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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. |
The equaton of the line in vector and cartesian from that passes through the point with position vector ` 2 hati - hatj + 4hatk ` and is in the direction ` hati + 2 hatj - hatk ` areA. ` r = ( 2hati - hatj + 4hatk ) + lamda ( hati + 2 hatj - hatk ) , ( x - 1 ) / ( 2 ) = ( y - 2 ) / ( - 1 ) = ( z + 1 ) / ( - 4 ) `B. ` r = ( hati + 2 hatj - hatk ) + lamda ( 2 hati - hatj + 4 hatk ) , ( x - 1 ) / ( 2 ) = ( y - 2 ) / ( - 1 ) = ( z + 1 ) / (4 ) `C. ` r = ( 2hati - hatj + 4 hatk ) + lamda ( hati + 2 hatj - hatk ), ( x - 2 ) / ( 1 ) = ( y + 1 ) / ( 2 ) = ( z - 4 ) / ( - 1 ) `D. |
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Answer» Correct Answer - D The equation of the line in vector from ` therefore r = 2 hati - hatj + 4hatk + lamda ( hati + 2 hatj - hatk ) ` Now, put ` r = x hati + y hati + z hatk ` ` therefore x hati + y hatj + z hatk = ( 2 + lamda ) hati + ( - 1 + 2 lamda ) hatj + ( 4 + lamda ) hatk ` ` rArr x = 2 + lamda, = - 12lamda , 2 = 4 - lamda ` ` rArr ( x - 2 ) / ( 1 ) = ( y+ 1 ) /( 2 ) = ( z - 4 ) / ( - 1 ) ` which the required equation of the given line in cartesian form. |
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| 2. |
Find the length of the perpendicular from ( 3,2,1) to the line `(x - 7) /(-2) = (y - 7)/(2) = (z - 6 ) /(3)`.A. (9,5,)B. (9,-5,3)C. (9,5,-3)D. (-9,5,3) |
| Answer» Correct Answer - A | |
| 3. |
The cartesian equation of a line passing through the point having position vector `2hati +hatj - hatk` and parallel to the line passing joining the points `-hati + hatj + 4hatk` and `hati + 2hatj + 2hatk`, isA. `(x-2)/2 = (y-1)/1 = (z+1)/-2`B. `(x+2)/2 = (y+1)/1 = (z-1)/-2`C. `(x-2)/-2 = (y-1)/1 = (z+1)/2`D. `(x+2)/-2 = (y+1)/1 = (z-2)/2` |
| Answer» Correct Answer - A | |
| 4. |
Find the length of the perpendicular from ( 3,2,1) to the line `(x - 7) /(-2) = (y - 7)/(2) = (z - 6 ) /(3)`.A. 3B. 2C. 7D. 6 |
| Answer» Correct Answer - C | |
| 5. |
Find the length of the perpendicular from ( 3,2,1) to the line `(x - 7) /(-2) = (y - 7)/(2) = (z - 6 ) /(3)`.A. `sqrt(531/14)`B. `sqrt(531/98)`C. `sqrt(1779/98)`D. `sqrt(1779/49)` |
| Answer» Correct Answer - A | |
| 6. |
The equation of a line passing through (a, b, c) and parallel tp z-axis isA. `(x-a)/l - (y-b)/m = (z-c)/0`B. `(x-a)/0 = (y-b)/0 =(z-c)/n`C. `(x-a)/l = (y-b)/m = (z-c)/1`D. `(x-a)/0 = (y-b)/m = (z-c)/0` |
| Answer» Correct Answer - A | |
| 7. |
The foot of the perpendicular drawn from point (2,-3,1) to the line `(x+1)/2 = (y-3)/3 = (z+2)/-1` isA. `(-22/14, -3/14, -13/14)`B. `(-22/7, -3/14, -13/4)`C. `(22/14, 3/14, 13/14)`D. `(22/7, 3/14, 13/14)` |
| Answer» Correct Answer - B | |
| 8. |
The equation of a line passing through point (2,3,-1) and perpendicular to XY 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 - C | |
| 9. |
Find the equation of the perpendicular from point `(3,-1,11)` to line `x/2=(y-2)/3=(z-3)/4`. Also, find the coordinates of foot of perpendicular and the length of perpendicular.A. `(x+1)/1 = (y-3)/-6 = (z+11)/3`B. `(x-3)/2 = (y-1)/-6 = (z+11)/3`C. `(x-3)/1 = (y+1)/-6 = (z-11)/4`D. `(x-3)/2 = (y+1)/6 = (z-11)/7` |
| Answer» Correct Answer - C | |
| 10. |
Find the equation of the line passing through the point (3,1,2) and perpendicular to the lines `(x-1)/1=(y-2)/2=(z-3)/3` and `x/(-3)=y/2=z/5`A. `(x-3)/2 = (y-1)/-7 = (z-2)/2`B. `(x+3)/2 = (y+1)/7 = (z-2)/2`C. `(x-3)/2 = (y-1)/-7 = (z-2)/4`D. `(x+3)/2 = (y+1)/-7 = (z+2)/4` |
| Answer» Correct Answer - C | |
| 11. |
Find the equation of the line passing through the point (3,1,2) and perpendicular to the lines `(x-1)/1=(y-2)/2=(z-3)/3` and `x/(-3)=y/2=z/5`A. `(x-2)/2 = (y-1)/-7 = (z-3)/4`B. `(x+2)/2 = (y+1)/-7 = (z+3)/4`C. `(x-2)/2 = (y-1)/-7 = (z-3)/2`D. `(x+2)/2 = (y+1)/-7 = (z+3)/2` |
| Answer» Correct Answer - A | |
| 12. |
Find the perpendicular distasnce of the point (1,0,0) from the lines (x-1)/2=(y+1)/(-3)=(z+10)/8`A. `4sqrt(3)`B. `3sqrt(6)`C. `6sqrt(2)`D. `2sqrt(6)` |
| Answer» Correct Answer - D | |
| 13. |
If OA=a, OB=b, OC=c are the co-terminus edges of regular parallelopiped, then the shortest distance between the diagonal and the side OB not meeting the diagonal isA. `(bc)/sqrt(b^(2) + c^(2))`B. `(ca)/sqrt(c^(2) + a^(2))`C. `(ab)/sqrt(a^(2) + b^(2))`D. `(abc)/sqrt(a^(2) + b^(2) + c^(2))` |
| Answer» Correct Answer - B | |
| 14. |
If `l_(1), m_(1), n_(1)` and `l_(2),m_(2),n_(2)` are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are `m_(1)n_(2)-m_(2)n_(1),n_(1)l_(2)-n_(2)l_(1),l_(1)m_(2)-l_(2)m_(1)`.A. `m_(1)n_(2)+m_(2)n_(1).n_(1)l_(2)+n_(2)l_(1).l_(1)m_(2)+l_(2)m_(1)`B. `m_(1)n_(2)-m_(2)n_(1).n_(1)l_(2)-n_(2)l_(1),l_(1)m_(2)-l_(2)m_(1)`C. `m_(1)m_(2)-n_(1)n_(2).n_(1)n_(2)-l_(1)l_2,l_(1)l_(2)-m_(1)m_(2)`D. `m_(1)m_(2)+n_(1)n_(2).n_(1)n_(2)-l_(1)l_2,l_(1)l_(2)-m_(1)m_(2)` |
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Answer» Correct Answer - B Given lines are represented parallel to unit vectors `b_1=i_(1)hati+m_(1)hatj+n_(1)hatk....(i)` `and b_2=i_(2)hati+m_(2)hatj+n_(2)hatk....(ii)` Now, `b_(1) xx b_(2)` is a vector which is at right angles to both `b_1 and b_2` and is of magnitude unity. Henec, components of `b_(1) xx b_(2)` are directions cosines of a line which is at right angle to both `b_1 and b_2` `=(m_1n_2-m_2n_1)hati+(n_(1)l_(2)-n_(2)l_(1))hatj+(l_(1)m_(2)-l_(2)m_(1))hatk` Thus, the direction cosiness of the required line are `m_(1)n_(2)-m_(2)n_(1), n_(1)l_(2)-n_(2)l_(1),m_(2)-l_(2)m_(1)` |
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| 15. |
Shortest distance between line `2x+3y+4z-4=0=x+y+2z-3` and z-axis is -A. 1B. 2C. 4D. 3 |
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Answer» Correct Answer - B We have `x+y+2z-3=0` and `2x+3y+4z-4=0` Let a,b,c be the direction ratios of the line, then line lies on the both planes. `therefore a+b+2c=0 and 2a+3b+4c=0` On solving these equations by cross-multiplication method. `a/(4-6)=(b)/(4-4)=(c)/(3-2) Rightarrow (a)/(-2)=(b)/(0)=c/1` `therefore` Direction ratios of the line is (-2,0,1) and direction ratios of Z-axis (0,0,1) Hence, distance between Z-axis to the given line is `sqrt((-2-0)(2)+(0)^(2)+(1-1)^(2))=sqrt4=2` |
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| 16. |
The vector equation of the line `(x+2)/(3)=(1-y)/(-2)=(z-5)/(7) ` isA. `r=(3hati-2hatj+7hatk)+lambda(-2hati+hatj+5hatk)`B. `r=(-2hati+hatj+5hatk)+lambda(3hati-2hatj+7hatk)`C. `r=(3hati-2hatj+7hatk)+lambda(2hati+hatj-5hatk)`D. `r=(2hati+hatj+5hatk)+lambda(3hati+2hatj+7hatk)` |
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Answer» Correct Answer - B The direction ratios of the given line are (3,-2,7). `therefore` The given line is parallel to `3hati-2hatj+7hatk` From the cartesian equation of the given line, it is clear that is passes through the points (-2,1,5). `therefore` The vector equation of line is `r=(-2hati+hatj+5hatk)+lambda(3hati-2hatj+7hatk)` |
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| 17. |
Find the image of the point `(0,2,3)` in the line `(x+3)/(5)= (y-1)/(2) = (z+4)/(3)`.A. 21B. 42C. `sqrt(21)`D. `sqrt(42)` |
| Answer» Correct Answer - C | |
| 18. |
Find the equation of the perpendicular drawn from (2,4,-1) to the line `(x+5)/1=(y+3)/4=(z-6)/3`.A. ` ( x - 2 ) / ( 6 ) = ( y - 4 ) /( 3 ) = ( z+ 1 ) / ( 2 ) `B. ` ( x + 2 ) / ( 6 ) = ( y - 4 ) /( 3 ) = ( z + 1 ) /( 2 ) `C. ` ( x + 2 ) /( - 6 ) = ( y - 4 ) / ( 3) = ( z + 1 ) / ( 2 ) `D. ` ( x + 2 ) /( 6 ) = ( y + 4 ) / ( 3) = ( z+ 1 ) / ( 2 ) ` |
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Answer» Correct Answer - A Given equation of line in ` ( x + 5 ) / ( 1 ) = ( y+ 3 ) / ( 4 ) = ( z- 6 ) /( - 9 ) = lamda ` [say] Any point on the line is ` ( lamda - 5, 4lamda - 3, - 9 lamda + 6 ) ` Now, directions ratio of PQ are ` ( lamda - 5 - 2, 4lamda - 3 - 4, - 9 lamda + 6 + 1 ) ` i.e, ` ( lamda - 7, 4lamda - 7, 9lamda + 7 ) ` Since, PQ is perpendicular to the given line, ` therefore 1 ( lamda - 5) + 4 ( 4lamda - 7 ) - 9 ( - 9 lamda + 7 ) = 0 ` ` " "[ because a_ 1 a _ 2 + b _ 1 b _ 2 + c _ 1 c_ 2 = 0 ] ` ` rArr 9 lamda - 96 = 0 rArr lamda = 1 ` ` therefore ` Foot of perpendicular ` Q ( 1 - 5, 4 - 3 , - 9 + 6 ) ` i.e, `Q ( - 4, 1, - 3 ) ` The equation of the line passing through ` P ( 2, 4 , -1 ) and N ( -4, 1, - 3 ) ` is ` ( x - 2 ) / ( 6 ) = ( y - 4 ) /( 3 ) = ( z + 1)/( 2 ) ` or ` ( x - 2 ) / ( 6 ) = ( y - 4 ) / ( 3 ) = ( z + 1 ) / ( 2 ) ` |
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| 19. |
The point of the line `(x-2)/(1)=(y+3)/(-2)=(z+5)/(-2)` at a distance of 6 from the point (2,-3,-5) isA. (3,-5,-3)B. (4,-7,-9)C. (0,2,-1)D. (-3,5,3) |
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Answer» Correct Answer - B Direction of the given line is `(1)/(3)-(2)/(3),-(2)/(3)` Hence , the equation of line passing throught `(2,-3,-5)` and parallel to the given line is `(x-2)/(1//3) = (y+3)/(-2//3) =(z+5)/( - 2//3) = r` `therefore` Points is `(2+(r)/(3), - 3-(2r)/(3), -5-(2r)/(3))` But is given ` " " r = pm 6` `therefore` Points are (4,-7,-9) and (0,-1,-1) |
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| 20. |
The vector equationm of the line `(x-5)/(3)=(y+4)/(7)=(z-6)/(2)` isA. `r=3hati -7hatj -2hatk+lambda(5hati -4hatj +6hatk)`B. `r=5hati +4hatj +6hatk+lambda(3hati -7hatj +2hatk)`C. `r=3hati +7hatj +2hatk+lambda(5hati -4hatj +6hatk)`D. `r=5hati -4hatj +6hatk+lambda(3hati +7hatj +2hatk)` |
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Answer» Correct Answer - D The direction ratios of the given line are (3,7,2). `therefore` The given line is parallel to `3hati+7hatj+2hatk` From the cartesian equation of the given line, it is clear that is passes through the points (5,-4,6). The vector equation of the line is `r=5hati-4hatj+6hatk+lambda(3hati+7hatj+2hatk)` |
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| 21. |
The position vector of the foot of the perpendicular draw from the point `2hati-hatj+5hatk` to the line `vecr=(11hati-2hatj-8hatk)+lamda(10hati-4hatj-11hatk)` isA. (3,2,1)B. (1,2,3)C. (2,1,3)D. (1,3,2) |
| Answer» Correct Answer - B | |
| 22. |
Find the coordinates of apoint on the `(x-1)/2=(y+1)/(-3)=z`atg a distance `4sqrt(14)`from the point `(1,-1,0)dot`A. (9,-13,4)B. `(8sqrt14,-12,-1)`C. `(-8sqrt14,12,1)`D. `(-7,11,-4)` |
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Answer» Correct Answer - D Given equation of line in `(x-1)/(2) = (y +1)/(-3) = z = r` Any point on given lines is `p(2r + 1, -- 3r - 1,r)`, its distance form `(1,-1,0)` `rArr " "(2r)^(2) + (-3r)^(2) + r^(2) = (4 sqrt(14))^(2)` `rArr" "r = pm 4` , `rArr` Coordinates are `R(9,-14,4) and S(-7,11,-4)` and nearer to the origin is `(-7,11,-4)` `[{:(,OR = sqrt((9)^(2) +(-13)^(2) +(4))= sqrt(266)),(,and OS = sqrt((-7)^(2) +11^(2) +(-4)^(2))= sqrt(186)):}]` |
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| 23. |
The vector equation of the line passing through the points (1,-2,5) and (-2,1,3) isA. `r=-2hati +hatj + 3hatk + lambda (3hati-3hatj+2hatk)`B. `r=-2hati -hatj + 3hatk + lambda (hati+3hatj-5hatk)`C. `r=-hati -2hatj + 5hatk + lambda (-2hati-hatj+3hatk)`D. `r=-2hati +hatj + 3hatk + lambda (hati-2hatj+5hatk)` |
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Answer» Correct Answer - A Let `A-=(-2,1,3) and B-=(1,-2,5)` Then `a=-2hati+hatj+3hatk and b=hati-2hatj+5hatk` `therefore b-a=3hati-3hatj+2hatk` Let r be the position vector of any point on the line. Then, the vector equation of the line is `r=-2hati+hatj+3hatk+lambda(3hati-3hatj+2hatk)` |
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| 24. |
The line `(x+1)/(1)=(y-1)/(2)=(z-2)/(-1),(x-1)/(2)=y/1=(z+1)/(4)` areA. parallel linesB. intersecting linesC. perpendicularD. None of these |
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Answer» Correct Answer - C Direction ratios of given lines are (1,2,-1) and (2,1,4) Now, (1) (2)+(2)(1)+(-1)(4)=0 `therefore` Lines are perpendicular |
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| 25. |
The lines `(x-1)/(2)=(y+1)/(2)=(z-1)/(4) and (x-3)/(1)=(y-k)/(2)=(z)/(1)` intersect each other at pointA. `(-2,-4,5)`B. `(-2,-4,-5)`C. `(2,4,-5)`D. `(2,-4,-5)` |
| Answer» Correct Answer - B | |
| 26. |
If a,b,c and `1/(bc), 1/(ca), 1/(ab)` are the direction ratio of two lines, then the lines areA. mutually perpendicularB. skewC. coincidentD. intersecting |
| Answer» Correct Answer - C | |
| 27. |
Equation of a line passing through the points (3,1,2) and (-1,2,1) isA. `(x+3)/(-4)=(y-1)/(1)=(z-2)/(1)`B. `(x-3)/(-4)=(y-1)/(1)=(z-2)/(1)`C. `(x-3)/(-4)=(y-1)/(1)=(z-2)/(-1)`D. `(x-3)/(-4)=(y-1)/(1)=(z-2)/(1)` |
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Answer» Correct Answer - C Equation of a line passing through the point (3,1,2) and (-1,2,1) is `(x-3)/(-1-3)=(y-1)/(2-1)=(z-2)/(-2) Rightarrow (x-3)/(-4)=(y-1)=(z-2)/(-1)` |
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| 28. |
The length of the perpendicular from P(1,0,2) on the line `(x+1)/(3)=(y-2)/(-2)=(z+1)/(-1)` isA. `(3sqrt6)/(2)`B. `(6sqrt3)/(5)`C. `3sqrt2`D. `2sqrt3` |
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Answer» Correct Answer - A `(x+1)/(3) = (y -2)/(-2) = (z+1)/(-1) = M` Now , any point on the given line is `M -= (3lambda -1,-2 lambda + 2, -lambda -1)` Direction ratio of `PM -=( 3 lambda -2, -2 ldmba +2, - lambda-3)` and ratio of the line (3,2,-1) . Since ,PM is perpendicular to the line, hence `3(3lambda -2)-2(2-2lambda) - 1(-lamda-3)=0" "[because a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2) =0]` `rArr " " 14lambda - 7 = 0 rArr=(1)/(2)` Hence , coordinate of `M = (3xx(1)/(2) -1,1-2xx (1)/(2) +2,-(1)/(2)-1) = ((1)/(2) , 1-(3)/(2))` Now , perpendicular length `Pm = sqrt((1-(1)/(2))^(2) +(0-1)^(2) +(2+(3)/(2))^(2))` ` = sqrt((1)/(4) +1+(49)/(4)) = sqrt((54)/(4)) =(3sqrt(6))/(2)` |
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| 29. |
The equation of a line passing through the point (4,-2,5) and parallel to the vector `3hati -hatj + 2hatk` isA. `(4hati - 2hatj + 5hatk) + lambda(2hati - hatj + 3hatk)`B. `(4hati - 2hatj + 5hatk) + lambda(hati - 2hatj + 3hatk)`C. `(4hati - 2hatj - 5hatk) + lambda(3hati - hatj - 2hatk)`D. `(4hati - 2hatj + 5hatk) +lambda(3hati -hatj + 2hatk)` |
| Answer» Correct Answer - D | |
| 30. |
The cartesian equation of a line is `(x - 6) /(2) = ( y + 4)/(7) = (z -5)/(3)` , find its vector equation .A. `(-6hati + 4hatj - 5hatk) + lambda(2hati + 7hatj + 3hatk)`B. `(6hati - 4hatj + 5hatk) + lambda(2hati + 7hatj + 3hatk)`C. `(2hati + 7hatj + 3hatk) + lambda(-6hati + 4hatj - 5hatk)`D. `(2hati +7hatj + 3hatk) + lambda(6hati - 4hatj + 5hatk)` |
| Answer» Correct Answer - B | |
| 31. |
The vector equation of a line `(x+5)/3 = (y+4)/5 = (z+5)/6` isA. `(3hati + 5hatj + 6hatk) + lambda(5hati + 4hatj + 5hatk)`B. `(5hati + 4hatj + 5hatk) + lambda(3hati + 5hatj + 6hatk)`C. `(-5hati - 4hatj - 5hatk) + lambda(3hati + 5hatj + 6hatk)`D. `(3hati + 5hatj + 6hatk) + lambda(-5hati - 4hatj - 5hatk)` |
| Answer» Correct Answer - C | |
| 32. |
If AB=21, B`-=(-2,1,-8)` and the direction cosines of AB are `6/7, 2/7, 3/7` , then the co-ordinates of points in line PQ nearer to the origin at a distance of 14 units from A areA. `(-16,-7,-1)`B. `(-20, -5,17)`C. `(16,7,1)`D. `(20, 5,17)` |
| Answer» Correct Answer - C | |
| 33. |
The distance of the point P(1,2,3) from the line which passes through the point (4,2,2) and parallel to the vector `2hati+3hatj+6hatk` isA. `sqrt10`B. `sqrt7`C. `sqrt5`D. 1 |
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Answer» Correct Answer - A The equation of the of line which passess through the point `A (4,2,2)` and parallel to the vector `2hati + 3hatj + 6hatk` is `(x-4)/(2) = (y -2)/(3) = (x-2)/(6)` Distance of point `P(1,2,3)` form the line `sqrt((1-4)^(2) +(2-2)^(2) (3-2)^(2) - {2(1-4)+3(2-2) +6(3-2)^(2)})` ` = sqrt(9+0+1-(-6 + 0 + 6)^(2)) = sqrt(10)` |
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| 34. |
The angle made by line AB with perpendicular of co-ordinates axes if `A-=(0,sqrt(3),0)` and `B-=(0,0,-1)` areA. `pi/3, pi/4, pi/2`B. `(3pi)/4, pi/2, pi/3`C. `pi/2, (5pi)/6, (2pi)/3`D. `pi/2, (2pi)/3, (5pi)/6` |
| Answer» Correct Answer - C | |
| 35. |
The vector equation of the line `(x-2)/(2)=(2y-5)/(-3),z=-1` is `r=(2hati+(5)/(2)hatj-hatk)+lambda(2hati-3/2hatj+xhatk)` where x is equal to |
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Answer» Correct Answer - A The given equation of line is `(x-2)/(2)=(2y-5)/(-3),z=-1` `Rightarrow (x-2)/(2)=(y-(5)/(2))/(-(3)/(2))=(z+1)/(0)` So its vector is `vecr=(2hati+5/2hatj-hatk)+lambda(2hati-3/2hatij+0hatk)` Hence, x=0 |
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| 36. |
The cartesian equation of a line is 3x - 1 = 6y + 2 = 1 - z . Find the vector equation of the line .A. `(-1/3hati + 1/3hati +hatk) +lambda(2hati + hatj - 6hatk)`B. `(1/3hati - 1/3hatj + hatk) + lambda(2hati +hatj - 6hatk)`C. `(hati + 2hatj + hatk) +lambda(2hati + hatj - 6hatk)`D. `(hati - 2hatj - hatk) + lambda(2hati + hatj - 6hatk)` |
| Answer» Correct Answer - B | |
| 37. |
The cartesian equation of a line is ` 3x + 1 = 6 y - 2 = 1 -z`. Find the vector equation of the line.A. `(hati + hatj -hatk) + lambda(2hati + hatj - hatk)`B. `(-hati + hatj + hatk)+ lambda(2hati + hatj - hatk)`C. `(1/3hati + 1/6hatj - hatk) +lambda(2hati + hatj - 6hatk)`D. `(-1/3hati + 1/3hatj +hatk) + lambda(2hati + hatj - 6hatk)` |
| Answer» Correct Answer - D | |
| 38. |
The vector equation of the line `(x+2)/(3)=(1-y)/(-2)=(z-5)/(7) ` isA. `(-2hati + 5/2hatj - hatk)+lambda(4/5hati + 3/5hatj)`B. `(-2hati + 5/2hatj - hatk) + lambda(2/5hati + 3/5hatj)`C. `(-2hati + 5/2hatj -hatk)+ lambda(2hati -3/5hatj)`D. `(-2hati + 5/2hatj -hatk) + lambda(4hati - 3/5hatj)` |
| Answer» Correct Answer - A | |
| 39. |
The vector equations of a line passing through the origin and the point (5,-2,3) isA. `lambda(-5hati + 2hatj - 3hatk)`B. `lambda(5hati - 2hatj + 3hatk)`C. `-5hati + 2hatj - 3hatk`D. `5hati - 2hatj + 3hatk` |
| Answer» Correct Answer - B | |
| 40. |
The shortest distance between the straight lines `(x-6)/(1)=(2-y)/(2)=(z-2)/(2)" and "(x+4)/(3)=(y)/(-2)=(1-z)/(2)` isA. 9B. `(25)/(3)`C. `(16)/(3)`D. 4 |
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Answer» Correct Answer - B Given lines can be rewritten as ` (x -6)/(1) = (y-2)/(-2) = (Z-2)/(2) and (x+4)/(3) =(y)/(-2) =(z-1)/(-2)` Here ,`" "x_(1) = 6, y_(1) = 2 , z_(1) =2 ` ` x_(2) = - 4 , y_(2) 0 , z_(2) = 1` `a_(1)=1, b_(1) = - 2 , c_(1) = 2` and `" "a_(2) = 3,b_(2) = - 2 2,c_(2) = -2` Now |
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| 41. |
The equation `r=lambda hati` representsA. the X-axisB. the yoz-planeC. the y-axisD. the Z-axis |
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Answer» Correct Answer - A The given equation is `vecr=lambda hati` `Rightarrow xhati+yhatj+zhatk=lambdahati Rightarrow x =lambda, y=0, z=0` Which are equations of X-axis as they they can be written as `(x-0)/(1)=(y-0)/(0)=(z-0)/(0)=lambda` |
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| 42. |
The cartesian equation of line passing through the points (5,4,3) and having direction ratios `-3,4,2` isA. `(x+5)/-3 =(y+4)/-4 = (z+3)/2`B. `(x+5)/-3 = (y+4)/4 = (z+3)/2`C. `(x-5)/-3 = (y-4)/-4 = (z-3)/2`D. `(x-5)/-3 = (y-4)/4 = (z-3)/2` |
| Answer» Correct Answer - D | |
| 43. |
The Cartesian equationof a line are `3x+1=6y=2=1-z`. Find the direction ratios and write down its equation in vector form.A. 2,1,-6B. `-2,1,-3`C. `6,2,-1`D. `3,6,-1` |
| Answer» Correct Answer - A | |
| 44. |
The Cartesian equationof a line are `3x+1=6y=2=1-z`. Find the direction ratios and write down its equation in vector form.A. `-2, 1,-3`B. `6,2,-1`C. `2,1,-6`D. `3,6,-1` |
| Answer» Correct Answer - C | |
| 45. |
The cartesian equation of a line passing through the origin and point (5,-2,3) isA. `(x-5)/5 = (y+2)/-2 = (z-3)/3`B. `(x-5)/-5 = (y-2)/-2 = (z-3)/3`C. `x/5 = y/-2 = z/3`D. `x/-5 = y/2 = z/3` |
| Answer» Correct Answer - C | |
| 46. |
Find `lamda`, if the points `(-1, 3 ,2),(-4, 2 ,-2)and(5, 5 ,lambda)` are collinear.A. -6B. 5C. 6D. 10 |
| Answer» Correct Answer - D | |
| 47. |
The equation of a line passing through the points `A(4,2,1)` and B(2,-1,3) isA. `(x-4)/-2 = (y-2)/3 = (z-1)/2`B. `(x+4)/-2 = (y-2)/0 = (z-5)/11`C. `(x-4)/2 = (y-2)/3 = (z-1)/-2`D. `(x+4)/2 = (y+2)/3 = (z+1)/-2` |
| Answer» Correct Answer - C | |
| 48. |
The equation of a line passing through the points `A(2,-1,1)` and `B(3,1,1)` isA. `(x-2)/1 =(y+1)/2 , z=1`B. `(x-2)/1 = (y-1)/2, z=1`C. `(x-2)/1 = (y+1)/2, z=-1`D. `(x+4)/2 = (y+2)/3 = (z+1)/-2` |
| Answer» Correct Answer - A | |
| 49. |
The vector equation of the straight line `(1-x)/(3)=(y+1)/(-2)=(3-z)/(-1)` isA. `r=(hati-hatj+3hatk)+lambda(3hati+2hatj-hatk)`B. `r=(hati-hatj+3hatk)+lambda(3hati-2hatj-hatk)`C. `r=(3hati-2hatj-hatk)+lambda(hati-hatj+3hatk)`D. `r=(3hati+2hatj-hatk)+lambda(hati-hatj+3hatk)` |
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Answer» Correct Answer - A Given line is `(x-1)/(-3)=(y+1)/(-2)=(z-3)/(1)` `Rightarrow` Line is passing through (1,-1,3) and having direction ratios (-3,-2,1) i.e, (3,2,-1) `therefore` Vector equation of the line is `tau=(hati-hatj+3hatk)+lambda(3hati+2hatj-hatk)` |
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| 50. |
The points with position vectors `5hati + 5hatk, -4hati + 3hatj - hatk` and `2hati +hatj + 3hatk`A. collinearB. non-collinearC. non-collinear and non planarD. non-collinear but coplanar |
| Answer» Correct Answer - A | |