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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.
| 51. |
Statement 1: The plane `5x+2z-8=0` contains the line `2x-y+z-3=0` and `3x+y+z=5`, and is perpendicular to `2x-y-5z-3=0`. Statement 2: The plane `3x+y+z=5`, meets the line `x-1=y+1=z-1` at the point (1,1,1)A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - C The equation of the family of planes containing the line `2x-y+z-3=0,3x+y+z=5` is `2x-y+z-3+lamda(3x+y+z-5)=0` For `lamda=1` this reduces to `5x+2z-8=0` So the plane `5x+2z-8=0` contains the given line. Also `2xx5-1xx0-5xx2=0` So, the plane `5x+2z-8=0` is perpendicular to `2x-y-5z-3=0` Hence, statement -1 is true. The coordinate of any point on line `(x-1)/1=(y+1)/1=(z-1)/1` are `(r+1,r-1,r+1)`. If this point lies on the plane `3x+y+z=5`. Then, `3r+3r-1+r+1=5impliesr=2/5` Thus, the line meets the plane at `(7/5,-3/5,7/5)` So, Statemnet -2 is not true. |
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| 52. |
A plane II passes through the point (1,1,1).If `b,c,a` are the direction ratios of a normal to the plane where `a,b,c(altbltc)` are the prime factors of 2001, then the equation of the plane II isA. `29x+31y+z=63`B. `23x+29y-29z=23`C. `23x+29y+3z=55`D. `31x+37y+3z=71` |
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Answer» Correct Answer - C The equation of the plane is `b(x-1)+c(y-1)+a(z-1)=0`……………i Now, `2001=3xx23xx29` `:.altbltcimpliesa=3,b=23` and `c=29` Subsituting the values a,b,c in i we obtain `23x+29y+3z=55` as the equation of the required plane. |
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| 53. |
The dr. of normal to the plane through `(1,0,0), (0,1,0)` which makes an angle `pi/4` with plane , `x+y=3` areA. `1,sqrt(2),1`B. `1,1,sqrt(2)`C. `1,1,2`D. `sqrt(2),1,1` |
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Answer» Correct Answer - B Let the direction ratios of the normal to the plane be proportional to a,b,c. Then, the equation of the plane is `a(x-1)+b(y-0)+c(z-0)=0`……………i It passes through (0,1,0) `:.a(-1)+b(1)+c(0)=0impliesa=b`…………….ii It is given that the plane i makes an angle `pi//4` with the plane `x+y=3`. `:."cos"(pi)/4=(axx1+bxx1+cxx0)/(sqrt(a^(2)+b^(2)+c^(2))sqrt(1+1))` `implies1/(sqrt(2))=(a+b)/(sqrt(a^(2)+b^(2)+c^(2))sqrt(2))` `impliessqrt(a^(2)+b^(2)+c^(2))=a+b` `impliesa^(2)+b^(2)+c^(2)=a^(2)+b^(2)+2abimpliesc^(2)=2ab`............iii From ii and iii, we have `a:b:c=a:a:sqrt(2)a=1:1:sqrt(2)` |
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| 54. |
The equation of the plane through the point (1,2,3) and parallel to the plane `x+2y+5z=0` isA. `(x-1)+2(y-2)+5(z-3)=0`B. `x+2y+5z=14`C. `x+2y+5z=6`D. none of these |
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Answer» Correct Answer - A Required planeis parallel to the palne `x+2y+5z=0`.Therefore direction of a vector normal to the plane are proportional to 1,2,5. Hence the equation of the required plane is `(x-1)+2(y-2)+5(z-3)=0` |
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| 55. |
Find the distance of the point `(21,0)`from the plane `2x+y+2z+5=0.`A. `10/3`B. `5/3`C. `10/9`D. none of these |
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Answer» Correct Answer - A We know that the distance of the point `(x_(1),y_(1),z_(1))` from the plane `ax+by+cz+d=0` is `(|ax_(1)+by_(1)+cz_(1)+d|)/(sqrt(a^(2)+b^(2)+c^(2))` So, required distance `=(|2xx2+1+2xx0+5|)/(sqrt(2^(2)+1^(2)+2^(2)))=10/3` |
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| 56. |
The equation of the plane through the intersection of the planes `vecr.(2hati+6hatj)+12=0` and `vecr.(3hati-hatj+4hatk)=0` and at a unit distance from the origin, isA. `vecr.(2hati+hatj+2hatk)+3=0`B. `vecr.(hati-2hatj+2hatk)+3=0`C. `vecr.(hati-2hatj-2hatk)+3=0`D. `vecr.(2hati+hatj+2hatk)-3=0` |
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Answer» Correct Answer - A The equation of the through the intersection of the given planes is `{vecr.2hati+6hatj+12}+lamda{vecr.3hati-hatj+4hatk}=0` or `vecr.{(2+3lamda)hati+(6-lamda)hatj+4lamda hatk+12}=0`……………i It is at a unit distance from the origin `:.|12/(sqrt((2+3lamda)^(2)+(6-lamda)^(2)+16lamda^(2)))|=1implieslamda=+-2` Hence the equations of required planes are `vecr.(2hati+hatj+2hatk)+3=0` and `vecr.(hati-2hatj+2hatk)-3=0` |
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| 57. |
The lines `x/1=y/2=z/3` and `(x-1)/(-2)=(y-2)/(-4)=(z-3)/(-6)` areA. coincidentB. skewC. intersectingD. parallel |
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Answer» Correct Answer - D Clearly direction ratios of two lines are proportional i.e, `1/(-2)=2/(-4)=3/(-6)` so given lines are parallel. |
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| 58. |
The angle between the lines `2x=3y=-z and 6x=-y=-4z` is (A) `0^0` (B) `90^0` (C) `45^0` (D) `30^0`A. `0^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)` |
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Answer» Correct Answer - D The equations of the given lines can be written in the from `x/3=y/2=2/(-6)` and `x/1=y/(-6)=z/(-3//2)` Clearly `3xx1+2(-6)-6xx(-3/2)=0` So given lines are perpendicular to each other. |
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| 59. |
The equation `3y+4z=0` represents aA. plane containing z-axisB. plane containing x-axisC. plane containing y-axisD. line with direction numbers 0,3,4 |
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Answer» Correct Answer - B The equation `3y+4z=0` or `0x+3y+4z=0` represents a plane through the origin and contains the line`x/1=y/0=z/0` i.e. `x-` axis. |
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| 60. |
Equation of the plane containing the straight line `x/2=y/3=z/4` and perpendicular to the plane containing the straight lines `x/2=y/4=z/2` and `x/4=y/2=z/3` isA. `x+2y-2z=0`B. `3x+2y-2z=0`C. `x-2y+z=0`D. `5x+2y-4z=0` |
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Answer» Correct Answer - C The equation of a plane containing the straight line `x/2=y/3=z/4` is `ax+by+cz=0`………….i where `2a+3b+4c=0`………….ii Let the direction ratios of the normal to the plane containing the straight lines `x/3=y/4=z/2` and `x/4=y/2=z/3` be proportional to `alpha, beta, gamma`. Then `3alpha+4beta +2gamma=0` and `4alpha+2beta+3gamma=0` `:. (alpha)/8=(beta)/(-1)=(gamma)/(-10)` It is given that the plane (i) is perpendicular to the plane the direction ratios of normal to which are proportional to 8,-1,-10. `:.8a-b-10c=0`.............iii From i and iii we have `a/(-30+4)=b/(32+20)=c/(-2-24)` `impliesa/(-26)=b/52-c/(-26)` `implies a/a=b/(-2)=c/1` Substituting the values of `a,b,c ` in (ii) we obtain `x-2y+z=0` as the required plane. ALITER As the line `x/2=y/3=z/4` passes through the origin. So plane containing it also passes through the origin. Let `vecn` be a vector normal to the required plane. Then `vecn` is perpendicular to `veca=2hati+3hatj+4hatk` an `vecbxxvecc` where `vecb=3hati+4hatj+2hatk` and `vecc=4hati+2hatj+3hatk`. `:.vecn=vecaxx(vecbxxvecc)` `impliesvecn=(veca.vecc)vecb-(veca.vecb)vecc` `impliesvecn=26(3hati+4hatj+hatk)-26(4hati+2hatj+3hatk)=26(-hati+hatj-hatk)` Hence the equation of the plane is `vecr.vecn=0` `impliesvecr.(-hati+2hatj-hatk)=0` or `x-2y+z=0` |
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| 61. |
The plane `2x-(1+lambda)y+3lambdaz=0` passes through the intersection of the planeA. `2x-y=0` and `y-3z=0`B. `2x+3z=0` and `y=0`C. `2x-y+3z=0` and `y-3z=0`D. none of these |
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Answer» Correct Answer - A The equation of the family of planes can be written as `(2x-y)+lamda(-y+3z)=0` Clearly it represents a family of planes passing through the intersection of the planes `2x-y=0` and `-y+3z=0` or `2x-y=0` and `y-3z=0` |
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| 62. |
If the distance between the plane Ax 2y + z = d and the plane containing the lines2 1x=3 2y=4 3zand3 2x=4 3y=5 4zis 6 , then |d| isA. `3`B. 4C. 6D. 1 |
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Answer» Correct Answer - C The equation of the plane containing the given lines is `|(x-1,y-2,z-3),(2,3,4),(3,4,5)|=0impliesx-2y+z=0` This plane is at a distance of `sqrt(6)` units from the plane `Ax-2y+z=d`. `:.A=1` and `(|d-0|)/(sqrt(1+4+1))=sqrt(6)` `implies|d|=6` |
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| 63. |
The plane `2x-(1+lambda)y+3lambdaz=0` passes through the intersection of the planeA. `2x-y=0` and `y+3z=0`B. `2x-y=0` and `y-3z=0`C. `2x+3yz=0` and `y=0`D. none of these |
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Answer» Correct Answer - B The equation of the given plane can be written in the form `(2x-y)+lamda(-y+3z)=0` which is a plane passing through the intersection of the plane `2x-y=0` and `-y+3z=0` |
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| 64. |
The equation `ax+by +c=0` represents a plane perpendicular to theA. xy-planeB. yz-planeC. zx-planeD. none of these |
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Answer» Correct Answer - A The equation of xy-plane is `z=0` i.e.`0x+0y+z=0` Clearly, the given plane is perpendicular to this plane. |
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| 65. |
The equation of the plane containing the two lines `(x-1)/2=(y+1)/(-1)=z/3` and `x/(-1)=(y-2)/3=(z+1)/(-1)` isA. `8x+y-5x-7=0`B. `8x+y+5z-7=0`C. `8x-y-5z-7=0`D. none of these |
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Answer» Correct Answer - D Given lines pass throuhg `(1-1,0)` and `(0,2,-1)` respectively and parallel to the vectors `2hati-hatj+3hatk` and `-2hati-3hatj-hatk`. `:.[(veca_(2)-veca_(1),vecb_(1),vecb_(2))]=|(-1,3,-1),(2,-1,3),(-2,-3,-1)|=-14!=0` So the given lines are non-coplanar. |
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| 66. |
Prove that the lines `(x+1)/3=(y+3)/5=(z+5)/7a n d(x-2)/1=(y-4)/4=(z-6)/7`are coplanar . Aslo, find the plane containing these two lines.A. `x-2y+z=0`B. `x+2y-z=0`C. `x-2y+z=1`D. none of these |
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Answer» Correct Answer - A The equation of the plane containig the given lines is `|(x+1,y+3,z+5),(3,5,7),(1,4,7)|=0impliesx-2y+z=0` |
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| 67. |
Equation of the line passing through `(1, 1, 1)` and parallel to the plane `2x +3y + z + 5 =0` isA. `(x-1)/1=(y-1)/2=(z-1)/1`B. `(x-1)/(-1)=(y-1)/1=(z-1)/(-1)`C. `(x-1)/3=(y-1)/2=(z-1)/1`D. `(x-1)/2=(y-1)/3=(z-1)/1` |
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Answer» Correct Answer - B Clearly, all lines in the given optics pass through (1,1,1). We know that a line is parallel to a plane if the normal to the plane is perpendicular to the line. We obsere that the vector normal to the given plane is `vecn=2hati+3hatj+hatk` and the line in option b is parallel to the vector `vecb=-hati+hatj-hatk` such that `vecb.vecn=-2+3-1=0` i.e. `vecb_|_vecn`. Hence option b is correct. |
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| 68. |
The equation of a line passing through (1,-1,0) and parallel to `(x-2)/3=(2y+1)/2=(5-z)/(-1)` isA. `(x-1)/3=(y+1)/3=(z-0)/(-1)`B. `(x-1)/3=(y+1)/1=(z-0)/(-1)`C. `(x-1)/3=(y+1)/1=(z-0)/1`D. `(x-1)/3=(y+1)/2=(z-0)/1` |
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Answer» Correct Answer - C The equation of the give line can be are written as `(x-2)/3=(y+1//2)/1=(z-5)/1` Clearly, its direction ratios are proportional to 3,1,1. So direction ratios of parallel line are also proportional to 3,1,1. Hence the equation of the required line is `(x-1)/3=(y+1)/1=(z-0)/1` |
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| 69. |
The line `(x-3)/1=(y-4)/2=(z-5)/2` cuts the plane `x+y+z=17` atA. (3,4,5)B. (4,6,7)C. (4,5,8)D. (8,4,5) |
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Answer» Correct Answer - B The coordinates of any point of the given line are given by `(x-3)/1=(y-4)/2=(z-5)/2=lamda` Suppose the given line cuts the given plane at `(lamda+3,2lamda+4,2lamda+5)` This lies on the plane. `:.lamda+3+2lamda+4+2lamda+5=17implieslamda=1` Hence the coordinates of the required point are (4,6,7). |
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| 70. |
The distance between the point (3,4,5) and the point where the line `(x-3)/1=(y-4)/2=(z-5)/2` meets the plane `x+y+z=17` isA. 1B. 2C. 3D. none of these |
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Answer» Correct Answer - C The the intersects the given plane at `(4,6,7)` `:.` Required distance `=sqrt((4-3)^(2)+(6-4)^(2)+(7-5)^(2))=3` |
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| 71. |
Thedistance of the point (1, 0, 2) from the point of intersection of the line `(x-2)/3=(y+1)/4=(z-2)/(12)`and the plane x y + z = 16, is :(1) `2sqrt(14)`(2) 8 (3) `3sqrt(21)`(4) 27A. `3sqrt(21)`B. `13`C. `2sqrt(14)`D. 8 |
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Answer» Correct Answer - B The coordinates of an arbitrary point on the line `(x-2)/3=(y+1)/4=(z-2)/12` are given by `(x-2)/3=(y+1)/4=(z-2)/12=lamda` where `lamda` is a parameter `implies x=3lamda+2,y=4lamda-1,z=12lamda+2` Suppose the given line intersects the plane `x-y+z=16` at `(3lamda+2,4lamda-1,12lamda+2)`. Then `(3lamda+2)-(4lamda-1)+(12lamda+2)=16implies11lamda=11implieslamda=1` so the coordinate of the point of intersection are (5,3,14). Hence required distance `=sqrt((5-1)^(2)+(3-0)^(2)+(14-2)^(2))=13` |
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| 72. |
If `veca,vecb` and `vecc` are three non-coplanar vectors, then the vector equation `vecr-(1-p-q)veca+pvecb+qvecc` represents aA. straight lineB. planeC. plane pasing through the originD. sphere |
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Answer» Correct Answer - B We have `vecr=(1-p-1)veca+pvecb+qvecc` `impliesvecr=veca+p(vecb-veca)+(vecc-veca)` Clearly, it represents a plane passing through a point having position vector `veca` and parallel to the vector `vecb-veca` and `vecc-veca`. |
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| 73. |
If the three planes `x=5,2x-5ay+3z-2=0` and `3bx+y-3z=0` contain a common line, then `(a,b)` is equal toA. `(-1/5,8/15)`B. `(1/5,-8/15)`C. `(-8/15,1/5)`D. `(8/15,-1/5)` |
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Answer» Correct Answer - B The line of intersection of first two planes is `(x-5)/0=y/(-3)=(z+7//3)/(-5a)` It must lie on third plane. `:.3bxx0(-3)xx1+(-3)(-5a)=0` and `3bxx5+0xx1+(-3)(-8//3)=0` `impliesa=1/5` and `15b+8impliesa=1/5`and `b=-8/15` |
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| 74. |
Find the equation of the plane through the points `A(2,2,-1),B(3,4,2)a n dC(7,0,6.)`A. `5x+2y+3z=17`B. `5x+2y-3z=17`C. `5x-2y+3z=17`D. none of these |
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Answer» Correct Answer - B The general equation of a plane passing through (2,2,-1) is `a(x-2)+b(y-2)+c(z+1)=0`………………i It will pass through `B(3,4,2)` and `C(7,0,6)`, if `a+2b+3c=0` ……………….ii and `5a-2b+7c=0`…………….iii Solving ii and iii by cross multiplication we have `a/(14+6)=b/(15-7)=c/(-2-10)` `impliesa/5=b/2=c/(-3)=lamda` (say) `impliesa=5lamda, b=2lamda` and `c=-3lamda` Substituting the values of a,b, and c in i we get `5lamda(x-2)+2lamda(y-2)-3lamda(z+1)=0` `implies5(x-2)+2(y-2)-3(z+1)=0` `implies5x+2y-3z=17` which is the required equation of the plane. |
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| 75. |
Write the equation of the plane whose intercepts on the coordinate axesare `-4,2a n d3`respectively.A. `3x+6y+4z=12`B. `-3x+6y+4z=12`C. `-3x-6y-4z=12`D. none of these |
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Answer» Correct Answer - B We know that the equation of a plane whose intercepts on the coordinate axes are a,b, and c respectively, is `x/a+y/b+z/c=1` Here `a=-4,b=2` and `c=3` So the equation of the required plane is `x/(-4)+y/2+z/3=1` or `-4x+6y+4z=12` |
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| 76. |
The sine of the angle between the straight line `(x-2)/3=(y-3)/4=(z-4)/5` and the plane `2x-2y+z=5` isA. `10/(6sqrt(5))`B. `4/(5sqrt(2))`C. `(sqrt(2))/10`D. `(2sqrt(3))/5` |
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Answer» Correct Answer - C The line is parallel to the vector `vecb=3hati+4hatj+5hatk` and the plane is normal to the vector `vecn=2hati-2hatj+hatk`. If `theta` is the angle between the line and the plane. Then `sin theta=(vecb.vecn)/(|vecb||vecn|)impliessin theta=(6-8+5)/(sqrt(50)sqrt(9))=1/(5sqrt(2))=(sqrt(2))/10` |
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| 77. |
A plane meets the coordinate axes in A,B,C such that the centroid of triangle ABC is the point `(p,q,r)`. If the equation of the plane is `x/p+y/q+z/r=k` then `k=`A. 1B. 2C. 3D. none of these |
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Answer» Correct Answer - C Let the equation of the required plane be `x/a+y/b+z/c=1`…………….i Then the coordinates of A,B and C are `A(a,0,0),B(0,b,0)` and `C(0,0,c)` are respectively. So, the centroid of triangle ABC is `(a//3,b//3,c//3)` But the coordinates of the centroid are `(p,q,r)` `:.p=a/3,q=b/3` and `r=c/3impliesa=3p,b=3q` and `c=3r` Substituting the values of a,b and c in i , we obtain the required plane as `x/(3p)+y/(3q)+z/(3r)=1impliesx/p+y/q+z/r=3` |
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| 78. |
The intercepts of the plane `5x-3y+6z-60=0` on the coordinate axes areA. `10,20,-10`B. `10,-20,12`C. `12,-20,10`D. `12,20,-10` |
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Answer» Correct Answer - B The equation of the plane is `5x-3y+6z-60=0` `implies5x-3y+6z=60impliesx/12+y/(-20)+z/10=1` Hence the intercepts are 12,-20,10. |
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| 79. |
The plane `x / 2 + y / 3 + z / 4 = 1` cuts the co-ordinate axes in `A, B, C`: then the area of the `DeltaABC` isA. `sqrt(29)` sq. unitsB. `sqrt(41)` sq. unitsC. `sqrt(61)` sq. unitsD. none of these |
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Answer» Correct Answer - C The given plane cuts the coordinate axes in `A(2,0,0),B(0,3,0)` and `C(0,0,4)`. `:.` Area of `DeltaABC=1/2ABxxACxxsin/_BAC` Now, `AB=sqrt(4+9+0)=sqrt(13),AC=sqrt(4+0+16)=sqrt(20)` `cos /_BCA(vec(AB).vec(AC))/(|vec(AB)||vec(AC)|)=((-2hati+3hatj).(-2hati+4hatk))/(sqrt(4+9)sqrt(4+16))` `impliescos /_BAC=(4+0+0)/(sqrt(13)sqrt(20))=4/(sqrt(13)sqrt(20))=2/(sqrt(65))` `impliessin /_BAC=sqrt(1-4/65)=sqrt(61/65)` Hence Area of `DeltaABC=1/2xxsqrt(13)xxsqrt(20)xxsqrt(61/65)=sqrt(61)` sq. units. |
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| 80. |
The distance of the point (1,-5,9) from the plane x-y+z = 5 measured along the line x = y = z isA. `5sqrt(3)`B. `3sqrt(10)`C. `3sqrt(5)`D. `10sqrt(3)` |
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Answer» Correct Answer - D The equation of the line pasing through the point `A(1,-5,9)` and parallel to the line `x=y=z` is `(x-1)/1=(y+5)/1=(z-9)/1` Suppose it cuts the plane `x-y+z=5` at `P(lamda+1,lamda-5,lamda+9)` As `P` lies on `x+y+z=5` `:. lamda+1-lamda+5+lamda+9=5implieslamda=-10.` So, the coordinates of `P` are `(-9,-15,-1)`. Hence, required distance `-AP-sqrt(100+100+100)=10sqrt(3)` |
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| 81. |
Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]` Let `omega` be a solution of `x^3-1=0` with `Im(omega)gt0. I fa=2` with b nd c satisfying (E) then the vlaue of `3/omega^a+1/omega^b+3/omega^c` is equa to (A) -2 (B) 2 (C) 3 (D) -3 |
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Answer» Correct Answer - D We have `[(a,b,c)][(1,9,7),(8,2,7),(7,3,7)]=[(0,0,0)]` `implies a+8b+7c=0,9a+2b+3c=0,7a+7b+7c=0` `impliesa=1,b=6,c=-7` Clearly `P(a,b,c)` lies on the plane `2x+y+z=1` `:.7a+b+c=7+6-7=6` |
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| 82. |
The locus of a point `P(x,y,z)` which moves in such a way that `z=c` (constant), is aA. line parallel to z-axisB. plane parallel to xy-planeC. line parallel to y-axisD. line parallel to x-axis |
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Answer» Correct Answer - B Since `z=0` represents the xy-plane. Therefore, `z=c` represents a plane parallelto xy-plane. |
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| 83. |
The locus of a point `P(x,y,z)` which moves in such away that `x=a`and `y=b` is aA. plane parallel to xy-planeB. line parallel to x-axisC. line parallel to y-axisD. line parallel to z-axis |
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Answer» Correct Answer - D Since`x=0` and `y=0` together represents z-axis, Therefore `x=a` and `y=b` represent a line parallel to z-axis. |
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| 84. |
In a three dimensional space the equation `x^2-5x+6=0`representsa. points b.planes c. curves d. pair of straight linesA. two pointsB. two parallel planesC. two parallel linesD. a pair of non -parallel lines |
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Answer» Correct Answer - B We have `x^(2)-5x+6=0implies(x-2)(x-3)=0impliesx=2,x=3` Clearly these two equations represent two parallel planes parallel to YOZ plane. |
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| 85. |
A non vector `veca` is parallel to the line of intersection of the plane determined by the vectors `hati,hati+hatj` and thepane determined by the vectors `hati-hatj,hati+hatk` then angle between `veca and hati-2hatj+2hatk` is = (A) `pi/2` (B) `pi/3` (C) `pi/6` (D) `pi/4`A. `pi//3`B. `pi//4`C. `pi//6`D. none of these |
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Answer» Correct Answer - B Clearly `veca` is perpendicular to the normals to the two planes determined by the given pairs of vectors. We have, `vecn_(1)=` normal vector to the plane determined by `hati` and `hati+hatj` `impliesvecn_(1)=hatix(hati+hatj)=hatk` `vecn_(2)=` normal vector to the plane determined by `hati-hatj` and `hati+hatk` `impliesvecn_(2)=(hati-hatj)xx(hati+hatk)=-hati-hatj+hatk` Since `veca` is perpendicular to `vecn_(1)` and `vecn_(2)`. Therefore, `veca=lamda(vecn_(1)xxvecn_(2))=lamda{hatixx(-hati-hatj+hatk)}=lamda(-hatj+hati)` Let `theta` be the angle between`veca` and `hati-2hatj+2hatk`. Then, `cos theta=(lamda(1+2+0))/(lamdasqrt(2)sqrt(1+4+4))=1/(sqrt(2))impliestheta=pi//4`. |
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| 86. |
A non vector `veca` is parallel to the line of intersection of the plane determined by the vectors `hati,hati+hatj` and thepane determined by the vectors `hati-hatj,hati+hatk` then angle between `veca and hati-2hatj+2hatk` is = (A) `pi/2` (B) `pi/3` (C) `pi/6` (D) `pi/4`A. `(pi)/2`B. `(pi)/3`C. `(pi)/6`D. `(pi)/4` |
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Answer» Correct Answer - D The line of intersection of the two planes is perpendicular to their normals. So, it parallel to the vector `veca={(hatixx(hati+hatj)}xx{(hati-hatj)xx(hati=hatk)}` `impliesveca=hatkxx(-hatj+hatk-hatk)=hati-hatj` Let `theta` be the angle between `veca` and `vecb-hati-2hatj+2hatk`. Then `cos theta=(veca.vecb)/(|veca||vecb|)=3/(sqrt(2)xx3)=1/(sqrt(2))impliestheta=(pi)/4` |
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| 87. |
If `alpha+beta+gamma =2 and veca=alphahati+betahatj+gammahatk, hatkxx (hatkxxveca)=vec0` then gamma= (A) 1 (B) -1 (C) 2 (D) none of these |
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Answer» Correct Answer - C Since `(alpha, beta,gamma)` lies on the plane `x+y+z=2`. `:.alpha+beta+gamma=2`……………i We have `hatkxx(hatkxxveca)=vec0` `implies(hatk.hatk)veca-(hatk.veca)hatk=vec0` `implies veca-gammahatk=vec0impliesalpha hati+betahatj=vec0impliesalpha=beta=0` Putting `alpha=beta=0` i (i) we get `gamma=2`. |
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| 88. |
If the plane `7x+11y+13z=3003` meets the axes in A,B and C then the centorid of `DeltaABC` isA. `(143,91,77)`B. `(143,77,91)`C. `(91,143,77)`D. `(143,66,91)` |
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Answer» Correct Answer - A Given plane meets the coordinate axes at `A(429,0,0),B(0,273,0)` and `C(0,0,231)` So the coordinates of the centroid of `DeltaABC` are `(429/3,273/3,231/3)` i.e. `(143,92,77)` |
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