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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.
| 201. |
Co - ordinates of the foot of the perpendicular from the point (a,b,c) on the Z - axis areA. `(a,b,0)`B. `(0,0,c)`C. `(a,0,0)`D. `(0,b,0)` |
| Answer» Correct Answer - b | |
| 202. |
In which of plane does the point (4,-3,0) lie ? |
| Answer» Correct Answer - xy-plane | |
| 203. |
If a line is equally inclines to co - ordinate to co - ordinate axes , then its each direction angle is of measureA. `sin^(-1)sqrt(3)`B. `cos^(-1)sqrt(3)`C. `cos^(-1)sqrt(3)`D. `sec^(-1)sqrt(3)` |
| Answer» Correct Answer - d | |
| 204. |
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. `ne1`B. `+-1//3`C. `+-sqrt5`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)` `implies(x-2)/(3)=1and(y+1)/(2)=1` `impliesx=5andy=1` Putting these valese in `xy=c^(2)` we get `5=c^(2)orc=+-sqrt5` |
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| 205. |
In a three-dimensional `x y z`space , the equation `x^2-5x+6=0`representsa.Points b. planes c. curves d. pair of straight linesA. pointsB. planesC. curvesD. pair of straight lines |
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Answer» Correct Answer - b `x^(2)-5x+6=0` `impliesx-2=0,x-3=0` which represents planes. |
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| 206. |
The direction cosines of a line parallel to the line `(x+1)/(2)=(y)/(-3)=(z-5)/(6)` areA. `1, (-3)/(2), 3`B. `(-2)/(3), 1, -2`C. `(1)/(3), (-1)/(2), 1`D. `(2)/(7), (-3)/(7), (6)/(7)` |
| Answer» Correct Answer - D | |
| 207. |
If the line segment joining the points A(7, p, 2) and B(q, -2, 5) be parallel to the line segment joining the points C(2, -3, 5) and D(-6, -15, `11), find the value of p and q.A. a=4, b = -3B. a=-4, b=3C. a=4, b=3D. a=-4, b=3 |
| Answer» Correct Answer - C | |
| 208. |
The line joining the points`-2,1,-8)a n d(a ,b ,c)`is parallel to the linewhose direction ratios are `6,2,a n d3.`Find the values of `a ,ba n d cdot`A. a=0, b=5, c=-5B. a=4, b=3, c=-5C. `a=3, b=5, c=11`D. `a=1, b=2, c=-6 |
| Answer» Correct Answer - B | |
| 209. |
The points (-7, 4, -2), (-2, 1, 0) and (3, -2, 2) areA. non-collinearB. non-coplanarC. non-collinear and non-coplanarD. collinear |
| Answer» Correct Answer - D | |
| 210. |
If the points `(6, -1, 2), (8, -7, lambda)` and (5, 2, 4) are collinear then `lambda=`A. 4B. 2C. -2D. -4 |
| Answer» Correct Answer - C | |
| 211. |
The mid-points of the sides of a triangle are `(1,5,-1),(0,4,-2)a n d(2,3,4)dot`Find its vertices. |
| Answer» Correct Answer - (1,2,3); (3,4,5); (-1,6,-7) | |
| 212. |
If the origin is the centroid of a triangle ABC having vertices `A(a ,1,3), B(-2, b ,-5)a n d C(4,7, c)`, find the values of `a , b , cdot` |
| Answer» Correct Answer - `a=-2, b=-8, c=2` | |
| 213. |
Consider a set of point R in which is at a distance of 2 units from the line `(x)/(1)= (y-1)/(-1)= (z+2)/(2)` between the planes `x-y+2z=3=0 and x-y+2z-2=0`.A. The volume of the bounded figure by points R and the planes is `(10//3sqrt3)pi` cube units.B. The area of the curved surface formed by the set of points R is `(20 pi //sqrt6)` sq. units.C. The volume of the bounded figure by the set of points R and the planes is `(20pi//sqrt6)` cubic units.D. The area of the curved surface formed by the set of points R is `(10//sqrt3)pi` sq. units. |
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Answer» Correct Answer - b,c Distance between the planes is `h= 5//sqrt6`. Also the figure formed is a cylinder, whose radius is `r=2` units. Hence, the volume of the cylinder is `" "pir^(2)h= pi(2)^(2)*(5)/(sqrt6)= (20pi)/(sqrt6)` cubic units. Also the curved surface area is `" "2pirh= 2pi(2)*(5)/(sqrt(6))= (20pi)/(sqrt6)` |
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| 214. |
Show that the distance between planes `2x-2y+z+3=0 and 4x-4y+2z+5=0 is 1/6` |
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Answer» Equation of first plane `3x+2y-z + 1 = 0` Let `(x_(1),y_(1),z_(1))` be any point on this plane. `:. 3x_(1)+2y_(1)-z_(1) + 1 = 0 "……"(1)` Perpendcicular distance from point `(x_(1),y_(1)z_(1))` to the second plane `3x+2y-z+5 = 0` `=|(3x_(1)+2y_(1)-z_(1)+5)/(sqrt(3^(2)+2^(2)+(-1)^(2)))|` , [From eq.(1)] `= |((-1)+5)/(sqrt(9+4+1))|=4/(sqrt(14))` units Therefore the distance between parallel planes `= (4)/(sqrt(14))` units. |
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| 215. |
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) 27 |
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Answer» `(x-2)/3 = (y+1)/4 = (z-2)/12 = lamda` `x= 3 lamba + 2; y= 4 lamda - 1; z= 12 lamda + 2` `(x,y,z) -> x-y+z=16` `3 lamda + 2 - ( 4 lamda - 1) + 12 lamda + 2 = 16` `3 lamda + 2 - 4 lamda + 1 + 12 lamda + 2 = 16` `11 lamda = 11` `lamda = 1` `(x,y,z) = (5,3,14)` `(1,0,2) & (5,3,14)` `d= sqrt((5-1)^2 + (3-0)^2 + (14-2)^2)` `= sqrt(16 + 9 + 144) = sqrt(169) = 13` option 3 is correctAnswer |
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| 216. |
If the line, `(x-3)/2=(y+2)/(-1)=(z+4)/3`lies in the place, `l x+m y-z=9`, then `l^2+m^2`is equal to:(1) 26(2) 18(3) 5(4) 2 |
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Answer» Given line is `(x-3)/2 = (y+2)/-1 = (z+4)/3` So, point `P(3,-2,-4)` lies on the given plane `lx+my-z = 9` `:. 3l-2m+4 = 9` `=>3l-2m = 5->(1)` Now, direction ratios of the line and direction ratios of the plane will be perpendicular. So, their product is `0`. `:.2l-m-3 = 0=>2l-m = 3->(2)` so, multiplying (2) with `2` and subtracting it from (1), `3l-2m-4l+2m = 5-6` `=>l = 1` Putting value of `l` in (2), `2(1)-m = 3` `=>-2-m = 3=> m = -1` `:. l^2+m^2 = 1^2+(-1)^2 = 1+1= 2` |
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| 217. |
Line `L_1`is parallel to vector ` vecalpha=-3 hat i+2 hat j+4 hat k`and passes through a point `A(7,6,2)`and line `L_2`is parallel vector ` vecbeta=2 hat i+ hat j+3 hat k`and point `B(5,3,4)dot`Now a line `L_3`parallel to a vector ` vec r=2 hat i-2 hat j- hat k`intersects the lines `L_1a n dL_2`at points `Ca n dD ,`respectively, then find `| vec C D|dot` |
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Answer» Line `L_(1)` is parallel to vector `vecalpha=-3hati+2hatj+4hatk` and passes through a point `A(7, 6, 2)`. Therefore, position vector of any point on the line is `7hati+6hatj+2hatk+lamda(-3hati+2hatj+4hatk),` line `L_(2)` is parallel to a vector `vecbeta=2hati+hatj+3hatk` and passes through a point `B(5, 3, 4)`. Position vector of any point on the line is `5hati+3hatj+4hatk+mu(2hati+hatj+3hatk)` `therefore" "vec(CD)=2hati+3hatj-2hatk+lamda(-3hati+2hatj+4hatk)-mu(2hati=hatj+3hatk)` Since it is parallel to `2hati-2hatj-hatk`, we have `" "(2-3lamda-2mu)/(2)=(3+2lamda-mu)/(-2)=(-2+4lamda-3mu)/(-1)` Solving these equations we get `lamda=2 and mu=1`. Therefore, `" "vec(CD)=-6hati+6hatj+3hatk` or `" "|vec(CD)|=9` |
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| 218. |
Find the equation of theline drawn through point (1, 0, 2) to meet the line `(x+1)/3=(y-2)/(-2)=(z1)/(-1)`at right angles. |
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Answer» Given line at `overset(harr)(AB)-=(x+1)/(3)=(y-2)/(-2)=(z+1)/(-1)" "`(i) Let `P-=(1, 0, 2)` Any point on line (i) is `Q-=(3r-1, -2r+2, -r-1)` Direction ratios of `PQ " are "3r-2, -2r+2, -r-3` Direction ratios of line `AB" are "3, -2, -1` `because" "PQbotAB` `therefore" "3(3r-2)-2(-2r+2)-1(-r-3)=0` or `" "14r=7 or r=(1)/(2)` Therefore, direction ratio of PQ are `-(1)/(2), 1, -(7)/(2) or 1, -2, 7` `" "`Equation of line `PQ-=(x-1)/(1)=(y)/(-2)=(z-2)/(7)` |
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| 219. |
If `vecr=(hati+2hatj+3hatk)+lamda(hati-hatj+hatk) and vecr=(hati+2hatj+3hatk)+mu(hati+hatj-hatk)` are two lines, then find the equation of acute angle bisector of two lines. |
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Answer» Lines are `vecr=(hati+2hatj+3hatk)+lamda(hati-hatj+hatk) and vecr=(hati+2hatj+3hatk)+mu(hati+hatj-hatk)` along vectors `(hati-hatj+hatk) and (hati+hatj-hatk)`, respectively. `" "` Angle between two lines = `cos^(-1)""(((1)xx(1)+(-1)(1)+(1)(-1))/(sqrt(3)sqrt(3)))=cos^(-1)(-(1)/(3))` which is an obtuse angle. `therefore" "` Vector along acute angle bisector = `lamda[(hati-hatj+hatk)/(sqrt(3))-(hati+hatj-hatk)/(sqrt(3))]=(2lamda)/(sqrt(3))(-hatj+hatk)` `therefore" "` Equation of acute angle bisector = `(hati+2hatj+3hatk)+t(hatj-hatk)` |
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| 220. |
Find the vector equationof the line passing through the point `(1," "2," "" "4)`and perpendicular to the two lines: `(x-8)/3=(y+19)/(-16)=(z-10)/7`and `(x-15)/3=(y-29)/8=(z-5)/(-5)` |
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Answer» Since the line to be determined is perpendicular to the given two straight lines, it is directed towards `" ""Vector "(3hati-16hatj+7hatk)xx(3hati+8hatj-5hatk)` `=|{:(hati,,hatj,,hatk),(3,,-16,,7),(3,,8,,-5):}|=24hati+36hatj+72hatk` Therefore, direction ratios of the line are 24, 36, 72 or 2, 3, 6. Hence, equation of line passing through the point (1, 2, -4) and parallel to the vector `2hati+3hatj+6hatk` is `" "vecr=(hati+2hatj-4hatk)+lamda(2hati+3hatj+6hatk)` This is the equation of the required line. |
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| 221. |
Find the equations of het planes parallel to the plane `x+2y-2z+8=0`which are at distance of 2 units from the point `(2,1,1)dot` |
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Answer» Let equation of plane parallel to plane `x+2y-2z+18=0` `x+2y-2z+lambda = 0 "….."(1)` Perpendicular distance from `(2,1,1)` to this plane `|(2+2-2+lambda)/(sqrt(1+4+4))|=|(lambda+2)/(3)|` Given that `|(lambda+2)/(3)| = 3 rArr (lambda+2)/(3) = +-3` `rArr lambda+2=+-9` `rArr lambda = 7` or ` lambda = - 11` From eq. (1) Required equations of planes `x+2y-2z+7=0` or `x+2y-2z-11 = 0`. |
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| 222. |
In a three-dimensionalcoordinate system, `P ,Q ,a n dR`are images of a point `A(a ,b ,c)`in the `x-y ,y-za n dz-x`planes,respectively. If `G`is the centroid of triangle `P Q R ,`then area of triangle `A O G`is (`O`is the origin)a. `0`b. `a^2+b^2+c^2`c. `2/3(a^2+b^2+c^2)`d. noneof these |
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Answer» Correct Answer - a Point A is `(a,b,c)implies"Points "P,Q,R" are"(a,b,-c),(-a,b,c)and(a,-b,c)`, respectively `implies` Centroid of triangle PQR is `((a)/(3),(b)/(3),(c)/(3))` `implies((a)/(3),(b)/(3),(c)/(3))` `impliesA,O,G" are"" collinear"` `implies` Area of triangle AOG is zero. |
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| 223. |
From the point `P(a ,b ,c),`let perpendicualars `P La n dP M`be drawn to `Y O Za n dZ O X`planes,respectively. Then the equation of the plane `O L M`isa. `x/a+y/b+z/c=0`b. `x/a+y/b-z/c=0`c. `x/a-y/b-z/c=0`d. `x/a-y/b+z/c=0`A. `(x)/(a)+(y)/(b)+(z)/(c)=0`B. `(x)/(a)+(y)/(b)-(z)/(c)=0`C. `(x)/(a)-(y)/(b)-(z)/(c)=0`D. `(x)/(a)-(y)/(b)+(z)/(c)=0` |
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Answer» Correct Answer - b Coordinates of L and M are (0,b,c) and (a,0,c), respectively. Therefore, the equation of the plane passing through (0,0,0), (0,b,c) and (a,0,c) is `|[x-0,y-0,z-0],[0,b,c],[a,0,c]|=0or(x)/(a)+(y)/(b)-(z)/(c)=0` |
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| 224. |
LetL be the line of intersection of the planes `2x""+""3y""+""z""=""1`and `x""+""3y""+""2z""=""2`. If L makes an angles ` alpha `withthe positive x-axis, then cos` alpha `equals |
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Answer» `P_1 = 2x+3y + z = 1` `P_2 = x+ 3y + 2z = 2` Dr of line`_|_`to `P_1 = (2,3,1)` Dr of line`_|_`to `P_2 = (1,3,2)` Dr of line L=`(a,b,c)` `L _|_ P_1, 2a+ 3b + c = 0` (1) `L _|_ P_2, a+ 3b+2c= 0` (2) subtracting eqn 2 from 1 `a=c` `3b= 3a` so,`a=b=c` `cos alpha = a/(sqrt(a^2 + b^2+ c^2))= a/sqrt(3a^2) = 1/sqrt3` option 1 is correct |
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| 225. |
Statement-1:The point A(3, 1, 6) is the mirror image of thepoint B(1, 3, 4) in the plane `x""""y""+""z""=""5`.Statement-2:The plane x `x""""y""+""z""=""5`bisects the line segmentjoining A(3, 1, 6) and B(1, 3, 4).(1)Statement-1 istrue, Statement-2 is true; Statement-2 is not the correct explanation forStatement-1(2)Statement-1 istrue, Statement-2 is false(3)Statement-1 isfalse, Statement-2 is true(4)Statement-1 istrue, Statement-2 is true; Statement-2 is the correct explanation forStatement-1 |
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Answer» MP of AB=`((3+1)/2 , (1+3)/2,(6+4)/2)` `= (2,2,5)` `P(2,2,5) = 2-2 +5-5 = 0` MP is on plane Drs of line `_|_` to plane `P_1` = (1,-1,1) Drs of line PQ = `(2,-2,2)` `2/1 = -2/-1 = 2/1 = 2` `PQ || `(line `_|_ `to plane) option 2 is correct |
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| 226. |
Find the equation of line `x+y-z-3=0=2x+3y+z+4`in symmetric form. Find the direction of theline. |
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Answer» In the section of planes we will see the equation of the form `ax+by+cz+d=0` is the equation of the plane in the space. Now equation of line in the form `x+y-z-3=0=0=2x+3y+z+4` means set of those points in space which are common to the planes `x+y-z-3=0 and 2x+3y+z+4=0`, which lie on the line of intersection of planes. For example, equation of x-axis is y=z=0 where xy-plane (z=0) and xz-plane (y=0) intersect. Now to get the equation of line in symmetric form, in above equations, first we eliminate any one of the variables, say z. Then adding `x+y-z-3=0 and 2x+3y+z+4=0`, we get `" "3x+4y+1=0 or 3x=-4y-1=lamda("say")` or `" "x=(lamda)/(3), y=(lamda+1)/(-4)` Putting these values in `x+y-z-3=0,` we have `(lamda)/(3)+(lamda+1)/(-4)-z-3=0` or `" "lamda=39+12z` Comparing values of `lamda`, we have equation of line as `" "3x=-4y-1=12z+39` or `" "(3x)/(12)=(-4y-1)/(12)=(12z+39)/(12) or (x)/(4)=(y+(1)/(4))/(-3)=(z+(13)/(4))/(1)` Hence, the line is passing through point `(0, -(1)/(4), -(13)/(4))` and have direction ratios 4, -3, 1. If we eliminate x or y first we will get equation of line having same direction ratio but with different point on the line. |
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| 227. |
If `1/2,1/3,n` are direction cosines of a line, then the value of `n` isA. `(7)/(36)`B. `(7)/(6)`C. `(sqrt(23))/(36)`D. `(sqrt(23))/(6)` |
| Answer» Correct Answer - D | |
| 228. |
Find the equation of theplane which passes through the point `(12,3)`and which is at the maxixumdistance from the point `(-1,0,2)dot` |
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Answer» The plane passes through the point `A(1, 2, 3)` and is at the maximum distance from the point `B(-1, 0, 2)`, then the plane is perpendicular to line `AB`. Therefore, the direction ratios of the normal to the plane are 2, 2 and 1. Hence, the equation of the plane is `" "2(x-1)+2(y-2)+1(z-3)=0` or `" "2x+2y+z=9` |
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| 229. |
If `A(3,2,-4),B(5,4,-6)a n dC(9,8,-10)`are three collinear points,then find the ratio in which point `C`divides `A Bdot` |
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Answer» Let C divides AB in the ratio `lamda:1`. Then `" "C-=((5lamda+3)/(lamda+1),(4lamda+2)/(lamda+1), (-6lamda-4)/(lamda+1))=(9, 8, -10)` Comparing, we get `" "5lamda+3=9lamda+9 or 4lamda=-6 or lamda=-3//2` Also, from `4lamda+2=8lamda+8 and -6lamda-4=-10lamda-10`, we get the same value of `lamda`. |
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| 230. |
A line passes through the points (3, 1, 2) and (5, -1, 1) . Then the direction ratios of the line areA. 2, -2, -1B. 2, 2, -1C. 2, 2, 1D. `-2, -2, 1 ` |
| Answer» Correct Answer - A | |
| 231. |
If the direction cosines of a line are `p,-p, (-p)/(2)`, then p =A. `pm(4)/(9)`B. `pm(9)/(4)`C. `pm (2)/(3)`D. `pm (3)/(2)` |
| Answer» Correct Answer - C | |
| 232. |
The reflection of the point (2,-1,-3) in the plane 3x - 2y -z = 9 is : |
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Answer» line AB=`(x-2)/3=(y+1)/-2=(z-3)/-1=lambda` `(x,y,z)=(3lambda+2,-2lambda-1,-lambda+3)` `3x-2y-z=9` `3(3lambda+2)-2(-2lambda-1)-3+lambda=9` `9lambda+6+4lambda+2-3+lambda=9` `14lambda=4` `lambda=2/7` `C(x,y,z)=(20/7,-11/7,19/7)` A(2,-1,3) C is MP of A and B `C=(A+B)/2, B=(2C-A)` `B=(26/7,-22/7+1,38/7-3)` `B=(26/7,-15/7,17/7)` option b is correct. |
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| 233. |
A line makes the some angle `theta` with each of the `x` and z-axes. If the angle `beta`, which it makes with y-axis, is such that `sin^(2)beta=3sin^(2)theta` then `cos^(2)theta` equalsA. `(4)/(5)`B. `(2)/(5)`C. `(3)/(5)`D. `(9)/(5)` |
| Answer» Correct Answer - C | |
| 234. |
Find the equation of a line passes through the point `(2,3,4)` and whose direction ratios are `3,-1,-2`. |
| Answer» Correct Answer - `(x-2)/(3)=(y-3)/(-1)=(z-4)/(-2)` | |
| 235. |
Find the equation of a line parallel to the line `(x-5)/(3) = (y+1)/(-2) = (z)/(1)` and passes through the point `(0,-1,2)`. |
| Answer» Correct Answer - `x/3 = (y+1)/(-2)=(z-2)/(1)` | |
| 236. |
Find the radius of the circular section of the sphere `|vecr|=5` by the plane `vecr*(veci+vecj-veck)=4sqrt(3)`. |
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Answer» The radius of the sphere is 5 The given plane is `x+y-z= 4sqrt(3)` The length of the perpendicular from the centre (0, 0, 0) of the sphere on the plane `= (4sqrt(3))/(sqrt(1+1+1))= 4`. Hence, radius of the circular section is `sqrt(25-16) = sqrt9 = 3 ` |
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| 237. |
The extremities of adiameter of a sphere lie on the positive y- and positive z-axes at distance 2and 4, respectively. Show that the sphere passes through the origin and findthe radius of the sphere. |
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Answer» We are given the extremities of the diameter as (0, 2, 0) and (0, 0, 4). Therefore, the equation of the sphere is `(x-0)(x-0)+(y-2)(y-0)+ (z-0)(z-4)=0` `or x^(2)+y^(2)+z^(2)-2y-4x=0`. This sphere clearly passes through the origin. |
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| 238. |
Find the equationfo the plane through points (2,1,0),(3,-2,-2), and (3,1,7). |
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Answer» We know that the equation of a plane passing through three non-collinear points `(x_(1),y_(1),z_(1)), (x_(2),y_(2),z_(2)` and `(x_(3),y_(3),z_(3))` 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` `implies |(x-2,y-1,z-0),(3-2,-2-1,-2-0),(3-2,1-1,7-0)|=0` `implies |(x-2,y-1,z),(1,-3,-2),(1,0,7)|=0` `implies (x-2)(-21+0)-(y-1)(7+2)=z(3)=0` `implies -21x+42-9y+9+3z=0` `implies -21x-9y+3z=-51` `:. 7x+9y-z=17` So, the required equation of plane is `7x+3y-z=17`. |
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| 239. |
A line makes an angle `alpha, beta, gamma, delta` with the four diagonals of a cube, then `sin^(2)alpha+sin^(2)beta+sin^(2)gamma+sin^(2)delta=`A. `(4)/(3)`B. `(8)/(3)`C. `(1)/(3)`D. `(2)/(3)` |
| Answer» Correct Answer - B | |
| 240. |
If the edges of a rectangular parallelepiped are a,b, c, prove that the angles between the four diagonals are given by `cos^(-1)"((pma^(2)pmb^(2)pmc^(2))/(a^(2)+b^(2)+c^(2)))`.A. `cos^(-1)((apm b pm c)/(a+b+c))`B. `(1)/(2)cos^(-1)((a pm b pm c)/(a+b+c))`C. `cos^(-1)((a^(2)pm b^(2)pm c^(2))/(a^(2)+b^(2)+c^(2)))`D. `(1)/(2) cos^(-1) ((a^(2)pm b^(2)pm c^(2))/(a^(2)+b^(2)+c^(2)))` |
| Answer» Correct Answer - C | |
| 241. |
The distance between the planes `3x+5y+z=8` and `3x+5y+z+27 = 0` is :A. `(8)/(sqrt(35))`B. `(27)/(sqrt(35))`C. `sqrt(35)`D. `2sqrt(35)` |
| Answer» Correct Answer - C | |
| 242. |
Find the distance of thez-axis from the image of the point `M(2-3,3)`in the plane `x-2y-z+1=0.` |
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Answer» Correct Answer - `1` If image of point `(2, -3, 3)` in the plane `x-2y-z+1=0` is `(a, b, c)`, then `(a-2)/(1)= (b+3)/(-2)= (c-3)/(-1)= (-2(2-2(-3)-3+1))/((1)^(2)+ (-2)^(2)+ (-1)^(2)) = -2` Hence, the image is `(0, 1, 5)` Obviously distance of image of the point from the `z`-axis is 1. |
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| 243. |
Consider a plane `x+y-z=1` and point `A(1, 2, -3)`. A line L has the equation `x=1 + 3r, y =2 -r and z=3+4r`.A. `x-3y+5=0`B. `x+3y-7=0`C. `3x-y-1=0`D. `3x+y-5=0` |
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Answer» Correct Answer - b The equation of plane containing the line L is `" "A(x-1)+ B(y-2) + C(z-3)=0`, where `3A-B+4C=0" "` (i) (i) also contains point `A(1, 2, -3)`. Hence, C =0 and `3A= B`. The equation of plane is `x-1+ 3(y-2) =0 or x+ 3y-7=0` |
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| 244. |
Find the equation of theplane such that image of point `(1,2,3)`in it is`(-1,0,1)dot` |
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Answer» Since the image of `A(1, 2, 3)` in the plane is `B(-1, 0, 1)`, the passes through the midpoint (0, 1, 2) of `AB` and is normal to the vector `vec(AB)-=-2hati-2hatj-2hatk`. Hence, the equation of the plane is `-2(x-0)-2(y-0)-3(z-2)=0 or x+y=z=3.` |
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| 245. |
Find the equation of theimage of the plane `x-2y+2z-3=0`in plane `x+y+z-1=0.` |
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Answer» The image of the plane `x-2y+2z-3=0" "` (i) in the plane `x+y+z-1=0" "`(ii) passes through the line of intersection of the given planes. Therefore, the equation of such a plane is `(x-2y+2z-3) + t(x+y+z-1)=0, t in R` `(1+t)x+ (-2+t) y + (2+t) z-3 - t =0" "` (iii) Now plane (ii) makes the same angle with plane (i) and image plane (iii). Thus, `" "(1-2+2)/(3sqrt(3))=pm(1+t-2+t+2+t)/(sqrt(3)sqrt((t+1)^(2)+ (t-2)^(2)+ (2+t)^(2)))` `" "(1)/(3)=pm (3t+1)/(sqrt(3t^(2)+2t+9))` `" "3t^(2)+2t+9=9(9t^(2)+6t+1)` `" "3t^(2)+2t+9=81t^(2)+54t+9` `" "78t^(2)+52t=0` `" "t=0 or t=-(2)/(3)` For `t=0`, we get plane (i) , hence for image plane, `t=-(2)/(3)` The equation of the image plane is `" "3(x-2y+2z-3)-2(x+y+z-1)=0` or `" "x-8y+4z-7=0` |
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| 246. |
Find the image of the point `(1,3,4)`in the plane `2x-y+z+3=0.`A. `(3,5,-2)`B. `(2,3,-5)`C. `(-3,5,2)`D. `(-2,3,5)` |
| Answer» Correct Answer - C | |
| 247. |
Find the value of `m`for which thestraight line `3x-2y+z+3=0=4x=3y+4z+1`is parallel to the plane `2x-y+m z-2=0.`A. -2B. 8C. -18D. 11 |
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Answer» Correct Answer - a Vector `((3hati-2hatj+hatk)xx(4hati-3hatj+4hatk))` is perpendicular to `2hati-hatj+mhatk`. Thus, `|[3,-2,1],[4,-3,4],[2,-1,m]|=0orm=-2` |
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| 248. |
Consider a plane `x+y-z=1` and point `A(1, 2, -3)`. A line L has the equation `x=1 + 3r, y =2 -r and z=3+4r`. The coordinate of a point B of line L such that AB is parallel to the plane isA. `(10, -1, 15)`B. `(-5, 4, -5)`C. `(4, 1, 7)`D. `(-8, 5, -9)` |
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Answer» Correct Answer - d The line `(x-1)/(3)= (y-2)/(-1) = (z-3)/(4)= r` Any point say `B-= (3r+1, 2-r, 3+4r)` (on the line L) `vec(AB) = 3r, -r, 4r+6` Hence, `vec(AB)` is parallel to `x+y-z=1` `rArr" "3r-r-4r-6=0 or r=-3` `B` is `(-8, 5,-9)` |
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| 249. |
Find the value of `m`for which thestraight line `3x-2y+z+3=0=4x=3y+4z+1`is parallel to the plane `2x-y+m z-2=0.` |
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Answer» Given line is the line intersection of planes `3x-2y+z+3=0 and 4x-3y+4z+1=0`. Now vector along line of intersection is `" "|{:(hati,,hatj,,hatk),(3,,-2,,1),(4,,-3,,4):}|=-5hati-8hatj-hatk` Line is parallel to plane `2x-y+mz-2=0` So plane is perpendicular to vector `-5hati+8hatj-hatk`. Thus, `" "(2)(-5)+ (-1)(-8)+ (m)(-1)=0` or `" "m = -2` |
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| 250. |
Find the equation of the plane parallel to the plane x - 2y + 2z - 3 = 0, which is at a unit distance from (1,2,3) |
| Answer» Correct Answer - `x-2y+2z=0,z-2y+2z-6=0` | |