the plane `vecr.(2hati-2hatj+4hatk)+5=0` |
| Answer» Correct Answer - `(13sqrt(6))/12` units, `(-1,12,25/12,-1/6)` | |
| 6. |
Find the coordinates of the image of the point P`(1, 3, 4)` in the plane `2x-y+z+3=0`. |
| Answer» Correct Answer - `(-3,5,2)` | |
| 7. |
Find the equationof the plane passing through each group of points. i) A(2,2,-1), B(3,4,2) and C(7,0,6) ii) A(0,-1,-1), B(4,5,1) and C(3,9,4) iii) A(-2,6,-6), B(-3,10,-9) and C(-5,0,-6). |
| Answer» Correct Answer - i) `5x+2y-3z=17`, ii) `5x-7y+11z+4=0`, iii) `2x-y-2z=2` | |
| 8. |
Find the coordinates of the foot of the perpendicular and the perpendicular distance from the point P(3,2,1) to the plane `2x-y+z+1=0`. |
| Answer» Correct Answer - (1,3,0) units,`sqrt(6)` units, `(-1,4,-1)` | |
| 9. |
Find the distance of the point `(-1,-5,-10)` from the point of the intersection of the line `vecr = 2hati-2hatk+lambda(3hati+4hatj+2hatk)` and the plane `vecr.(hati-hatj+hatk) = 5`. |
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Answer» Correct Answer - 13 units. The cartesian equation of the planes is `(x-2)/3=(y+1)/4=(z-2)/2=lambda` (say). A general point on line (i) is `N(3lambda+2,4lambda-1,2lambda+2)`. The Cartesian equation of the given plan is `x-y+z=5`. If this point lies on the given plane, we have `(3lambda+2)-(4lambda-1)+(2lambda+2)=5 rArr lambda=0`. `therefore` point P is `P(2,-1,2)` and the other point is `Q(-1,-5,-10)`. `therefore` PQ =13 units. |
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| 10. |
Find the distance of the point `(2hati-hatj-4hatk)` from the plane `vecr.(3hati-4hatj+12hatk)=9`. |
| Answer» Correct Answer - `47/13` units | |
| 11. |
From the point `P(1,2,4)` a perpendicular is drawn on the plane `2x+y-2z+3=0`. Find the equation the length and the coordinates of the foot of perpendicular. |
| Answer» Correct Answer - `(x-1)/2=(y-1)/1=(z-4)/-2; 1/3` unit; `(11/9,19/9, 34/9)` | |
| 12. |
Find the equation of the plane passing through (a,b,c) and paralle toteh plne `vecr.(hati+hatj+hatk)=2`. |
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Answer» Correct Answer - `vecr.(hati+hatj+hatk)=a+b+c` We know that any plane parallel to `vecr.vecn=q` is `vecr.vecn=q_(1)`. Let the required plane parallel to the given plane be `vecr.(hati+hatj+hatk)=q_(1)` `rArr (xhati+yhatj+zhatk).(hatk+hatj+hatk)=q_(1)` `rArr (ahati+bhatj+chatk).(hati+hatj+hatk)=q_(1)` `rArr q_(1)=(a+b+c)`. Hence, the required equation is `vecr.(hati+hatj+hatk)=(a+b+c)`. |
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| 13. |
Find the distance of the point (2,3,4) from the plane `3x+2y+2z+5=0` measured parallel to the line `(x+3)/3=(y-2)/6=z/2`. |
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Answer» Equation of a line through A(2,3,4) and parallel to the given line is `(x-2)/3=(y-3)/6=(z-4)/2=lambda` (say). A general point on line (i) is `N(3lambda+2,6lambda+3,2lambda+4)`. For some value of `lambda`, it lies on the plane `3x+2y+2z=5`. `therefore 3(3lambda+2)+2(2lambda+4)+5 =0 rArr 25lambda=-25 rArr lambda=-1`. `therefore` the coordinates of N are `(-3+2,-6+3,-2+4)`, i.e.,`(-1,-3,2)`. Now, find AN=7 units. |
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| 14. |
Find the equation of the linepassing through the point `"P"(4,6,2)`and the point of intersection of theline `("x"-1)/3="y"/2=("z"+1)/7`and the plane `"x"+"y"-"z"=8.` |
| Answer» Correct Answer - 10 units | |
| 15. |
find the coordinates of point where the line through (3,-4,-5) and (2,-3,1) crosses the plane `2x+y+z=7`. |
| Answer» Correct Answer - `(1,-2,7)` | |
| 16. |
Find the distance of the point `(0,-3,2)` from the plane `3x+2y+2z+5=0`, measured parallel to the line `(x+1)/3=(y+1)/2=z/3`. |
| Answer» Correct Answer - 7 units | |
| 17. |
Find the angle betweenthe line `(x-2)/-1=(y+3)/2=(z+4)/3` and the plane `2x-3y+z=5`. |
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Answer» The given line is `(x-x_(1))/(a_(1))=(y-y_(1))/(b_(1))=(z-z_(1))/(c_(1))` where `a_(1)=-1, b_(1)=2 c_(1)=3` `The given plane is `a_(2)x+b_(2)y+c_(2)y+d=0` where `a_(2)=2, b_(2)=-3 c_(2)=1,d=-5` Let the required angle be `phi` Then `sin phi =(|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))}` `=|(-1)xx2+2xx(-3)+3xx1)|/({sqrt(-1)^(2)+2^(2)+3^(2))}.{sqrt(2^(2)+(-3)^(2)+1^(2))}` =|(-2-6+3)|/{sqrt(14xxsqrt(14))=|(-5)|/(14)=(5)/(14)` `rarr phi sin^(-1) (5/14)` Hence the anlge between the given line and the given plane is `sin^(-1)(5/14)` |
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| 18. |
Find the equation of the plane which passes through the point `(3,4,-1)`and is parallel to the plane `2x-3y+5z+7=0.`Also, find the distance between the two planes. |
| Answer» Correct Answer - `2x-3y+5z+11=0, 2/19sqrt(38)` units | |
| 19. |
Find the vector equation of a plane which is parallel to the plane `vecr.(2hati-hatj+2hatk)=5` and passes through the point whose position vector is `(hati+hatj+hatk)`. |
| Answer» Correct Answer - `vecr.(2hati-hatj+2hatk)=3` | |
| 20. |
Find the equation of the plane through the points (2,2,1) and (9,3,6)` and perpendicular to the plane `2x+6y+6z=1`A. `x+2y-3z+5=0`B. `2x-5y+4z-8=0`C. `4x+5y-6z+3=0`D. `3x+4y-5z-9=0` |
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Answer» Correct Answer - d The equation of a plane passing through the point A(2,2,1) is `a(x-2)+b(y-2)+c(z-1)0`…………………(i) Since it passes through the point B(9,3,6), we have `a(9-2)+b(3-2)+c(6-1)=0 rArr 7a+b+5c=0`………(ii) Also, it being perpendicular to the plane `2x+6y+6z=1`, we have `2a+6b+6c=0 rArr a+3b+3c=0`. ...............(iii) On solving (ii) and (iii) by cross multiplication, we have `a/(3-15)=b/(5-21)=c/(21-1) rArr a/-12=b/-16=c/20 rArr a/3=b/4=c/-5=k` `rArr a=3k,b=4k` and `c=-5k`. substituting these values in (i), we get `3k(x-2)+4k(y-2)-5k(z-1)=0` `rArr 3(x-2)+4(y-2)-5(z-1)=0 rArr 3x+4y-5z-9=0`. |
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| 21. |
Find the equation of the plane mid-parallel to the planes `2x-3y+6z+21=0` and `2x-3y+6z-14=0`. |
| Answer» Correct Answer - `4x-6y+12z+7=0` | |
| 22. |
Find the equations of the planes parallel to the plane `x-2y+2z-3=0`which is at a unit distance from the point `(1,2,3)dot` |
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Answer» The equation fo the given plane is `x-2y+2z-3=0`. Let the required plane parallel to the given plane be `x-2y+2z+lambda=0` for some real number `lambda`…………….(i) Now, the plane (i) is at unit distance from the point (1,2,3). `therefore (|1-(2xx2)+(2xx3)+lambda|)/(1^(2)+(-2)^(2)+2^(2)) = 1 rArr |lambda+3|=3`. `rArr lambda+3=3` or `-(lambda+3)=3`. `rArr lambda=0` or `lambda=-6`. Hence, the equations of the required planes are `x-2y+2z=0` and `x-2y+2z-6=0`. |
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| 23. |
Find the equations of the line passing through the point `(3,0,1)` parallel to the planes `x+2y=0` and `3y-z=0`. |
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Answer» Let the direction rations of the required line be a,b,c Then its equation is given by `(x-3)/(a)=(y-0)/(b) =(z-1)/(c )` Since the line (i) is parallel to ech of the planes xd+2y+oz=0 and 0x+3y-z=0 so it must be perpendicular to the normal of each of these planes `therefore axx1+bxx2+cxx0=0 rarr a+2b+0c =0` ltbvrgt `and `axx0 +bxx3+cxx(-1)=0 rarrr 0a + 3b -c =0` on solving (ii) and (iii) by cross multiplication we get `(a)/(-2-0)=(b)/(0+1)=(c )/(3-0)` `rarr (a)/(-2)=(b)/(1)=(c )/(3) =lambda(say) rarr a= 3 lambda , b = lambda` and ` c= 3lambda` putting these values of a,b,c in (i) we get `(x-3)/(-2lambda)=(y-0)/(lambda)=(z-1)/(3 lambda) rarr (x-3)/(-2)=(y)/(1)=(z-1)/(3)` Hence the required equation of the line is `(x-3)/(-2)=(y)/(1)=(z-1)/(3)` |
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| 24. |
The equation of the plane passing through the point A(2,3,4) and parallel to the plane `5x-6y+7z=3`, isA. `5x-6y+7z=20`B. `7x-6y+5z=72`C. `20x-18y+14z=11`D. `10x-18y+28z=13` |
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Answer» Correct Answer - a Let the prequired equation of the plane be `5x-6y+7z=k`. Since, it passes through the point A(2,3,4), we have `(5 xx 2)-6 xx 3 +7 xx 4 =k rArr k=(10-18+28)=20`. Hence, the required equation of the plane is `5x-6y+7z=20`. |
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| 25. |
Find the equation of the plane through the point (1,4,-2) and parallel to the plane `-2x+y-3z=7`. |
| Answer» Correct Answer - `2x-y+3z+8=0` | |
| 26. |
The equation of the plane passing through the points A(0,-1,0), B(2,1,-1) and C(1,1,1) is given byA. `(4x+3y-2z-3)=0`B. `4x-3y+2z+3=0`C. `4x-3y+2z-3=0`D. none of these |
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Answer» Correct Answer - c Here `(x_(1)=0, y_(1)=-1, z_(1)=0), (x_(2)=2,y_(2)=1,z_(2)=-1))` and `(x_(3)=1, y_(3)=1, z_(3)=1)`. So, the required equation of the plane is `|{:(x-x_(1),y-y_(1),z-z_(1)),(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(x_(3)-x_(1),y_(3)-y_(1),z_(3)-z_(1)):}|=0` `|{:(x-0,y+1,z-0),(2-0,1+1,-1-0),(1-0,1+1,1-0):}|=0 rArr |{:(x,y+1,z),(2,2,-1),(1,2,1):}|=0` `rArr x(2+2)-(y+1)(2+1)+z(4-2)=0`. `rArr 4x-3(y+1)+2z=0 rArr 4x-3y+2z-3=0`. |
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| 27. |
Find the equation of the plane mid-parallel to the planes `2x-2y+z+3=0` and `2x-2y+z+9=0`. |
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Answer» Let the required equation of the plane be `2x-2y+z+k=0`. This plane equidistant from each of the given planes. Lety `P(x_(1),y_(1),z_(1))` be any point on the plane `2x-2y+z+k=0`. Then, `2x_(1)-2y_(1)+z_(1)+k=0 rArr 2x_(1)-2y_(1)+z_(1)=-k`…………….(i) `therefore P(x_(1),y_(1),z_(1))`is equidistant from the planes `2x-2y+z+3=0` and `2x-2y+z+9=0`. `therefore |2x_(1)-2y_(1)+z_(1)+3|/sqrt(2^(2)+(-2)^(2)+1^(2))=|2x_(1)-2y_(1)+z_(1)+9|/sqrt(2^(2)+(-2)^(2)+1^(2))` `rArr |(2x_(1)-2y_(1)+z_(1)+3)|=|(2x_(1)-2y_(1)+z_(1)+9|` `rArr |-k+3|=(-k+9)` or `-(-k+3)=-k+9` `rArr k-3 =-k+9 rArr 2k=12 rArr k=6 [therefore (-k+3) ne (-k+9)]`. Hence, the required equation of the plane is `2x-2y+z+6=0`. |
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| 28. |
Find the equation of the plane passing through the origin and parallel to the plane `5x-3y+7z+13=0`. |
| Answer» Correct Answer - `5x-3y+7z=0` | |
| 29. |
Findthe equation of the plane through the intersection of the planes `3x" "" "y" "+" "2z" "" "4" "=" "0`and `x" "+" "y" "+" "z" "" "2" "=" "0`and the point (2, 2, 1).A. `7x+5y-4z-8=0`B. `7x-5y+4z-8=0`C. `5x-7y+4z-8=0`D. `5x+7y-4z+8=0` |
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Answer» Correct Answer - b Let the required equation of the plane be `(3x-y+2z-4)+ lambda(x+y+z-2)=0` `rArr (3+lambda)x+(lambda-1)y+(2+lambda)z-(4+2lambda)=0`………….(ii) Since it passes through the point A(2,2,1), we have `(3+lambda) xx 2+(lambda+1) xx 2 +(2+lambda) xx 1(4 -4/3)=0`. `rArr (7x)/3-(5y)/3+(4z)/3-8/3=0 rArr 7x-5y+4z-8=0`. |
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| 30. |
Find the vector equation of the plane through the points (2,1,-1) and (-1,3,4) and perpendicular to the plane `x-2y+4z=10`. |
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Answer» The general equation of a plane passing through the point A(2,1,-1) is given by `a(x-2)+b(y-1)+c(z+1)=0`……………(i) If the point B(-1,3,4) lies on plane (i), then we have `a(-1-2)+b(3-1)+c(4+1)=0` `rArr -3a+2b+5c=0`………….(ii) If the plane (i) is perpendicular to the plane `x-2y+4z=10`, then we have `(1 xx a) -(2 xx b)+(4 xx c)=0` `rArr a-2b+4c=0`.............(iii) On solving (ii) and (iii) by cross multiplication, we have `a/(8+10)=b/(5+12)=c/(6-2)` `rArr a/18=b/17=c/4= lambda` (say) `rArr a=18lambda, b=17lambda` and `c=4lambda` Substituting these values of a,b and c in (i), we get `18lambda(x-2)+17lambda(y-1)+4lambda(z+1)=0` `rArr 18(x-2)+17lambda(y-1)+4lambda(z+1)=0` `rArr 18(x-2)+17(y-1)+4(z+1)=0` `rArr 18x+17y+4z=49`. Required equation of the plane in vector form is given by `vecr.(18hati+17hatj+4hatk)=49`. |
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| 31. |
Find the equation of the plane passing through thepoint (-1, 3, 2) and perpendicular to each of the planes`x+2y+3z=5 `and `3x+3y+z=0` |
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Answer» Any plane passing through the point (-1,3,2) is given by `a(x+1)+b(y-3)+c(z-2)=0`………….(i) Now, (i) being perpendicular to each of the planes `x+2y+3z-5=0` and `3x+3y+z=0`, we have, `(a xx 1)+(b xx 2)+(c xx 3)=0 rArr a+2b+3c=0`..........(ii) `(a xx 3) + (b xx 3) + (c xx 1) =0 rArr 3a+3b+c=0`.............(iii) On solving (ii) and (iii) by cross multiplication, we get `a/(2-9)=b/(9-1)=c/(3-6)` `rArr a/-7=b/8 =c/-3 rArr a/7=b/-8=c/3=lambda` (say) `rArr a=7lambda, b=-8lambda` and `c=3lambda`. Putting, `a=7lambda, b=-8lambda` and `c=3lambda` in (i), we get `7lambda(x+1)-8lambda(y-3)+3lambda(z-2)=0` `rArr 7(x+1)-8(y-3)+3(z-2)=0` `rArr 7x-8y+3z+25=0`, which is the required equation of the plane. |
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| 32. |
Find the equation of the plane passing through the line of intersection of the planes `2x-y=0` and `3z-y=0`, and perpendicular to the plane `4x+5y-3z=9`. |
| Answer» Correct Answer - `28x-17y+9z=0` | |
| 33. |
Find the vector equation of the plane passing through the point (1,1,1) and passing through the intersection of the planes `vecr.(hati-hatj+3hatk)+1=0` and `vecr.(2hati+hatj-hatk)-5=0`. |
| Answer» Correct Answer - `vecr.(11hati+hatj+5hatk)-17=0` | |
| 34. |
Find the equation of the plane through the points `A(2,1,-1)` and `B(-1,3,4)` and perpendicular to the plane `x-2y+4z=10`. Also, show that the plane thus obtained contains the line `vecr=(-hati+3hatj+4hatk)+lambda(3hati-2hatj-5hatk)`. |
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Answer» Correct Answer - `18x+17y+4z=49` Obtain the required equation of the plane as `18x+17y+4z=49`………(A) The given line is `vecr=(3lambda-1)hati+(3-2lambda)hatj+(4-5lambda)hatk`. Coordinates of any points on this line are `(3lambda-1,3-2lambda,4-5lambda)`. These coordinates satisfy (A), as shown below: LHS =`18(3lambda-1)+17(3-2lambda)+4(4-5lambda)=49` =RHS. So, `(3lambda-1,3-2lambda,4-5lambda)` lies on plane (A). Hence, the plane (A) obtained above equation the given line. |
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| 35. |
Write the equation of the plane parallel to YZ-plane and passing through the point (-3,2,0). |
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Answer» Correct Answer - `x=-3` Any plane parallel to YZ-plane, is x=k. Sinece, it passes through `(-3,2,0)`, we have `-3=k`. Hence, the required equation of the plane is `x=-3`. |
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| 36. |
Find the equation of the plane through the line of intersection of the planes `vecr.(2hati-3hatj+4hatk)=1` and `vecr.(hati-hatj)+4=0` and perpendicular to the plane `vecr.(2hati-hatj+hatk)+8=0`. |
| Answer» Correct Answer - `vecr.(-5hati+2hatj+12hatk)=47` | |
| 37. |
Findthe equation of the plane through the intersection of the planes `3x" "" "y" "+" "2z" "" "4" "=" "0`and `x" "+" "y" "+" "z" "" "2" "=" "0`and the point (2, 2, 1). |
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Answer» Let the required equation of the plane be `(3x-y+2z-4)+lambda(x+y+z-2)=0` for some real value of `lambda` `rArr (3+lambda)x+(-1+lambda)y+(2+lambda)z -(4 +2lambda)=0`…………..(i) It is given that the point A(2,2,1) lies in (i) `therefore (3+lambda) xx 2+(-1+lambda) xx 2+(2 +lambda) xx 1-(4+2lambda)=0` `rArr (6+2lambda)+(-2+2lambda)+(2+lambda)-4-2lambda=0` `rArr 3lambda=-2 rArr lambda=-2/3`. Putting, `lambda=-2/3` in (i), we get `(3-2/3)x+(-1-2/3)y+(2-2/3)z-4+4/3=0` `rArr (7x)/3-(5y)/3+(4z)/3-8/3=0 rArr 7x-5y+4z-8=0`. Hence, the required equation of the plane is `7x-5y+4z-8=0`. |
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| 38. |
Find the equation of a plane passing through the intersection of the planes `vecr.(2hati-7hatj+4hatk)=3` and `vecr.(3hati-5hatj+4hatk) + 11 - 0` and passes through the point `(-2hati+hatj+3hatk)`. |
| Answer» Correct Answer - `vecr.(15hati-47hatj+28hatk)=7` | |
| 39. |
Find the vector equation to the plane through the point `(2,1,-1) ` passing through the line of intersection of the planes `vecr.(hati+3hatj-hatk)=0 ad vecr.(hatj+2hatk)=0` |
| Answer» Correct Answer - `vecr.(hati+9hatj+11hatk)=0` | |
| 40. |
Find the direction cosines of the normal to the plane `y=3`. |
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Answer» Correct Answer - 0,1,0 Direction ratios of the normal to the plane are 0,1,0 and `sqrt((-1)^(2)+0^(2)+0^(2))=1`. Hence,the required direction cosines are 0,1,0. |
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| 41. |
Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati+3hatj)-6=0` and `vecr.(3hati-hatj-4hatk)=0`, whose perpendicular distance from the origin is unity. |
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Answer» Taking `vecr=(xhati+yhatj+zhatk)`, the equations of the given planes are `(xhati+yhatj+zhatk). (hati+3hatj)-6=0 rArr x+3y-6=0`……….. (i) (xhati+yhatj+zhatk).(3hati-hatj-4hatk)=0 rArr 3x-y-4z=0`………(iii) Length of the perpendicular from the origin to plane (iii) is given as 1. `therefore |0+0-0-6|/sqrt((1+3lambda)^(2)+(3-lambda)^(2)+(-4lambda)^(2))=1` `rArr (1+3lambda)^(2)+(3-lambda)^(2)+)-4lambda)^(2)=36` `rArr 26lambda^(2)=26 rArr lambda^(2)=1 rArr lambda=+1`. Putting, `lambda=1` in (iii), we get `4x+2y-4z-6=0 rArr 2x+y-2z-3=0`. Putting, `lambda=-1`, in (iii), we get `-2x+4y+4z-6=0 rArr 2x+y-2z-3=0`. Hence, the required equations of the planes are `2x+y-2z-3 =0` and `x-2y-2z+3=0`. Note: The vector equations of these planes are `vecr.(2hati+hatj-2hatk)=3` and `vecr.(hati-2hatj-2hatk)+3=0`. |
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| 42. |
Find the vector equation of a plane passing through a point having position vector `(2hati-hatj+hatk)` and perpendicular to the vector `(4hati+2hatj-3hatk)`. Also, reduce it to Cartesian form. |
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Answer» Here, `veca=(2hati-hatj+hatk)` and `vecn=(4hati+2hatj-3hatk)`. Clearly, the required vector equation is `(vecr-veca).vecn=0`. `rArr vecr.vecn=veca.vecn` `rArr vecr.(4hati+2hatj-3hatk)=(2hati-hatj+hatk).(4hati+2hatj-3hatk)` =`(8-2-3)=3`. Hence, the vector equation of the given plane is `vecr.(4hati+2hatj-3hatk)=3`..............(1) Reduction to Cartesian form: Putting `vecr=(xhati+yhatj+zhatk)`, we get `(xhati+yhatj+zhatk).(4hati+2hatj-3hatk)=3 rArr 4x+2y-3z=3`. Hence, the equation ofthe given plane in Cartesian form is `4x+2y-3z=3`. |
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| 43. |
Find the equation of the plane through the line of intersection of the planes `x+y+z=1` and `2x+3y+4z=5`, which is perpendicular to the plane `x-y+z=0`. Also find the distance of the plane so obtained from the origin. |
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Answer» Let the required equation of the plane be `(x+y+z-1)+lambda(2x+3y+4z-5)=0` For some real value of `lambda`. `rArr (1+2lambda)x+(1+3lambda)+(1+3lambda) + (1+ 4lambda)z-(1+5lambda)=0`………………..(i) Since, this plane is perpendicular to the plane `x-y+z=0`, we have `(1+2lambda) xx 1 + (1 + 3lambda) xx (-1) +(1+4lambda) xx 1=0`. `rArr (1+2lambda)-1/3z+2/z=0 rArr x-z+2=0`. Hence, the required equation of the plane is `x-z+2=0`. Distance of the plane from the origin. = Length of perpendicular from the origin to the plane `=|0-0+2|/sqrt(1^(2)+0^(2)+(-1)^(2))=2/sqrt(2)=sqrt(2)`. |
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| 44. |
Distance between the two planes: `2x + 3y + 4z = 4`and `4x + 6y + 8z = 12`is(A) 2 units (B) 4 units (C) 8 units (D) `2/(sqrt(29))`units |
| Answer» Correct Answer - `2/29sqrt(29)` units | |
| 45. |
The equation of the plane which makes with coordinate axes, a triangle witih centroid `(alpha,beta,gamma)` is given byA. `alphax+betay+gammaz=1`B. `alphax+betay+gammaz=3`C. `x/alpha+y/beta=z/gamma=1`D. `x/alpha+y/beta+z/gamma=3` |
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Answer» Correct Answer - d Let the equation of the plane be `x/a+y/b+z/c=1`. Then, it makes the axes at A(a,0,0), B(o,b,0) and C(0,0,c). `therefore` centroid of `triangleABC` is `G((a+0+0)/3, (0+b+0)/3, (0+0+c)/3)`, i.e. `G(a/b,b/3,c/3)`. `therefore (a/3=alpha, b/3=beta "and" c/3=lambda) rArr a=3alpha, b=3beta "and" c=3gamma`. `therefore` the required equation of the plane is `x/(3alpha)+y/(3beta)+z/(3gamma)=1 rArr x/alpha+y/beta+z/gama=3`. |
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| 46. |
The angle between the line `vecr.(hati+hatj-3hatk)+lambda(2hati+2hatj+hatk)` and the plane `vecr.(6hati-3hatj+2hatk)=5`, is |
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Answer» We know that the angle `phi` between the line `vecr=veca+lambdavecb` and the plane `vecr.vecn=q` is given by `sinphi=|vecb.vecn|/(|vecb||vecn|)`. Here, `vecb=(2hati+2hatj+hatk)` and `vec=(6hati-3hatj+2hatk)`. `therefore sinphi=|(2hati+2hatj+hatk).(6hati-3hatj+2hatk)|/(sqrt(2^(2)+2^(2)+1^(2))sqrt(6^(2)+(-3)^(2)+2^(2))` `=|(12-6+2)|/(sqrt(9) xx sqrt(49)) = 8/(3 xx 7)=8/21`. `rArr phi=sin^(-1)(8/21)`. Hence, the angle between the given line and the given plane is `sin^(-1)(8/21)`. |
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| 47. |
Find ten equation of the plane passing through the point `(0,7,-7)`and containing the line `(x+1)/(-3)=(y-3)/2=(z+2)/1`. |
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Answer» The general equation of the plane passing through the point (0,7,-7) is given by `a(x-0)+b(y-7)=0 If (i) contains the given line then it must pass through the point *-1,3,-2) and must be parallel to the given line If (i) passes through the point (-1,3,-2) we have a(-1-0)+b(3-7)+c(-2+7)=0 rarr a+4b-5c=0 If (i) is parallel to the given line then its normal should be perpendicular to htis line `therefore (-3) a+2b+1xxc=0 rarr -3a+2b+c=0` on solving (ii) and (iii) by cross multiplication we get `rarr (A)/(1)=(b)/(1)=(c )/(1) = lambda`(say) Then `a=lambda, b = lambda` and `c = lambda` Putting `a=lambda , b = lambda` and `c = lambda` in we get `lambda x+lambda(y-7) + lambda(z+7) =0 rarr x + (y-7) + (z+7)=0` `rarr x+y+z=0` Hence the required equation of the plane is x + y Z=0 |
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| 48. |
Find the vector equation of a plane which is at a distance of 6 units from the origin and which is normal to the vector `(hati+2hatj-2hatk)`. |
| Answer» Clearly, the required equation of the plane is `vecr. Hatj=5`. | |
| 49. |
A plane meets the coordinate axes in `A ,B ,C`such that eh centroid of triangle `A B C`is the point `(p ,q ,r)dot`Show that the equation of the plane is `x/p+y/q+z/r=3.` |
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Answer» Let the required equation of the plane be `x/a+y/b+z/c=1`……………(i) Then, clearly the plane meets the coordinates axis at `A(a,0,0),B(0,b,0)` and C(0,0,c). Since, the centroid of `triangleABC` isG(p,q,r), we have `(a+0+0)/(3)=p, (0+b+0)/(3)` and `(0+0+c)/3=r` `rArr a=3p, b=3q` and `c=3r`. Putting, these values of a,b,c in (i), we get `x/(3p)+y/(3q)+z/(3r)=1 rArr x/p+y/q+z/r=3`. Hence, the required equation of the plane is `x/p +y/q+z/r=3`. |
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| 50. |
A variable plane moves in such a way that the sum of the reciprocals ofits intercepts on the three coordinate axes is constant. Show that the planepasses through a fixed point. |
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Answer» Let the equation of the variable plane be `x/a+y/b+z/c=1`…………….(i) Then, it makes intercepts a,b,c with the coordinate axes. `therefore 1/a+1/b+1/c=k,` where k is a constant (given) `rArr 1/(ka)+1/(kb)+1/(kc)=1 rArr 1/a(1/k)+1/b(1/k)+1/c(1/k)=1`. `rArr (1/k,1/k,1/k)` satisfies (i). Hence, the given plane passes through a fixed point `(1/k,1/k,1/k)`. |
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