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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. |
Find the direciton ratio of the line `(3-x)/(1)=(y-2)/(5) =(2z-3)/1`A. `(1,5,1/2)`B. (-5,5,1)C. `(-1,5,1/2)`D. (1,5,1) |
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Answer» Correct Answer - C Given equation of line is `(3-x)/(1)=(y-2)/(5)=(2z-3)/(1)` `Rightarrow (x-3)/(-1)=(y-2)/(5)=(z-(3)/(@))/((1)/(2))` `therefore` Direction ratios of line are `(-1,5,(1)/(2))` |
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| 52. |
The angle between the line `x=1=y=2 =z/1 and x/1=y=-1= z=0` isA. `30^(@)`B. `60^(@)`C. `90^(@)`D. `0^(@)` |
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Answer» Correct Answer - C Given lines are `(x-1)/(0)=(y-2)/(0)=z/1` `and x/1=(y+1)/(0)=z/0` `"Here", l_1=0, m_(1)=0, n_(1)=1` `and l_2=1, m_2=0n_(2)=0` `therefore cos theta =l_(1)l_(2)+m_(1)m_2+n_1n_2=0.1+0.0+0.1=0` `Rightarrow theta=90^(@)` |
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| 53. |
A line makes an angle `theta` both with x-axis and y-axis. A possible range of `theta` isA. `[0,pi/4]`B. `[0,pi/2]`C. `[pi/4,pi/2]`D. `[pi/3,pi/6]` |
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Answer» Correct Answer - C We know that `cos^(2)alpha=cos^(2)beta+cos^(2)gamma=1` `Rightarrow cos^(2)theta=cos^(2)theta+cos^(2)gamma=1" "[therefore alpha=beta=theta]` `Rightarrow 2 cos^(2)theta+cos^(2)gamma=1` `Rightarrow cos^2gamma=1-2cos^(2)theta` `Rightarrow cos^2gamma=-cos^(2)theta` `Rightarrow cos 2theta le theta` `therefore theta in [pi/4,pi/2]` |
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| 54. |
The angle between the lines `(x+4)/1=(y-3)/2=(z+2)/3` and `x/3=(y-1)/(-2)=z/1` isA. `sin^(-1)((1)/(7))`B. `cos^(-1)((2)/(7))`C. `cos^(-1)((1)/(7))`D. None of these |
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Answer» Correct Answer - C Angle between two lines is given by `cos theta=(1xx3+2xx-2+3xx1)/(sqrt(1^(2)+2^2+3^2)sqrt(3^(2)+(-2)^2+1^2))=2/(sqrt14 sqrt14)` `therefore theta=cos^(-1) ((1)/(7))` |
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| 55. |
Find the acute angle between the two straight lines whose direction cosines are given by `l+m+n=0` and `l^2+m^2-n^2=0`A. `(pi)/(3)`B. `pi/4`C. `pi/6`D. `pi-2` |
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Answer» Correct Answer - A We know that, angle between two line is `cos theta=(a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)c^(2)sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2))))` `therefore l+m+n=0 Rightarrow l=-(m+n)` `Rightarrow (m+n)^(2)=l^(2)` `Rightarrow m^(2)+n^(2)+2mn=m^(2)+n^(2)" "[therefore l^(2)=m^(2)+n^(2)]` `Rightarrow 2mn=0` when `m=0 Rightarrow l=-n` Hence, (l,m,n) is (1,0,-1) when, n=0, then l=-m Hence, (l,m,n) is (1,0,-1) `therefore cos theta=(1+theta+0)/(sqrt2xxsqrt2)=1/2` `Rightarrow theta=cos^(-1) ((1)/(2))=pi/3` |
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| 56. |
Find the angle between the pair of lines ` (x -1 )/(4) = ( y - 3)/(1) = (z) /(8) ` and `(x - 2)/(2) = ( y + 1) /(2) = (z - 4)/(1). `A. `cos^(-1)(14/27)`B. `cos^(-1)(-14/27)`C. `cos^(-1) (2/3)`D. `cos^(-1) (-2/3)` |
| Answer» Correct Answer - C | |
| 57. |
The acute angle between the line joining the points (2,1,-3) and (-3,1,7)and a line parallel to `(x-1)/3=y/4=(z+3)/5` through the point (-1,0,4) isA. `cos^(-1)((1)/(sqrt(10)))`B. `cos^(-1)((1)/(5sqrt(10)))`C. `cos^(-1)((7)/(5sqrt(10)))`D. `cos^(-1)((3)/(5sqrt(10)))` |
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Answer» Correct Answer - C Direction ratio of the line joining the points (2,1,-3) and (-3,1,7) are `(a_(1),b_1,c_(1))i.e, (-5,0,10)` Direction ratio of the line parallel to line `(x-1)/(3)=(y)/(4)=(z+3)/(5)"are"(a_(2),b_(2),c_(2))i.e, (3,4,5)` Angle between two lines is given by `cos theta=((-5xx3)+(0xx4)+(10xx5))/(sqrt(25+0+100)sqrt(9+16+25))=(35)/(25sqrt10)` `therefore theta=cos^(-1)((7)/(5sqrt10))` |
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| 58. |
The acute angle between the lines `(x+4)/-1 = (y-2)/2 = (z-3)/2` and `x/2 = (y-1)/-2 = (z+1)/1` isA. `cos^(-1) (4/9)`B. `cos^(-1)(-4/9)`C. `cos^(-1)(2/9)`D. `cos^(-1)(8/9)` |
| Answer» Correct Answer - A | |
| 59. |
The angle between the lines `(x+1)/2 = (y+3)/2 = (z-4)/-1` and `(x-4)/1 = (y+4)/2 = (z+1)/2` isA. `cos^(-1) (1/9)`B. `cos^(-1)(2/9)`C. `cos^(-1)(3/9)`D. `cos^(-1)(4/9)` |
| Answer» Correct Answer - D | |
| 60. |
The angle between the lines `(x-1)/1 = (y-1)/1 = (z-1)/2` and `(x-1)/(-sqrt(3)-1) =(y-1)/(sqrt(3)-1) = (z-1)/4` isA. `pi/6`B. `pi/3`C. `pi/4`D. `pi/2` |
| Answer» Correct Answer - B | |
| 61. |
The line `(x-2)/3=(y+1)/2=(z-1)/1`intersects the curve `x y=c^(I2),z=0`if `c`is equal toa. `+-1`b. `+-1//3`c. `+-sqrt(5)`d. none of theseA. ` pm 1 `B. ` pm ( 1 ) / ( 3 ) `C. ` pm sqrt 5 `D. None of these |
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Answer» Correct Answer - C We have, z = 0 for the point where the line intersects the curve ` therefore (x - 2 ) / (3 ) = ( y + 1 ) / ( 2 ) = ( 0 - 1 ) / ( - 1 ) ` ` rArr ( x - 2 ) / ( 3 ) = 1 and ( y +1 )/ ( 2 ) = 1 ` ` rArr x = 5 and y = 1 ` Put these values in ` xy = c^ 2 ` , we get ` 5 = c ^ 2 rArr c = pm sqrt 5 ` |
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| 62. |
the line passing through the points ` (3, -2--5),` and `(3, -2,6)`A. `(x+3)/0 = (y-2)/0 =(z-15)/-11`B. `(x+3)/0 = (y-2)/0 = (z-5)/11`C. `(x-3)/0 = (y+2)/0 = (z+5)/-11`D. `(x-3)/0 = (y+2)/0 = (z+5)/11` |
| Answer» Correct Answer - D | |
| 63. |
the line passing through the points ` (3, -2--5),` and `(3, -2,6)`A. `(3hati-2hatj - 5hatk) +lambda(-11hatk)`B. `(3hati - 2hatj - 5hatk) + lambda(11hatk)`C. `(3hati - 2hatj - 5hatk) + lambda(-4hatj)`D. `(3hati - 2hatj - 5hatk) + lambda(6hati)` |
| Answer» Correct Answer - B | |
| 64. |
Find the vector wuation of the line passing through the points A(3,4, -7) and B (6,-1, 1)A. `(x+3)/-3 = (y+4)/5 = (z-7)/8`B. `(x-3)/-3 = (y-4)/5 = (z+7)/8`C. `(x+3)/3 = (y+4)/-5 = (z-7)/8`D. `(x-3)/3 = (y-4)/-5 = (z+7)/8` |
| Answer» Correct Answer - D | |
| 65. |
Find the vector wuation of the line passing through the points A(3,4, -7) and B (6,-1, 1)A. `(6hati -hatj +hatk) + lambda(3hati - 5hatj + 8hatk)`B. `(3hati + 4hatj - 7hatk) + lambda(3hati - 5hatj + 8hatk)`C. `(6hati - hatj +hatk) + lambda(-3hati + 5hatj - 8hatk)`D. `(3hati + 4hatj - 7hatk) + lambda(-3hati + 5hatj - 8hatk)` |
| Answer» Correct Answer - B | |
| 66. |
The vector equation of a line passing through the points `(-2,3,4)` and (1,1,2) isA. `(-2hati + 3hatj + 4hatk) + lambda(3hati - 2hatj - 2hatk)`B. `-(2hati + 3hatj + 4hatk) + lambda(3hati - 2hatj + 2hatk)`C. `(-2hati + 3hatj + 4hatk) + lambda(3hati - 2hatj + 2hatk)`D. `(-2hati + 3hatj + 4hatk) + lambda(3hati + 2hatj + 2hatk)` |
| Answer» Correct Answer - A | |
| 67. |
The vector equation of a line passing through the points A(4,2,1) and B(2,-1,3) isA. `(-4hati - 2hatj - hatk) + lambda(2hati + 3hatj - 2hatk)`B. `(4hati + 2hatj + hatk) + lambda(-2hati - 3hatj + 2hatk)`C. `(-4hati - 2hatj - hatk) + lambda(2hati +3hatj - 4hatk)`D. `(4hati + 2hatj +hatk) + lambda(-2hati - 3hatj + 4hatk)` |
| Answer» Correct Answer - B | |
| 68. |
A line passing through the points A (-2 , -1 , 5) and B ( 1, 3 , -1) ,find the equation of the line in parametric form. Also, write the equation in non - parametric form.A. `vecr xx (3hati + 4hatj - 6hatk) = 14hati + 3hatj + 5hatk`B. `vecr xx (3hati + 4hatj - 6hatk) = 14hati - 3hatj - 5hatk`C. `vecr xx (3hati + 4hatj - 6hatk) = 14hati - 3hatj + 5hatk`D. `vecr xx (3hati + 4hatj - 6hatk) = -14hati + 3hatj - 5hatk` |
| Answer» Correct Answer - D | |
| 69. |
The lines `(x-2)/(1)=(y-3)/(1)=(z-4)/(-k) and (x-1)/(k)=(y-4)/(2)=(z-5)/(1)` are coplanar, ifA. `k = +-3`B. `k = +-1`C. `k=0, -1`D. `k=0,-3` |
| Answer» Correct Answer - D | |
| 70. |
Find the angle between the pair of line: vecr=3hati+2hatu-4hatk+lamds(hati+2hatj+2hatk), vecr=5hati-2hatk+mu(3hati+2hatj+6hatk)`A. `cos^(-1) (21/16)`B. `cos^(-1)(19/21)`C. `cos^(-1)(21/15)`D. `cos^(-1) (15/21)` |
| Answer» Correct Answer - B | |
| 71. |
The symmetric equation of lines 3x+2z-5=0 and x+y-2z-3=0 isA. `(x-1)/(5)=(y-4)/(7)=(z-0)/(1)`B. `(x+1)/(5)=(y+4)/(7)=(z-0)/(1)`C. `(x+1)/(-5)=(y+4)/(7)=(z-0)/(1)`D. `(x-1)/(-5)=(y-4)/(7)=(z-0)/(1)` |
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Answer» Correct Answer - C Let a,b,c be the direction ratios of required line. `therefore 3a+2b+c=0` and a+b-2c=0 `Rightarrow (a)/(-4-1)=(b)/(1+6)=(c)/(3-2)` `Rightarrow (a)/(-5)=(b)/(7)=(c)/(1)` In order to find on the required line, we put z=0 in the two given equations to obtain 3x+2y=5 and x+y=3. On solving these two equations, we get x=-1, y=4 `therefore` Coordintaes of point on required line are (-1,4,0) Hence, required line is `(x+1)/(-5)=(y-4)/(7)=(z-0)/(1)` |
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| 72. |
The line `(x+1)/(-10)=(y+3)/(-1)=(z-4)/1 and (x+10)/(-1)=(y+1)/(-3) =(z-1)/(4)` interest at the point.A. (11,-4,5)B. (-11,-4,5)C. (11,4,-5)D. (-11,-4,-5) |
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Answer» Correct Answer - B Let `(x+1)/(-10)=(y+3)/(-1)=(z-4)/(1)=lambda_(1)` `and (x+10)/(-1)=(y+1)/(-3)=(z-1)/(4)=lambda_(2)` `"Then", (-10lambda_(1)-1-lambda_(2)-3,lambda_(1)+4)` and `(-lambda_(2)-10,-3lambda_1-1, 4lambda_(2)+1)` are identical. `-10lambda_(1)-1=-lambda_(2)-10,-lambda,-3=-3lambda_2-1` and `lambda_(1)+4=4lambda_(2)+1` `Rightarrow -10lambda_(1)+lambda_(2)=-9-lambda_(1)+3lambda_(2)=2` `lambda_(1)-4lamda2=-3` On solving, we get `lambda_(1)=lambda_(2)=1` Intersection points is `(-10xx1-1, -1-3,1+4)i.e, (-11,-4,5)` On solving, we get `x=-11, y=-4 and z=5`. |
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| 73. |
Consider the lines ` L _ 1 : ( x - 1 )/ ( 3 ) = ( y + 2 ) / ( 1 ) = ( z + 1 ) / ( 2 ) ` ` L _ 2 : ( x - 2) / ( 1 ) = ( y + 2 ) / ( 2 ) = ( z - 3 )/ ( 3 ) ` The unit vector perpendicular to both ` L _ 1 and L_ 2 ` isA. ` ( 1 ) / ( sqrt (99)) ( - hati + 7 hatj +7hatk ) `B. ` ( 1 ) / ( 5 sqrt3) ( - hati - 7 hatj + 5 hatk ) `C. ` ( 1 ) / ( 5sqrt 3 ) ( - hati+ 7 hatj + 5 hatk ) `D. ` ( 1 ) /( sqrt ( 99)) ( 7 hati - 7hatj - hatk ) ` |
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Answer» Correct Answer - B Lines `L _ 1 and L _ 2 ` are parallel to the vectors ` b _ 1 = 3 hati + hatj + 2 hatk and b _ 2 = hati + 3 hatj + 3 hatk ` respectively. Therefore, a unit vector perpendicular to both `L _ 1 and L _ 2 ` is ` hatn = ( b _ 1 xx b _ 2 ) / ( | b _ 1 xx b _ 2 | ) ` ` b _ 1 xx b _ 2 = |{:( hati , hatj , hatk ) , ( 3, 1, 2 ) , (1, 2, 3 ) :}| = - hati - 7 hatj + 5hatk ` ` therefore hatn = ( 1 ) / ( 5sqrt 3 ) ( - hati - 7 hatj + 5 hatk ) ` |
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| 74. |
If the lines ` x = 1 + s, y = - 3 - lamda s , z = 1 + lamda s and x = ( t ) / ( s ) , y = 1 + t , z = 2 - t ` are coplanar, then ` lamda ` is equal toA. ` - 2 `B. `- 1 `C. ` - ( 1 ) / ( 2 ) `D. 0 |
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Answer» Correct Answer - A Given, line are ` ( x - 1 ) / ( 1 ) = ( y + 3 ) / ( - lamda ) = ( z - 1) / ( lamda ) ` and ` ( x ) / ( 1//2 ) = ( y - 1 ) / ( 1 ) = ( z - 2 ) / ( - 1 ) ` If lines are coplanar, then ` |{:( 1-0, -3-1,1-2 ) , ( 1, - lamda, lamda ), ( 1, 2, 2 ) :}| = 0 ` ` rArr |{:( 1 , - 4, - 1 ) , ( 1, - lamda,lamda ) , ( 1, 2, - 2) :}| = 0 rArr lamda = - 2 ` |
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| 75. |
If the lines `(x-1)/2=(y+1)/3=(z-1)/4`and `(x-3)/1=(y-k)/2=z/1`intersect, then k is equal to(1) `-1`(2) `2/9`(3) `9/2`(4) 0A. ` - 1 `B. ` ( 2 ) / ( 9 ) `C. ` ( 9 ) / ( 2 ) `D. ` 0 ` |
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Answer» Correct Answer - C ` L _ 1 : ( x - 1 ) / ( 2 ) = ( y + 1 ) / ( 3 ) = ( z - 1 ) / ( 4 ) = p ` ` L _ 2 : ( x - 3 ) / ( 1) = ( y - k) /( 2 ) = ( z - 0 ) / ( 1 ) = q ` ` rArr ` Anypoint P on line ` L _ 1 ` is of type ` P ( 2p + 1, 3p -1, 4p + 1 ) ` and any point Q on line ` L _ 2 ` is type Q ` ( q + 3, 2q + k , q ) ` Since, ` L _ 1 and L _ 2 ` are intersecting each other, hence both point P and Q should coincide at the point of intersection i.e., corresponding coordinates of P and Q should be same. ` 2p + 1 = q + 3 and 4 p + 1 = q `, we get the value of p and q as ` p = ( - 3 ) / ( 2 ) and q = - 5 ` On substituting the values of p and q in the third equation ` 3p - 1 = 2q + k `, we get ` therefore 3 ( ( - 3 ) / ( 2 ) ) - 1 = 2 ( - 5) + k` ` rArr k = ( 9 ) / ( 2 ) ` |
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| 76. |
Consider the line L 1 : x 1 y 2 z 1 312 +++ ==, L2 : x2y2z3 123A. ` 0 `B. ` 17//sqrt 3 `C. `41//5sqrt3 `D. `17//5 sqrt 2 ` |
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Answer» Correct Answer - D The shortest distance between ` L _ 1 and L _ 2 ` is ` (|{:( x _2 - x _ 1 , y _ 2 - y _ 1 , z _ 2- z_ 1 ) , ( a_ 1, b _ 1, c _ 1), ( a_ 2, b _ 2, c _ 2):}| ) /( sqrt ( ( a _ 1 b _ 2 - a _ 2 b _ 1 ) ^ 2 + ( b _ 1 c _ 2 - b _ 2 c _ 1 ) ^ 2 + ( c _ 1 a _ 2 - c _ 2 a _ 1 ) ^ 2 ) ) ` ` (|{:( 3,0,4) ,( 3,1,2) ,(1,2,3 ):}| ) / ( sqrt ( 25 + 1 + 49)) = (3 ( 3- 4 ) + 0 + 4 ( 6 - 1) ) / ( sqrt ( 75 )) ` ` = ( 17 ) / ( sqrt ( 2 ) ) ` |
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| 77. |
If the lines `(x-1)/(-3)=(y-2)/(2k)=(z-3)/(-2)a n d(x-1)/(3k)=(y-5)/1=(z-6)/(-5)`are at right angel, thenfind the value of `kdot`A. `-10`B. `-7/10`C. `-10/7`D. `10/7` |
| Answer» Correct Answer - C | |
| 78. |
Find the two points on the line ` ( x - 2 ) / ( 1 ) = ( y + 3) / ( - 2 ) = ( z + 5) / ( 2 ) ` on either side of ` ( 2,-3, - 5 ) ` which are at a distance of 3 units from it.A. `(-3,-5,3` and `(1,-1,-7)`B. `(3,-5,-3)` and `(1,-1,-7)`C. `(3,-5,-3)` and `(-1,1,-7)`D. `(-3,-5,3)` and (`-1,1,-7)` |
| Answer» Correct Answer - B | |
| 79. |
Match the following columns. A. `{:(A,B,C,D),(q,r,s,p):} `B. `{:(A,B,C,D),(s,p,q,r):} `C. `{:(A,B,C,D),(p,q,r,s):} `D. `{:(A,B,C,D),(r,p,s,q):} ` |
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Answer» Correct Answer - B A-s, B- p , C - q, D - r |
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| 80. |
The straight line `(x-3)/3=(y-2)/1=(z-1)/0`isParallel to x-axisParallel to the y-axisParallel to the z-axisPerpendicular to the z-axisA. parallel to X - axisB. parallel to Y - axisC. parallel to Z- axisD. perpendicular to Z- axis |
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Answer» Correct Answer - D Equation of x, y and z axes are ` (x ) / (1 ) = ( y ) / (0 ) = ( z ) / ( 0 ) , ( x ) / ( 0 ) = ( y ) / ( 1 ) = ( z ) / ( 0 ) and ( x ) / ( 0 ) = ( y ) / ( 0 ) = ( z ) / ( 1 ) ` respectively. given equation is ` ( x - 3 ) / ( 3) = ( y - 2 ) / ( 1 ) = ( z - 1 ) / ( 0 ) ` Here, ` 3 xx 0 + 1 xx 0 + 0 xx 1 = 0 ` Hence, the line is perpendicular to Z - axis. |
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| 81. |
If the lines `(x-4)/15 = (y-17)/9 = (z-11)/8` and `(x-15)/4 = (y-9)/17 = (z-8)/11` are intersecting at point A, then `OA^(2)=`A. `sqrt(1398)`B. `sqrt(1390)`C. 1398D. 1390 |
| Answer» Correct Answer - C | |
| 82. |
The points (-2,3,4), (1,1,2) and (4,-1,0) areA. collinearB. non-collinearC. non-coplanarD. non-collinear but coplanar |
| Answer» Correct Answer - A | |
| 83. |
Find the two points on the line ` ( x - 2 ) / ( 1 ) = ( y + 3) / ( - 2 ) = ( z + 5) / ( 2 ) ` on either side of ` ( 2,-3, - 5 ) ` which are at a distance of 3 units from it.A. `(-3,-5,3` and `(1,-1,-7)`B. `(3,-5,-3)` and `(1,-1,-7)`C. `(3,-5,-3)` and `(-1,1,-7)`D. `(-3,-5,3)` and (`-1,1,-7)` |
| Answer» Correct Answer - B | |
| 84. |
If the lines `(x-1)/(-3)=(y-2)/(2k)=(z-3)/2`and `(x-1)/(3k)=(y-1)/1=(z-6)/(-5)`are perpendicular, find the value of k.A. `(-10)/(7)`B. `(10)/(7)`C. `(-10)/(11)`D. `(10)/(11)` |
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Answer» Correct Answer - A Given lines can be rewritten as `(x-1)/(-3)=(y-2)/(2alpha)=(z-3)/(2)` and `(x-1)/(3alpha)=(y-1)/(1)=(z-6)/(-5)` `therefore` The direction ratio of given lines are `(-3,2,alpha,2) and (3alpha,1,-5)` Since, lines are perependicular `therefore a_(1)a_(2)=b_(1)b_(2)+c_(1)c_(2)=0` `Rightarrow (-3)(3alpha)+2alpha(1)+2(-5)=0` `Rightarrow -9alpha+2alpha-10=0` `Rightarrow alpha=-10/7` |
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| 85. |
Find the equation of a line passing through `(1,-1,0)`and parallel to the line `(x-2)/3=(2y+1)/2=(5-z)/1`A. ` ( x - 1 ) / ( 3 ) = ( y + 1 ) / ( 2 ) = ( 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 given line can be rewritten as ` ( x - 2 ) / ( 3) = ( y + ( 1 ) / ( 2 )) / ( 1 ) = ( z - 5 )/ ( 1 ) ` direction ratios are 3, 1, 1 Hence, equation of required line ` ( x- 1) / ( 3 ) = ( y + 1 ) / ( 1 ) = ( z - 0 ) / ( 1 ) ` |
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| 86. |
The length of theperpendicular drawn from `(1,2,3)`to the line `(x-6)/3=(y-7)/2=(z-7)/(-2)`isa. `4`b. `5`c. `6`d. `7`A. 5 unitsB. 7 unitsC. 4 unitsD. None of these |
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Answer» Correct Answer - B Let L be the foot the perpendicular drawn from the point ` P ( 1, 2 , 3 ) ` to the given line. Let the co - ordinates of L be ` ( 3 lamda + 6, 2 lamda + 7, - 2lamda + 7 ) " " `…(i) ` therefore ` Direction ratios of the given line are proportional to 3, 2, - 2 Since PL is perpendicular to the given line. ` therefore 3 ( 3 lamda + 5 ) + 2 ( 2 lamda + 5 ) + ( - 2 ) ( - 2 lamda + 4 ) = 0 ` ` rArr lamda = - 1 ` Hence, co - oridinates of L are ` ( 3, 5, 9 ) ` therefore PL = sqrt ( ( 3 - 1 ) ^ 2 + ( 5 - 2 ) ^ 2 + ( 9 - 3 ) ^ 2 )) = ` 7 units |
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| 87. |
The straight line `(x-3)/3=(y-2)/1=(z-1)/0`isParallel to x-axisParallel to the y-axisParallel to the z-axisPerpendicular to the z-axisA. parallel to X-axisB. parallel to Y-axisC. parallel to Z-axisD. perpendicular to Z-axis |
| Answer» Correct Answer - D | |
| 88. |
The point of intersectionof the lines `(x-5)/3=(y-7)/(-1)=(z+2)/1a n d=(x+3)/(-36)=(y-3)/2=(z-6)/4`isa. `(21 ,5/3,(10)/3)`b. `(2,10 ,4)`c. `(-3,3,6)`d. `(5,7,-2)`A. (2,10,4)B. (-3,3,6)C. `(5,7,-2)`D. `(21,5/3, 10/3)` |
| Answer» Correct Answer - D | |
| 89. |
The equation to the straight line passing through the points (4,-5,2) and (-1,5,3) isA. `(x-4)/(1)=(y+5)/(-2)=(z+2)/(-1)`B. `(x+1)/(1)=(y-5)/(2)=(z-3)/(-1)`C. `(x)/((-1))=(y)/(5)=z/3`D. `x/4=y/-5=z/-2` |
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Answer» Correct Answer - A Equation of straight line passing through (4,-5,-2) and (-1,5,3) is `(x-4)/(-1-4)=(y+5)/(5+5)=(z+2)/(3+2)` `Rightarrow (x-4)/(-5)=(y+5)/(10)=(z+2)/(5)` `Rightarrow (x-4)/(1)=(y+5)/(-2)=(z+2)/(-1)` |
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| 90. |
If`(x-1)/l=(y-2)/m=(z+1)/n` is the equation of the line through `(1,2,-1)`and`(-1,0,1)`, then`(l,m,n)`A. (0,1,0)B. (-1,0,1)C. (1,1,-1)D. (1,2,-1) |
| Answer» Correct Answer - C | |
| 91. |
The direction cosines of the line `x-y+2z=5, 3x+y+z=6` areA. ` ( - 3 ) / ( 5 sqrt ( 2 ) ), ( 5 ) / ( 5sqrt2 ) , ( 4 ) / ( 5sqrt 2 ) `B. ` ( 3 ) /( 5sqrt 2 ) , ( - 5 ) /(5sqrt 2 ) , ( 4 ) / ( 5sqrt2 ) `C. ` ( 3 ) /( 5sqrt 2 ) , ( 5 ) / ( 5sqrt 2) , ( 4 ) / ( 5sqrt 2 ) `D. None of these |
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Answer» Correct Answer - A Vector normal to given planes are ` n_ 1 = hati - hatj + 2 hatk and n _ 2 = 3 hati + hatj + hatk ` . So , their line of intersection is parallel to the vector ` n = n _ 1 xx n _ 2 = |{:( hati , hatj , hatk ) , ( 1, -1, 2 ) , ( 3, 1 , 1 ) :}| = - 3 hati +5 hatj + 4hatk ` ` rArr hatn = ( - 3 ) / ( 5 sqrt 2 ) hati + ( 5 ) / ( 3 sqrt 2 ) hatj + ( 4 ) /( 5sqrt 2 ) hatk ` Hence, direction cosines of the line are ` ( - 3 ) / ( 5sqrt2 ) , ( 5 ) / ( 5sqrt2) , ( 4 ) / ( 5sqrt2 ) ` |
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| 92. |
For the lines `L_1 ; vec a+t(vec b+vec c) and L_2 ; vec r=vec b+s(vec c+vec a)` then `L_1 and L_2` 1ntersect atA. aB. bC. a b+cD. a 2b |
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Answer» Correct Answer - C Since the lines intersect `therefore a+t(b+c)=b+s(c+a)` We easily get t=1, s=1 Hence point of intersection is a+b+c. |
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| 93. |
The lines `x/1=y/2=z/3` and `(x-1)/(-2)=(y-2)/(-4)=(z-3)/(-6)` areA. perpendicularB. coincidentC. intersectingD. skew |
| Answer» Correct Answer - B | |
| 94. |
Direction cosines of the line `(x+2)/(2)=(2y-5)/(3),z=-1` areA. `4/5, 3/5,0`B. `3/5, 4/5, 1/5`C. `-3/5, 4/5, 0`D. `4/5, -2/5, 1/5` |
| Answer» Correct Answer - A | |
| 95. |
The shortest distance between the lines `(x-5)/4 = (y-7)/-5 = (z+3)/-5` and `(x-8)/7 = (y-7)/1 = (z-5)/3` isA. `342/sqrt(3830)`B. `171/sqrt(3830)`C. `282/sqrt(3830)`D. `141/sqrt(3830)` |
| Answer» Correct Answer - C | |
| 96. |
Find the shortest distancebetween the lines `(x-1)/2=(y-2)/3=(z-3)/4a n d(x-2)/3=(y-4)/4=(z-5)/5`.A. `2/sqrt(6)`B. `1/sqrt(6)`C. `2sqrt(6)`D. `sqrt(6)` |
| Answer» Correct Answer - B | |
| 97. |
The distance between two parallel lines can be taken out by the formulaA. `d=|(b.(a_(2)-a_(1)))/(|b|)|`B. `d=|(b.(a_(2)-a_(1)))/(|a_(2)-a_(1)|)|`C. `d=|(b.(a_(2)-a_(1)))/(|a_(2)-a_(1)|)|`D. `d=|(bxx(a_(2)-a_(1)))/(|b|)|` |
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Answer» Correct Answer - D The distance between the given parallel lines is `d=|(b xx(a_(2) -a_(1)))/(|b|)|` |
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| 98. |
A symmetrical form of the line of intersection of the planes `x=ay+b` and `z=cy+d` isA. `(x-b)/a = y/1 = (z-d)/c`B. `(x-b)/a = y/b = (z-d)/c`C. `(x-a)/b= y/1 = (z-c)/d`D. `(x+a)/b = y/1 = (z+c)/d` |
| Answer» Correct Answer - A | |
| 99. |
Show that the line`(x-1)/2=(y-2)/3=(z-3)/4a n d(x-4)/5=(y-1)/2`intersect. Find their point of intersection.A. (0,0,0)B. (1,1,1)C. (-1,-,1,-1)D. (1,2,3) |
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Answer» Correct Answer - C Given lines are `(x-1)/(2)=(y-2)/(3)=(z-3)/(4)=r["say"]......(i)` `and (x-4)/(5)=(y-1)/(2)=z.....(ii)` Any point on the line (i) is (2r+1, 3r+2, 4r+3) If they interest, then the point satisfies the second line we get `(2r+1-4)/(5)=(3r+2-1)/(2)=4r+3` `Rightarrow (2r-3)/(5)=(3r-1)/(2) Rightarrow r=-1` `therefore` Required point is [(2(-1)+1,3(-1)+2,4(-1)+3] `=(-2+1,-3+2,-4+3)` =(-1,-1,-1) |
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| 100. |
The equation of line is ` 2x - 2 = 3y + 1 = 6z - 2 ` find its direction ratios and also find the vector equation of the line .A. 1,2,3B. 3,2,1C. 6,3,2D. 2,3,6 |
| Answer» Correct Answer - B | |