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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.
| 101. |
The shortest distance between the skew lines `bar(r) = bar(a_(1)) + lambda bar(b_(1)) "and" bar(r) = bar(a_(2)) + mu bar(b_(2))` isA. `d=|((b_(1)-b_(2)).(a_(1)xxa_2))/(a_(1)xxa_(2))|`B. `d=|((b_(1).b_(2)).(a_(1)-a_2))/(|b_(1)||b_2|)|`C. `d=|((b_(1)xxb_(2)).(a_(2)-a_1))/(|b_(1)|xx|b_2|)|`D. `d=|((b_(1)xxb_(2)).(a_(2)xxa_1))/(|b_(1)xxb_2||a_1xxb_2)|` |
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Answer» Correct Answer - C `therefore ` The shortest distance between two skew lines is `d=|((b_(1) xxb_(2)).(a_(1)xxa_(1)))/(|b_(1) xx b_(2)|)|` |
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| 102. |
The point of intersection of the lines ` ( x + 1 )/( 3 ) = ( y + 3 ) /( 5 ) = ( z+ 5 ) /( 7 ) and ( x - 2 )/( 1) = ( y - 4 ) /( 3 ) = ( z - 6 )/ ( 5 ) ` isA. ` ( ( 1 )/ ( 2), ( 1 ) /(2), - ( 3 ) / ( 2)) `B. ` ( - ( 1 ) / ( 2 ), - ( 1 ) / ( 2 ), ( 3 ) / ( 2 )) `C. ` (( 1 )/ ( 2 ), - (1)/ ( 2 ) , - ( 3 ) / ( 2)) `D. ` ( - ( 1 ) / ( 2) , ( 1 ) / ( 2 ), ( 3 ) / ( 2) ) ` |
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Answer» Correct Answer - C Let ` ( x + 1 ) /( 3 ) = ( y + 3 )/ ( 5 ) = ( z + 5 ) / ( 7 ) = lamda " "`…(i) Any point on the line is ` ( 3lamda - 1 , 5lamda - 3, 7 lamda - 5 ) ` Again, let ` ( x - 2 ) / ( 1 ) = ( y- 4 ) / ( 3) = ( z- 6 ) / ( 5) = mu " " `...(ii) Any point on the line is ` ( mu + 2, 3mu + 4, 5 mu + 6 ) ` For intersection , they have a common point ` therefore ( 3 lamda - 1 ) = ( mu + 2 ) `, ` ( 5lamda - 3 ) = ( 3 mu + 4 ) `, and ` ( 7 lamda - 5 ) = ( 5mu + 6 ) ` From first two , we have ` mu = 3 lamda -3 " " `...(iii) and ` 3 mu = 5lamda - 7 " " `...(iv) From Eqs (iii) and (iv) ,we have ` 3 ( 3 lamda - 3 ) = 5 lamda - 7 rArr lamda = ( 1 ) / ( 2 ) ` Point of intersection is ` (( 3 ) / (2 ) - 1, (5 ) / ( 2 )- 3, ( 7 )/ ( 2 ) - 5 ) = (( 1) / ( 2 ) , - ( 1 ) / (2 ) , - ( 3 ) / (2 )) ` |
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| 103. |
The vector equation of a line passing through the point (5,4,3) and having directions ratios `-3,4,2` isA. `-(5hati - 4hatj - 3hatk) +lambda(-3hati + 4hatj + 2hatk)`B. `(5hati + 4hatj + 3hatk) + lambda(-3hati + 4hatj + 2hatk)`C. `(-3hati + 4hatj + 2hatk) + lambda(-5hati - 4hatj - 3hatk)`D. `(-3hati + 4hatj + 2hatk) + lambda(-5hati - 4hatj - 3hatk)` |
| Answer» Correct Answer - B | |
| 104. |
If the lines `x=a_(1)y + b_(1), z=c_(1)y +d_(1)` and `x=a_(2)y +b_(2), z=c_(2)y + d_(2)` are perpendicular, thenA. `a_(1)a_(2) + b_(1)b_(2) +1=0`B. `a_(1)a_(2) +c_(2)c_(2)+1=0`C. `a_(1)a_(2) + b_(1)b_(2)=0`D. `a_(1)c_(2) + b_(1)c_(1)=0` |
| Answer» Correct Answer - B | |
| 105. |
Find the coordinates of the points on the line `(x +1)/(2) = (y - 2)/(3) = (z + 3)/(6)`, which are at a distance of 3 units from the point ( -1 , 2 , -3).A. `(1/7,-23/7, 3/7), (13/7, -5/7, 39/7)`B. `(-1/7, 23/7, -3/7), (13/7, -5/7, 39/7)`C. `(-1/7, 23/7, -3/7), (-13/7, 5/7, -39/7)`D. `(1/7, -23/7, 3/7), (-13/7, 5/7, -39/7)` |
| Answer» Correct Answer - C | |
| 106. |
The equation of a line passing through the point (4,-2,5) and parallel to the vector `3hati -hatj + 2hatk` isA. `(x-4)/3 = (y+2)/-1 = (z-5)/2`B. `(x+4)/3 = (y-2)/-1 = (z+5)/2`C. `(x-3)/4 = (y+1)/-2 = (z-2)/5`D. `(x+3)/4 = (y-1)/-2 = (z+2)/5` |
| Answer» Correct Answer - A | |
| 107. |
बिंदुओं `(-1, 0, 2 )` और `(3, 4, 6) ` से होकर जाने वाली रेखा का सदिश समीकरण ज्ञात कीजिए |A. ` - hati + 2hatk + lamda ( 4 hati + 4 hatj + 4 hatk ) `B. ` 2 hati - hatk + lamda (4 hati + 4hatj + 4 hatk ) `C. ` - hati + 2 hatk + lamda ( - hati + 4hatj + 2hatk ) `D. ` - hati + 2 hatk + lamda ( 3 hati + 4hatj + 6 hatk ) ` |
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Answer» Correct Answer - A Let a and b be the position vectors of the point ` A ( - 1 , 0 , 2 ) and B ( 3, 4 , 6 ) ` Then, ` a= - hati + 2 hatk and b = 3 hati + 4hatj + 6 hatk ` Let r be the position vector of any point on the line. Then the vector equation of the line is ` r = - hati + 2 hatk + lamda ( 4hati + 4hatj + 4 hatk ) ` |
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| 108. |
A line passes through two points `A(2,-3,-1)` and `B(8,-1,2)`. The coordinates of a point on this lie at distance of 14 units from a areA. (14,1,5), (-10,7,-7)B. (14,1,5), (10,-7,-7)C. (14,1,5), (-10,-7,7)D. (14,1,5), (-10, -7,7) |
| Answer» Correct Answer - D | |
| 109. |
If the line passing through the points (5,1,p) and (3,q,1) crosses the YZ plane at the point `(0,17/2, -13/2)`, thenA. p=6, q=4B. `p=-6, q=4`C. `p=6, q=-4`D. `p=-6, q=-4` |
| Answer» Correct Answer - A | |
| 110. |
The cartesian equation of a line passing through the point (3,2,1) and is parallel to the vector `2hati + 2hatj - 3hatk` isA. `(x+3)/2 =(y+2)/2 = (z+1)/-3`B. `(x-3)/ 2 = (y-2)/2 = (z-1)/-3`C. `(x+2)/-3 = (y+2)/2 = (z-3)/1`D. `(x-2)/3 = (y-2)/2 =(z+3)/1` |
| Answer» Correct Answer - B | |
| 111. |
बिंदु `(1,2,-4)` से जाने वाली और दोनों रेखाओ `(x-8)/(3)=(y+19)/(-16)=(z-10)/(7) ` और`(x-15)/(3)=(y-29)/(8)=(z-5)/(-5)`पर लंब रेखा का सदिश समीकरण ज्ञात कीजिए lA. ` r = ( hati + 2 hatj - 4 hatk ) + lamda ( 2 hati + 3 hat j + 6 hatk ) `B. ` r = ( 2 hati + 3 hatj - 6 hatk ) + lamda ( hati + 2hatj - 4hatk ) `C. ` r = ( hati + 2 hatj - 4hatk ) + lamda ( 3hati + 8 hatj - 5hatk ) `D. ` r = ( hati + 2hatj - 4hatk ) + lamda ( 3 hati - 16 hatj - 7 hatk ) ` |
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Answer» Correct Answer - A Any line through ` ( 1, 2, - 4) ` can be written as ` ( x - 1 ) / ( a ) = ( y -2 ) / ( b ) = ( z+ 4 ) / ( c) " " `…(i) where a , b, c are the direction ratios of line (i). Now the line (i) be perpendicular to the lines ` ( x- 8 ) /(3 ) = ( y + 19) / ( -16 ) =( z - 10 ) /( 7 ) ` and ` ( x - 15 ) / ( 3 ) = ( y - 29) / ( 8 ) = ( z - 5 ) / ( - 5 ) ` ` therefore 3a - 16 b + 7 c = 0 " " `...(ii) and ` 3a +8 b - 5 c = 0 " " `...(iii) By cross - multiplication , we have ` ( a ) / ( 80 - 56 ) = ( b ) /( 21 + 15 ) = ( c ) /( 24 - 48 ) ` ` rArr ( a ) /( 24 ) = ( b ) /( 36 ) = ( c ) /( 72) ` ` rArr (a )/ ( 2 ) = ( b ) / ( 3 ) = (c ) / ( 6) = lamda " " ( say) ` ` therefore a = 2 lamda , b = 3lamda and c = 6lamda ` The equation of required line which passes through the point ` ( 1 , 2, -4 ) ` and perpendicular to vector ` 2 hati + 3 hatj + 6hatk ` is ` r = ( hati + 2 hatj - 4hatk ) + lamda ( 2 hati + 3 hatj + 6 hatk ) ` |
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| 112. |
Find the equation of line equally inclined to co-ordinate axes and passes through (-5, 1.-2).A. `x-5 = y +1 =z-2`B. `sqrt(3)(x+5) = y-1= sqrt(3)(z+2)`C. `x+5 = 1-y = z+2`D. `x+5 = y-1 = z+2` |
| Answer» Correct Answer - D | |
| 113. |
If x co-cordinate of a point on the line passing through the points (2,2,1) and (5,1,-2) is 2, then the y co-cordinate of the point isA. `-3`B. 3C. 2D. `-2` |
| Answer» Correct Answer - C | |
| 114. |
The cartesian equation of a line passing through the point (3,2,1) and is parallel to the vector `2hati + 2hatj - 3hatk` isA. `(2hati + 2hatj - 3hatk) + lambda(3hati + 2hatj +hatk)`B. `(3hati + 2hatj +hatk) + lambda(2hati + 2hatj - 3hatk)`C. `(2hati + 2hatj - 3hatk) + lambda(-hati - 4hatk)`D. `(3hati + 2hatj + hatk) + lambda(-hati - 4hatk)` |
| Answer» Correct Answer - B | |
| 115. |
The shortest distance between the lines ` r = ( - hati - hatj - hatk ) + lamda ( 7 hati - 6 hatj + hatk ) and r = ( 3 hati + 5 hatj + 7 hatk ) + mu ( hati - 2 hatj + hatk ) `A. ` sqrt ( 29 ) ` unitsB. 29 unitsC. ` ( 2 9 ) / ( 2 ) ` unitsD. ` 2 sqrt ( 29 ) ` units |
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Answer» Correct Answer - D The given lines are ` r = - hati - hatj - hatk+ lamda ( 7 hati - 6hatj + hatk ) ` and ` r = ( 3 hati + 5 hatj + 7 hatk ) + mu ( hati - 2hatj + hatk ) ` where, ` a _ 1 = - hati - hatj - hatk , b _ 1 = 7 hati - 6 hatj + hatk ` and ` a_ 2 = 3hati + 5 hatj + 7 hatk , b _ 2 = hati - 2 hatj + hatk ` Now, ` a _ 2- a _ 1 = ( 3 hati + 5hatj + 7 hatk ) - ( - hati - hatj - hatk ) ` ` = 4 hati + 6hatj + 8 hatk ` and ` b _ 1 xx b _ 2 = |{:( hati , hatj , hatk ) , ( 7, - 6, 1 ) , ( 1, -2, 1 ) :}| ` ` = hati ( - 6 + 2 ) - hatj ( 7 - 1 ) + hatk ( -14 + 6 ) ` ` = - 4hati - 6 hatj - 8 hatk ` ` rArr | b _ 1 xx b _ 2 | = sqrt(( - 4 ) ^ 2 + ( - 6 ) ^ 2 + ( -8 ) ^ 2 ) ` ` = sqrt ( 16 + 36 + 64 ) = sqrt ( 116 ) ` ` therefore ` Shortest distance between the given lines ` d= | (( b _ 1 xx b _ 2 ) * ( a _ 2 - a _ 1 ) ) / ( |b _ 1 xx b _ 2 | ) | ` ` (| ( - 4hati - 6 hatj - 8 hatk ) * ( 4hati + 6 hatj + 8 hatk )| ) / ( sqrt ( 116 ) ) ` ` = (| ( -4 ) xx 4 + ( - 6 ) xx 6 + ( - 8 ) xx 8 | ) /( sqrt ( 116 )) ` ` = (116 ) / ( sqrt (116 )) = sqrt (116) = 2 sqrt (29 ) ` units |
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| 116. |
Find the angle between the following pair of lines: A lines with direction ratios 2,2,1 A line joning (3,1,4)to (7,2,12)A. `cos^(-1)(1/3)`B. `cos^(-1)(2/3)`C. `cos^(-1)(3/3)`D. `cos^(-1)(4/3)` |
| Answer» Correct Answer - B | |
| 117. |
Equation of the line passing through the point (0,1,2) and equally inclined to the coordintate axes, areA. x=y-1=z-2B. x=y+1=z+2C. `x/0=y/3=z/2`D. None of these |
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Answer» Correct Answer - A Since, the line is equally inclined, therefore `l-m=n=cos theta` Now, equation of a line passing through the point (0,1,2) and having direction cosine `(cos theta, cos theta, cos theta)` is `(x-0)/(cos theta)=(y-1)/(cos theta)=(z-2)/(cos theta)` `Rightarrow x=y-1=z-2` |
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| 118. |
Find the vector equation of a line parallel to the vector `2hati-hatj+2hatk` and passing through a point A with position vector `3hati+hatj-hatk`.A. `(3hati + hatj -hatk) + lambda(2hati - hati + 2hatk)`B. `(2hati - hatj + 2hatk) +lambda(3hati + hatj- hatk)`C. `(hati + 2hatj - 3hatk) + lambda(2hati - hatj + 2hatk)`D. `(2hati - hatj + 2hatk) + lambda(hati + 2hatj - 3hatk)` |
| Answer» Correct Answer - A | |
| 119. |
The equation of a line passing through the points (a,b,c) and(a-b,b-c, c-a) isA. `(x-a)/a = (y-b)/b = (z-c)/c`B. `(x-a)/b = (y-b)/c = (z-c)/a`C. `(x-a)/(a-b) = (y-b)/(b-c) = (z-c)/(c-a)`D. `(x-a)/(2a-b) = (y-b)/(2b-c) = (z-c)/(2c-a)` |
| Answer» Correct Answer - B | |
| 120. |
Find the vector and the cartesian equations of the line that passes through the points`(3, 2, 5), (3, 2, 6)`.A. ` r = 3 hati - 2 hatj - 5 hatk , x - 3 = y + 2 = ( z + 5 ) / ( 11 ) `B. ` r = 3 hati - 2 hatj - 5 hatk , ( x - 3 ) / ( 0 ) = ( y + 2 ) /( 0 ) = ( z + 5 )/ ( 11 ) `C. ` r = 3hati - 2 hatj - 5 hatk + lamda ( 11 hatk ) , x - 3 = y + 2 = ( z + 5 ) / (11 ) `D. ` r = 3 hati - 2 hatj - 5 hatk + lamda ( 11 hatk ) , ( x -3 ) /( 0 ) = ( y + 2 ) /( 0 ) = ( z + 5 ) /( 0 ) ` |
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Answer» Correct Answer - D We have, ` a= 3hati - 2 hatj - 5hatk ` and ` b = 3hati - 2 hatj + 6 hatk ` we know that, the vector equation of a line passing through the point having position vectors a and b is ` r = 3 hati - 2hatj - 5 hatk + lamda [ ( 3 hati - 2 hatj + 6 hatk ) - ( 3 hati - 2hatj - 5 hatk ) ] ` ` rArr r = 3hati - 2 hatj - 5hatk + lamda ( 11 hatk ) " " ...(i)` [which is the vector equation] Putting ` r = x hati + y hatj + z` hatk , in Eq. (i) , we have ` x hati + y hatj + z hatk = 3 hati - 2 hatj + (11 lamda - 5 ) hatk ` On comparing coefficients of ` hati , hatj and hatk ` on both sides, we get ` x= 3, y = - 2 ` and ` z = 11 lamda - 5 ` ` therefore (x - 3 ) / ( 0 ) = ( y + 2 ) / 0 = ( z + 5 ) / ( 11 ) ` which is the cartesian form of required line. |
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| 121. |
The length of the perpendicular from P(1,6,3) to the line `x/1=(y-1)/(2)=(z-2)/(3) is `A. ` 3 `B. ` sqrt ( 11 ) `C. ` sqrt ( 13 ) `D. ` 5 ` |
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Answer» Correct Answer - C Given ` (x ) / ( 1 ) = ( y - 1 ) / ( 2 ) = ( z - 2 ) / ( 3 ) = lamda ` Any point on the line is ` P ( lamda , 2 lamda + 1 , 3 lamda + 2 ) ` Therefore, direction ratios of PQ are ` ( lamda - 1 , 2 lamda - 5, 3lamda - 1 ) ` ` because PQ ` is perpendicular to the given line Therefore, ` 1 ( lamda - 1 ) + 2 ( 2lamda - 5 ) + 3 ( 3lamda - 1 ) = 0 ` ` " "[ because a - 1 a _ 2 + b _ 1 b _ 2 + c _ 1 c _ 2 = 0 ] ` ` rArr lamda = 1 ` ` therefore ` The coordinates of P are ` ( 1, 3, 5 )` ` therefore ` Length of perpendicular , QP ` = sqrt (( 1 - 1) ^ 2 + (3 - 6 ) ^ 2 + ( 5 - 3) ^ 2) = sqrt ( 13 ) ` |
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| 122. |
A line `L_(1)` passes through the point `3hati` and is parallel to the vector `-hati+hatj+hatk` and another line `L_(2)` passes through `hati+hatj` and is parallel to the vector `hati+hatk`, then point of intersection of the lines issA. `hati+2hatj+hatk`B. `2hati+hatj+hatk`C. `hati-2hatj-hatk`D. `hati-2hatj+hatk` |
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Answer» Correct Answer - B Equation of `L_(1)` is `r=3hati+lambda(-hati+hatj+hatk)=(3-lambda)hati+lambdahatj+lambdahatk` Equation of `L_(2)` is `r=(hati+hatj)+mu(hati+hatk)=(1+mu)hati+hatj+muhatk` For point of intersection, we get `lambda=mu=1` `therefore r=2hati+hatj+hatk` |
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| 123. |
Find the cartesian equation of the line which passes through the point `(-2,4,-5)` and parallel and line are `(3,5,6)`. So, the equation of line is, `(x-(-2))/(3) = (y-4)/(5) = (z-(-5))/(6)`.A. `(x-2)/3 = (y+4) = (z-5)/6`B. `(x+2)/3 = (y-4)/5 = (z+5)/6`C. `(x-2)/3 = (y+3)/5 = (z+5)/6`D. `(x+2)/3 = (y+4)/5 = (z+5)/6` |
| Answer» Correct Answer - B | |
| 124. |
Equation of the line passing through the point ( 2,3, -4) and perpendicular to XZ-plane isA. `(x-2)/0 = (y-3)/1 =(z+1)/0`B. `(x-2)/1 = (y-3)/0 = (z+1)/0`C. `(x-2)/0 = (y-3)/0 = (z+1)/1`D. `(x-2)/1 = (y-3)/1 = (z+1)/1` |
| Answer» Correct Answer - B | |
| 125. |
The equation of a line which passes through the point ` ( 1, 2 , 3 ) ` and is parallel to the vector ` 3 hati + 2 hatj - 2 hatk ` , isA. ` r = ( 3 hati + 2hatj - 2 hatk ) + lamda ( hati + 2 hatj + 3 hatk ) `B. ` r = ( hati + 2 hatj + 3 hatk ) + lamda ( 3 hati + 2 hatj - 2 hatk ) `C. ` r = ( hati + 2 hatj + 3hatk ) + lamda ( 2 hati - 5hatk ) `D. ` r = ( hati + 2hatj + 3 hatk ) + lamda ( 4 hati + 4 hatj + hatk ) ` |
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Answer» Correct Answer - B It is given that the line passes through the point ` A ( 1, 2, 3 ) ` The position vector of A is a = ` hati + 2 hatj + 3hatk ` and parallel to a vector ` b = 3hati + 2hatj - 2 hatk ` The required equation of the line is ` r = hati + 2 hatj + 3 hatk + lamda ( 3hati + 2 hatj - 2 hatk ) ` |
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| 126. |
The acute angle between the line joining the points A(2,1,-3), B(1,-1,2) and the line `vecr = (2hati - hatk) + lambda(3hati + hatj)` isA. `cos^(-1)(1/(2sqrt(3)))`B. `cos^(-1)(1/sqrt(3))`C. `cos^(-1)(1/(5sqrt(10)))`D. `cos^(-1)(3/(5sqrt(10)))` |
| Answer» Correct Answer - A | |
| 127. |
The equation of a line passing through the point `(3,-2,1)` with the direction angle is `135^(@), 60^(@), 120^(@)` isA. `(x-3)/sqrt(2) = (y+2)/-1 = (z-1)/1`B. `(x-3)/-1 = (y+2)/sqrt(2) = (z-1)/sqrt(2)`C. `(x+3)/-sqrt(2) = (y+2)/1 = (z-1)/1`D. `(x+3)/1 = (y+2)/1 = (z-1)/sqrt(2)` |
| Answer» Correct Answer - A | |
| 128. |
Find the vector equation of the line passing through the point ` 2 hati + hatj - 3 hatk ` and perpendicular to the vectors ` hati + hatj + hatk and hati + 2 hatj - hatk`.A. `(2hati + hatj - 3hatk) + lambda(-3hati + 2hatj - hatk)`B. `(2hati + hatj - 3hatk) + lambda(3hati + 2hatj -hatk)`C. `(2hati + hatj - 3hatk) + lambda(-3hati - 2hatj + hatk)`D. `(2hati + hatj - 3hatk) + lambda(-3hati + 2hatj + hatk)` |
| Answer» Correct Answer - D | |
| 129. |
The angle between the lines `r = (2hati + hatj - 3hatk) + lambda(hati - hatj + hatk), (x-1)/1 = (y+2)/3 = (z+3)/2` isA. `pi/6`B. `pi/4`C. `pi/3`D. `pi/2` |
| Answer» Correct Answer - D | |
| 130. |
The vector equation of a line passing through the point `hati + 2hatj + 3hatk` and perpendicular to the vectors `hati +hatj + hatk` and `2hati - hatj + hatk` isA. `(hati + 2hatj + 3hatk) + lambda(-2hati - hatj + 3hatk)`B. `(hati + 2hatj + 3hatk) + lambda(2hati + hatj - hatk)`C. `(hati + 2hatj + 3hatk) +lambda(2hati + hatj - hatk)`D. `(hati + 2hatj + 3hatk) + lambda(2hati - hatj - 2hatk)` |
| Answer» Correct Answer - B | |
| 131. |
Find the shortest distance between the lines `vecr = hati+ hatj+hatk+lambda(3hati-hatj) and vecr=4hati-hatk+mu(2hati+3hatk)`A. (4,0,1)B. (4,0,-1)C. (-4,0,1)D. (-4,0,-1) |
| Answer» Correct Answer - B | |
| 132. |
The shortest distance between the lines `vecr = (-hati - hatj) + lambda(2hati - hatk)` and `vecr = (2hati - hatj) + mu(hati + hatj -hatk)` isA. `1/sqrt(7)`B. `1/sqrt(14)`C. `1/(2sqrt(7))`D. `1/(7sqrt(2))` |
| Answer» Correct Answer - B | |
| 133. |
The shortest distance between the lines `vecr = (2hati - hatj) + lambda(2hati + hatj - 3hatk)` `vecr = (hati - hatj + 2hatk) + lambda(2hati + hatj - 5hatk)`A. `1/(2sqrt(5))`B. `1/(4sqrt(5))`C. `1/sqrt(5)`D. `2/sqrt(5)` |
| Answer» Correct Answer - C | |
| 134. |
Find the shortest distance between the lines `vecr=(hatii+2hatj+hatk)+lamda(2hati+hatj=2hatk) and vecr=2hati-hatj-hatk+mu(2hati+hatj+2hatk)`.A. `3/sqrt(2)`B. `2/sqrt(3)`C. `1/(3sqrt(2))`D. `1/(2sqrt(3))` |
| Answer» Correct Answer - A | |
| 135. |
Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then. The value of `sin^(-1)sinlamda` is equal toA. 3B. `pi-3`C. 4D. `pi-4` |
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Answer» Correct Answer - D Given lines are `(x-3)/(2)=(y-2)/(3)=(z-1)/(lambda).....(i)` `and (x-2)/(3)=(y-3)/(2)=(z-2)/(3)......(ii)` These lines lie in the same plane. So, both lines are coplanar. `therefore |{:(,3-2,2-3,1-2),(,2,3,lambda),(,3,2,3):}|=0` `Rightarrow |{:(,1,-1,-1),(,2,3,lambda),(,3,2,3):}|=0` `Rightarrow 2(-3+2)-3(3+3)+lambda(2-3)=0` `Rightarrow -2-18+5lambda=0 Rightarrow 5lambda=20` `Rightarrow lambda=4` `therefore sin^(-1) sin lambda=sin^(-1)sin4=sin^(-1) sin(pi-4)=pi-4` |
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| 136. |
Distance of the point `P(x_2, y_2, z_2)` from the line `(x-x_1)/l=(y-y_1)/m=(z-z_1)/n`, where `l,m,n` are the direction cosines of the line, isA. `sqrt((x_(2)-x_(1))l+ (y_(2)-y_(1))m + (z_(2)-z_(1))n)`B. `sqrt((x_(2)-x_(1))^(2)l + (y_(2)-y_(1))^(2)m + (z_(2)-z_(1))^(2)n)`C. `sqrt((x_(2)-x_(1))^(2) + (y_(2)-y_(1))^(2) + (z_(2)-z_(1))^(2))`D. `sqrt((x_(1)-x_(2))^(2) + (y_(1)-y_(2))^(2) + (z_(1)-z_(2))^(2) -l(x_(1)-x_(2)) + m(y_(1)-y_(2)) + n(z_(1)-z_(2))^(2)` |
| Answer» Correct Answer - D | |