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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.
| 301. |
Find the co-ordinates of that point at which the lines joining the points `(1,1,2)` and `(3,5,-1)` meets the xy-plane. |
| Answer» Correct Answer - `(7/3,11/3,0)` | |
| 302. |
Find the co-ordinates of a point at which the line `(x+1)/(2) = (y-1)/(-2) = (z+5)/(6)`, meets the plane `x-2y+3z=8`. |
| Answer» Correct Answer - `(3,5,5)` | |
| 303. |
Find the co-ordinates of a point at which the line `(x+1)/(2) = (y-3)/(-2) = (z+5)/(6)`, meets the plane `3x-y+z=3`. |
| Answer» Correct Answer - `(1,1,1)` | |
| 304. |
Find the co-ordinates of that point at which the line joining the points `(-2,1,4)` and `(2,0,3)` meets the `yz-`plane. |
| Answer» Correct Answer - `(0,1/2,7/2)` | |
| 305. |
Let `A(1,1,1),B(23,5)a n dC(-1,0,2)`be three points, then equation of a planeparallel to the plane `A B C`which is at distance isa. `2x-3y+z+2sqrt(14)=0`b. `2x-3y+z-sqrt(14)=0`c. `2x-3y+z+2=0`d. `2x-3y+z-2=0`A. `2x-3y+z+2sqrt14=0`B. `2x-3y+z-2sqrt14=0`C. `2x-3y+z+2=0`D. `2x-3y+z-2=0` |
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Answer» Correct Answer - a A (1,1,1), B(2,3,5), C(-1,0,2) direction ratios of AB are `lt1,2,4gt`. Direction ratios of AC are `lt-2,-1,1gt`. Therefore, direction ratios of normal to plane ABC are `lt-2,-3,1gt` As a result, equation of the required plane be `2x-3y+z=0`. Let the equation of the requried plane is `2x-3y+z=k`. Then `|(k)/(sqrt(4+9+1))|=2ork=pm2sqrt14` Hence, equation of the required plane is `2x-3y+z+2sqrt14=0` |
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| 306. |
The line joining the points (4, 1, 2) and (5, p, 0) is parallel to the lne joining the points (2, 1, 1) and (3, 3, -1), then p =A. -1B. 1C. -3D. 3 |
| Answer» Correct Answer - D | |
| 307. |
Find the equation of theplane through the points `(23,1)a n d(4,-5,3)`and parallel to the x-axis. |
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Answer» Plane passes through the points `(2, 3, 1) and (4, -5, 3)` Therefore, plane is parallel to vector `" "(4-2)hati+(-5-3)hatj+ (3-1)hatk=2hati-8hatj+2hatk` Also plane is parallel to the x-axis, so it parallel to vector `hati` `therefore` Normal to the plane = `|{:(hati,,hatj,,hatk),(2,,-8,,2),(1,,0,,0):}|=2hatj+8hatk` Hence, equation of plane through point (2, 3, 1) and normal to vector `2hatj+8hatk` is `" "0(x-2)+2(y-3)+8(z-1)=0` or `" "y+4z-7=0` |
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| 308. |
Find the distance of the point `veca` from the plane `vecr*hatn=d` measured parallel to the line `vecr=vecb+vec(tc)`. |
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Answer» A line parallel to `vecc` and passing through `A(veca)` is `vecr=veca+lamdavecc` A point for some `lamda` satisfies plane `" "(veca+lamdavecc)*hatn=d` or `" "lamda=(vecd-veca*hatn)/(vecc*hatn)` `therefore" "` Distance `= (|vecd-hata*hatn|)/(|vecc*hatn|)|vecc|`. |
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| 309. |
If two vertices of a triangle are (1,3) and (4,-1) and the area of triangle is 5 sq. units, then the angle at the third vertex lies in : |
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Answer» `A=1/2*b*h` `5=1/2*5*h` `h=2 unit` from diagram AC=BC AD=Db=5/2 `tantheta=(5/2)/2=5/4` `theta=tan^(-1)5/4` `/_C=2tan^(-1)5/4` `C in (0,2tan^(-1)5/4)`. option a is correct. |
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| 310. |
Ifa line makes an angle of `pi/4`with the positive directions of each ofx-axis and y-axis, then the angle that the line makes with the positivedirection of the z-axis is(1) `pi/6`(2)`pi/3`(3) `pi/4`(4)`pi/2` |
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Answer» If`l,m,n`is direction of a line then `l^2+m^2+n^2=0` and if line make`(alpha,beta,gamma)`with Co-ordinate xis `l=Cosalpha` `m=Cosbeta` `n=Cosgamma` `Cos^2alpha+Cos^2beta+Cos^2gamma=1` `1/2+1/2+Cos^2gamma=1` `Cos^2gamma=0` `gamma=pi/2` option`2=pi/2` |
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| 311. |
If two distinct chords of a parabola `y^2=4ax` , passing through (a,2a) are bisected by the line x+y=1 ,then length of latus rectum can be |
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Answer» equation of chord with mid-point given: `y(1-t)-2a(x+t)=(1-t)^2-4at` `2a(1-t)-2a(a+t)=(1-t)^2-4at` `t^2-2t+1+2a^2-2a)=0` `4-4(1+2a^2-2a)>0` `1-1-2a^2+2a>0` `-2a(a-1)>0` `a<0` `a>0` and`0<1` `0lta<1` `0<4a<4` option a is correct. |
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| 312. |
Find the equation of the sphere passing through (0, 0, 0), (1, 0, 0) and (0, 0, 1). |
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Answer» Let the equation of the sphere be `" "x^(2)+y^(2)+z^(2)+2ux+2vy+2wx=0" "`(i) As (i) passes through (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1), we must have `d=0, 1+2u+d=0` `1+2v+d=0 and 1+2w+d=0` Since `d=0`, we get `2u=2v=2w=-1` Thus, the equation of the required sphere is `x^(2)+y^(2)+z^(2)-x-y-z=0` |
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| 313. |
Find the equation of a plane passes through the points `(0,0,0)` and `(1,3,5)` and parallel to the line `(x)/(-2) = y/1 = (z+3)/(4)`. |
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Answer» Let equation of the plane passing through the point `(0,0,0)` is `a(x-0)+b(y-0)+c(z-0) = 0` `rArr ax+ by+cz = 0 "….."(1)` This plane passes throough the point `(1,3,5)`. `:. a+3b+5+5c=0"……"(2)` This plane is parallel to the line `(x)/(-2) = (y)/(1) = (z+3)/(4)`. `:. -2a+b+4c = 0 "....."(3)` Solving eqs. (2) and (3) by cross multiplication method. `(a)/(12-5) = (b)/(-10-4) = (c)/(1+6)` `rArr a/7 = (b)/(-14) = c/1 = k` `:. a = k, b = - 2 k, c = k` From eq. (1) `kx - 2ky + kz = 0` `rArr x - 2y + z = 0` Which is the requried equation of the plane. |
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| 314. |
If the line `vecr = (2hati+ hatj - hatk) + lambda(hati+mhatj-2hatk)` is parallel to the plane `vecr.(2hati+hatj+mhatk) = 1` then find the value of m. |
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Answer» Here, `vecb=hati+mhatj-2hatk` and `vecn = 2hati+hatj+mhatk` If line is parallel to the plane, then `sin theta = 0` `rArr vecb.vecn = 0` `rArr (hati+mhatj-2hatk).(2hati+hatj+mhatk) = 0` `rArr 2+m - 2 m = 0` `rArr m = 2`. |
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| 315. |
Find the angle between the planes `vecr.(hati+hatj-2hatk)=3 and vecer.(2hati-2hastj+hatk)=2`2 |
| Answer» Correct Answer - (i) `2x+y-z=3` , (ii) `x+3y+2z+1=0` | |
| 316. |
Find the equation of thesphere which passes through `(10,0),(0,1,0)a n d(0,0,1)`and whose centre lies onthe plane `3x-y+z=2.` |
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Answer» Let the equation of the sphere be `x^(2)+y^(2)+z^(2)+2ux+2vy+2wz+d=0`. As the sphere passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1), we get `" "1+2u+d=0, 1+2v+d=0 and 1+2w+d=0` `rArr" "u=v=w=-(d+1)/(2)` Since the centre `(-u, -v, -w)` lies on the plane `3x-y+z=2`, we get `-3u+v-w=2` `rArr" "(3)/(2)(d+1)=2 or d+1=(4)/(3) or d=(1)/(3)` Thus, `u=v=w=-2//3` Therefore, the equation of the required sphere is `x^(2)+y^(2)+z^(2)-((2)/(3))x-((2)/(3))y-((2)/(3))z+(1)/(3)=0` or `" "3(x^(2)+y^(2)+z^(2))-2(x+y+z)+1=0` |
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| 317. |
Show that the lines `(x-a+d)/(alpha-delta)=(y-a)/alpha=(z-a-d)/(alpha+delta)`and `(x-b+c)/(beta-gamma)=(y-b)/beta=(z-b-c)/(beta+gamma)`are coplanar. |
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Answer» Given lines are ` (x-a+d)/(alpha-delta)= (y-a)/(alpha)= (z-a-d)/(alpha+delta), (x-b-c)/(beta-gamma)= (y-b)/(beta)=(z-b-c)/(beta+gamma)` Now `" "|{:(a-d-b+c,,a-b,,a+d-b-c),(alpha-delta,,alpha,,alpha+delta),(beta-gamma,,beta,,beta+gamma):}|` `" "=|{:(-d+c,,a-b,,d-c),(-delta,,alpha,,delta),(-gamma,,beta,,gamma):}|=0` `" " ("Applying " C_(1)to C_(1)-C_(2) and C_(3)to C_(3)-C_(2))` Hence, the lines are coplaner. Equation of plane containing these lines is `" "|{:(x-a+d,,y-a,,z-a-d),(alpha-delta,,alpha,,alpha+delta),(beta-gamma,,beta,,beta+gamma):}|=0` `" "|{:(x-2y+z,,y-a,,z-a-d),(0,,alpha,,alpha+delta),(0,,beta,,beta+gamma):}|=0` `" "("Applying "C_(1)to C_(1)+C_(3)-2C_(2))` or `" "(x-2y+z)(alphagamma-betadelta)=0` or `" "x-2y +z=0 " "[because (alphagamma ne betadelta)]` |
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| 318. |
Find the equation of asphere which passes through `(1,0,0)(0,1,0)a n d(0,0,1),`and has radius as small as possible. |
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Answer» Let the equation of the required sphere be `x^(2) +y^(2)+z^(2)+2ux+2vy+2wz+d=0" "` (i) As the sphere passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1), we get `" "1+2u+d=0, 1+2v+d=0 and 1+2w+d=0` `rArr" "u=v=w=-(1)/(2)(d+1)` If R is the radius of the sphere, then `" "R^(2)=u^(2)+v^(2)+w^(2)-d` `" "=(3)/(4)(d+1)^(2)-d` `" "=(3)/(4)[d^(2)+2d+1-(4)/(3)d]` `" "=(3)/(4)[d^(2)+(2)/(3)d+1]` `" "(3)/(4)[(d+(1)/(3))^(2)+1-(1)/(9)]` `" "=(3)/(4)[(d+(1)/(3))^(2)+(8)/(9)]` The last equation shows that `R^(2)` (and thus `R`) will be the least if and only if `d=-1//3`. Therefore, `" "u=v=w=-(1)/(2)(1-(1)/(3))=-(1)/(3)` Hence, the equation of the required sphere is `x^(2)+y^(2)+z^(2)-(2)/(3)(x+y+z)-(1)/(3)=0` `" "3(x^(2)+y^(2)+z^(2))-2(x+y+z)-1=0` |
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| 319. |
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. `4sqrt(26)`B. `20`C. `10 sqrt(13)`D. none of these |
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Answer» Correct Answer - d The distance of point `(1+3r, 2-r, 3+4r)` from the plane is `(|1+3r+2-r-3-4r-1|)/(sqrt(1+1+1))= (|2r+1|)/(sqrt3)= (4)/(sqrt3)` `rArr " "r= (3)/(2), -(5)/(2)` Hence, the points are `A((11)/(2), (1)/(2), (10)/(2)) and B ((-13)/(2), (9)/(2), (-14)/(2))` `rArr" "AB= sqrt(292)` |
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| 320. |
Let `P_(1)` denote the equation of a plane to which the vector `(hati+hatj)` is normal and which contains the line whose equation is `vecr=hati+hatj+veck+lamda(hati-hatj-hatk)andP_(2)` denote the equation of the plane containing the line L and a point with position vector j. Which of the following holds good?A. The equation of `P_(1)" is"x+y=2.`B. The equation of `P_(2)" is "vecr.(hati-2hatj+hatk)=2.`C. The acute angle the `P_(1)andP_(2)" is "cot^(-1)(sqrt3)`.D. The angle between the plnae `P_(2)` and the line L is `tan^(-1)sqrt3`. |
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Answer» Correct Answer - a, c Plane `P_(1)` contains the line `vecr=hati+hatj+hatk+lamda(hati-hatj-hatk)`, hence contains the point `hati+hatj+hatk` and is normal to vector `(hati+hatj)`. Hence, equation of plane is `" "(vecr-(hati+hatj+hatk))*(hati+hatj)=0` or `" "x+y=2` Plane `P_(2)` contains the line `" "vecr=hati+hatj=hatk+lamda(hati-hatj-hatk)` and point `hatj` Hence, equation of plane is `|{:(x-0,,y-1,,z-0),(1-0,,1-1,,1-0),(1,,-1,,-1):}|=0` or `" "x+2y-z=2` If `theta` is the acute angle between `P_(1) and P_(2)`, then `" "costheta= (vec(n_1)*vec(n_2))/(|vec(n_1)||vec(n_2)|)=|((hati+hatj)*(hati+2hatj-hatk))/(sqrt2*sqrt6)|` `" "= (3)/(sqrt2*sqrt6)= (sqrt(3))/(2)` `" "theta=cos^(-1)""(sqrt(3))/(2)= (pi)/(6)` As `L` is containec in `P_(2) rArr theta=0` |
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| 321. |
Find the length and thefoot of the perpendicular from the point `(7,14 ,5)`to the plane `2x+4y-z=2.` |
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Answer» The required length `=(2(7)+4(14)-(5)-2)/(sqrt(2^(2)+4^(2)+1^(2)))=(14+56-5-2)/(sqrt(4+16+1))=(63)/(sqrt(21))` Let the coordinates of the foot of the perpendicular from the point `P(7, 14, 5)` be `M(alpha, beta, gamma)`. Then the direction ratios of PM are `alpha-7, beta-14 and gamma-5`. Therefore, the direction ratios of the normal to the plane are `alpha-7, beta-14, and gamma-5`. But the direction ratios of normal to the given plane `2x+4y-z=2` are 2, 4 and -1. Hence, `" "(alpha-7)/(2)=(beta-14)/(4)=(gamma-5)/(-1)=k` `therefore" "alpha=2k+7, beta=4k+14 and gamma=-k+5" "`(i) Since`alpha, beta and gamma` lie on the plane `2x+4y-z=2, 2alpha+4beta-gamma=2` or `" "2(7+2k)+4(14+4k)-(5-k)=2` or `" "14+4k+56+16k-5+k=2` or `" "21k=-63` or `" " k=-3` Now, putting `k=-3` in (i), we get `" "alpha=1, beta=2, gamma=8` Hence, the foot of the perpendicular is (1, 2, 8). |
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| 322. |
Find the number of sphereof radius `r`touching thecoordinate axes. |
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Answer» Correct Answer - `8` Obviously one in each octant. |
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| 323. |
Find the ratio in which the`y-z`plane divides the join ofthe points `(-2,4,7)a n d(3,-5,8)dot` |
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Answer» Let the `y-z` plane divide the join of P(-2, 4, 7) and Q(3, -5, 8) in the ratio `lamda:1`. If `((3lamda-2)/(lamda+1), (-5lamda+4)/(lamda+1), (8lamda+7)/(lamda+1))` is in the y-z plane, then its x-coordinate is zero. Therefore, `" "(3lamda-2)/(lamda+1)=0or 3lamda-2=0` or `" "lamda=2//3` |
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| 324. |
A line passes through thepoints `(6,-7,-1)a n d(2,-3,1)dot`Find te direction cosinesoff the line if the line makes an acute angle with the positive direction ofthe x-axis. |
| Answer» Let `l, m and n` be the direction cosines of the given line. As it makes an acute angle with the x-axis, `l gt 0`. The line passes through (6, -7, -1) and (2, -3, 1), therefore, its direction ratios are (6, -2, -7+3, -1-1) or (4, -4, -2). Hence, the direction cosines of the given line are 2/3, -2/3 and -1/3. | |
| 325. |
Find the locus of a point,the sum of squares of whose distance from the planes `x-z=0,x-2y+z=0a n dx+y+z=0i s36`. |
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Answer» Given planes are `" "x-z=0, x-2y+z=0, x+y+z=0` Let the point whose locus is required be `P(alpha, beta, gamma)`. According to question, `" "(|alpha-gamma|^(2))/(2)+(|alpha-2beta+gamma|^(2))/(6)+(|alpha+beta+gamma|^(2))/(3)=36` or `" "3(alpha^(2)+gamma^(2)-2alphagamma)+alpha^(2)+4beta^(2)+gamma^(2)-4alphabeta-4betagamma+2alphagamma` `" "+2(alpha^(2)+beta^(2)+gamma^(2)+2alphabeta+2betagamma+2alphagamma)=36xx6` or `" "6alpha^(2)+6beta^(2)+6gamma^(2)=36xx6` or `" " alpha^(2)+beta^(2)+gamma^(2) = 36` Hence, the required equation of locus is `x^(2)+y^(2)+z^(2)=36`. |
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| 326. |
If the sum of the squaresof the distance of a point from the three coordinate axes is 36, then findits distance from the origin. |
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Answer» Let P(x, y, z) be the point. Now, under the given condition, `" "[sqrt(x^(2)+y^(2))]^(2)+[sqrt(y^(2)+z^(2))]^(2)+[sqrt(z^(2)+x^(2))]^(2)=36` or `" "x^(2)+y^(2)+z^(2)=18` Then distance from the origin to point (x, y, z) is `" "sqrt(x^(2)+y^(2)+z^(2))=sqrt(18)=3sqrt(2)` |
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| 327. |
If the planes `vecr.(hati+hatj+hatk)=q_(1),vecr.(hati+2ahatj+hatk)=q_(2)andvecr.(ahati+a^(2)hatj+hatk)=q_(3)`intersect in a line, the the value of a isA. 1B. `1//2`C. 2D. 0 |
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Answer» Correct Answer - a, b `vecr*vec(n_1)=vec(q_1) and vecr*vec(n_2)=vecq_(2), vecr*vecn_(3)= vecq_(3) ` intersect in a line if `[vecn_(1) vecn_(2) vecn_(3)]=0`. So, `" "|{:(1,,1,,1),(1,,2a,,1),(a,,a^(2),,1):}|=0` or `" "2a-a^(2)-1+a+a^(2)-2a^(2)=0` or `" "2a^(2)-3a+1=0` or `" "a=1//2, 1` |
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| 328. |
Statement 1: Line `(x-1)/1=(y-0)/2=(z2)/(-1)`lies in the plane `2x-3y-4z-10=0.`Statement 2: if line ` vec r= vec a+lambda vec b`lies in the plane ` vec rdot vec c=n(w h e r en`is scalar`),t h e n vec bdot vec c=0.`A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
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Answer» Correct Answer - c Statement 2 is true as when the line lies in the plane, vector `vecb` along which the line is directed is perpendicular to the normal `vecc` of the plane, but it does not explain Statement 1 as for `vecb * vecc= 0`, the line may be parallel to the plane. However, Statement 1 is correct as any point on the line `(t+1, 2t, -t-2)` lies on the plane for `t in R`. |
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| 329. |
Find the locus of appoint which moves such that the sum of the squares ofits distance from the points `A(1,2,3),B(2,-3,5)a n dC(0,7,4)i s120.` |
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Answer» Let `P(x,y, z)` be any point on the locus. Then `PA^(2)+PB^(2)+PC^(2)=120` `rArr " "(x-1)^(2)+(y-2)^(2)+(z-3)^(2)+(x-2)^(2)+(y+3)^(2)+(z-5)^(2) +(x-0)^(2)+(y-7)^(2)` `" "+(z-4)^(2)= 120` `" "3x^(2)+3y^(2)+3z^(2)-6x-12y-24z+117=120` `" "x^(2)+y^(2)+z^(2)-2x-4y-8z-1=0` This represents a sphere with centre at (1, 2, 4) and radius equal to `sqrt(1^(2)+2^(2)+4^(2)+1) =sqrt(22)`. |
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| 330. |
Find the angle between anytwo diagonals of a cube.A. `cos^(-1)((sqrt(2))/(3))`B. `cos^(-1)((1)/(sqrt(3)))`C. `cos^(-1)((2)/(3))`D. `cos^(-1)((1)/(3))` |
| Answer» Correct Answer - D | |
| 331. |
If he line drawn from the point `(-2,-1,-3)`meets a plane at right angle at the point `(1,-3,3),`find the equation of the plane. |
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Answer» Since the line drawn from the point `(-2,-1,-3)` meets a plane at right angle at the point `(1,-3,3)`. So the plane passes through the point `(1,-3,3)` and normal to plane is `(-3hati+2hatj-5hatk)` ltrbgt `implies veca=hati-3hatj+3hatk` and `vecN=-3hati+2hatj-6hatk` So the equation of required plane is `(vecr-veca).vecN=0` `implies[(xhati+yhatj+zhatk)-(hati-3hatj+3hatk)].(-3hatk+2hatj-6hatk)=0` `implies [(x-1)hati+(y+3)hatj+(z-3)hatk].(-3hati+2hatj-6hatk)=0` `implies -3x+3+2y+6-6z+18=0` `implies -3x+2y-6z=-27` `:. 3x-2y+6z-27=0` |
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| 332. |
Find the vector and cartesian equation of a line passes through the points `(1,3,2)` and origin. |
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Answer» Cartesian equation of a line passing through two given points `(x-x_(1))/(x_(2) - x_(1))= (y - y_(1))/(y_(2) - y_(1)) = (z - z_(1))/(z_(2) - z_(1))` Therefore equation of a line passing through `(0,0,0)` and `(1,3,2)` is `(x-0)/(1-0) = (y-0)/(3-0) = (z-0)/(2-0)` `rArr x/1 = y/3 = z/2` Position vectors of points `(0 ,0,0)` and `(1,3,2)` are respectively. `vecr_(1) = 0.hati+0.hatj+0.hatk = vec0` and `vecr_(2) = hati+3hatj+2hatk` Equation of a line passint through two points whose position vectros are `vecr_(1)` and `vecr_(2)` is `vecr = vecr_(1) + lambda(vecr_(2) + vecr_(1))` `rArr vecr = vec0+lambda(hati+3hatj+2hatk-hat0)` `rArr vecr= lambda(hati+3hatj+2hatk)` |
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| 333. |
Statement 1: Let `theta`be the angle between the line `(x-2)/2=(y-1)/(-3)=(z+2)/(-2)`and the plane `x+y-z=5.`Then `theta=sin^(-1)(1//sqrt(51))dot`Statement 2: The anglebetween a straight line and a plane is the complement of the angle betweenthe line and the normal to the plane.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
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Answer» Correct Answer - c `sin theta = |(2-3+2)/(sqrt(4+9+4)sqrt(3))|= (1)/(sqrt(51))` Therefore, Statement 1 is true and Statement 2 is also true by definition. |
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| 334. |
Statement 1: there exists aunique sphere which passes through the three non-collinear points and whichhas the least radius.Statement 2: The centre ofsuch a sphere lies on the plane determined by the given three points.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
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Answer» Correct Answer - c Obiviously the answer is (b). |
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| 335. |
Find the equation of thesphere described on the joint of points `Aa n dB`having position vectors `2 hat i+6 hat j-7 hat ka n d-2 hat i+4 hat j-3 hat k ,`respectively, as the diameter. Find the center and theradius of the sphere. |
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Answer» If point `P` with positive vector `vecr=xhati+yhatj+hatk` is any point on the sphere, then `" "vec(AP)*vec(BP)=0` `" "(x-2)(x+2)+(y-6)(y-4)+(z+7)(z+3)=0` or `" "(x^(2)-4)+(y^(2)-10y+24)+(z^(2)+10z+21)=0` or `" "x^(2)+y^(2)+z^(2)-10y+10z+41=0` The centre of this sphere is (0, 5, -5) and its radius is `sqrt(5^(2)+(-5)^(2)-41)=sqrt(9)=3` |
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| 336. |
A point `P(x ,y ,z)`is such that `3P A=2P B ,`where `Aa n dB`are the point `(1,3,4)a n d(1,-2,-1),`erespectivley. Find the equation to the locus of the point`P`and verify that the locus is a sphere. |
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Answer» Since `3PA= 2PB`, we get `9PA^(2)= 4PB^(2)` `" " 9[(x-1)^(2)+(y-3)^(2) + (z-4)^(2)]` `" "= 4[x-1^(2)+ (y+2)^(2)+ (z+1)^(2)]` `" "9[x^(2)+y^(2)+z^(2)-2x-6y--8z+26]` `" "=4[c^(2)+y^(2)+z^(2)-2x+4y+2z+6]` `" "5x^(2)+5y^(2)-10x-70y-80z+210=0` `" "x^(2)+y^(2)=z^(2)-2x-14y-16z+42=0` This represents a sphere with centre at (1, 7, 8) and radius equal to `sqrt(1^(2-)+7^(2)+8^(2)-42)= sqrt(72) = 6sqrt(2)` |
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| 337. |
The angle between diagonal of a cube and diagonal of a face of the cube will beA. `cos^(-1)((2)/(3))`B. `cos^(-1)((1)/(3))`C. `cos^(-1)sqrt((2)/(3))`D. `cos^(-1)sqrt((1)/(3))` |
| Answer» Correct Answer - C | |
| 338. |
A line makes an angel `theta`with each of thex-and z-axes. If the angel `beta,`which it makes with the y-axis, is such that`sin^2beta=3sin^2theta,t h e ncos^2theta`equalsa. `2/3`b. `1/5`c. `3/5`d. `2/5`A. `(2)/(3)`B. `(1)/(5)`C. `(3)/(5)`D. `(2)/(5)` |
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Answer» Correct Answer - c Here `sin^(2)beta=3sintheta" "(i)` By the question, `cos^(2)theta+cos^(2)theta+cos^(2)beta=1" "(ii)` `impliescos^(2)beta=1-2cos^(2)theta" "(iii)` Adding (i) and (iii), we get `1=1+3sin^(2)theta-2costheta` `=1+3(1-cos^(2)theta)-2cos^(2)theta` or `5cos^(2)theta=3` or `cos^(2)theta=(3)/(5)` |
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| 339. |
Statement 1: let `A( vec i+ vec j+ vec k)a n dB( vec i- vec j+ vec k)`be two points. Thenpoint `P(2 vec i+3 vec j+ vec k)`lies exterior to the sphere with `A B`as its diameter.Statement 2: If `Aa n dB`are any two points and `P`is a point in space such that ` vec P Adot vec P B >0`, then point `P`lies exterior to the sphere with `A B`as its diameter.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
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Answer» Correct Answer - b `vec(PA)*vec(PB)= 9 gt 0`. Therefore, P is exterior to the sphere. Statement 2 is also true (standard result). |
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| 340. |
A line is drawn through a fix point P(`alpha, beta`) to cut the circle `x^2 + y^2 = r^2` at A and B. Then PA.PB is equal to : |
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Answer» Point of power= PA*PB=PC*PC=`PT^2` In`/_PTO` `OT^2+PT^2=PO^2` `PT^2=2^2+B^2-r^2` `PA*PB=PT^2=2^2+B^2=r^2`. |
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| 341. |
IF the direction ratios of two vectors are connected by the relations `p + q + r = 0` and `p^(2) + q^(2) - r^(2) = 0.` Find the angle between them.A. `90^(@)`B. `60^(@)`C. `45^(@)`D. `30^(@)` |
| Answer» Correct Answer - B | |
| 342. |
The cartesian equation of a line is `(x+2)/(1) = (y+3)/(-2) = (z)/(3)`, find its vector equation. |
| Answer» Correct Answer - `vecr= (-2hati+3hatj)+lambda(hati-2hatj+3hatk)` | |
| 343. |
Prove that the lines through `A(0,-1,-1)a n dB(4,5,1)`intersecrs the line through `C(3,9,2)a n dD(-4,4,4)dot`Also, find their point of intersection. |
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Answer» We know that the cartesian equation of a line that passes through two points `(x_(1),y_(1),z_(1))` and `(x_(2),y_(2),z_(2))` is `(x-x_(1))/(x_(2)-x_(1))=(y-y_(1))/(y_(2)-y_(1))=(z-z_(1))/(z_(2)-z_(1))` Hence the cartesian equation of line passes through `A(0,-1,-1)` and `B(4,5,1)` is `(x-0)/(4-0)=(y+1)/(5+1)=(z+1)/(1+1)` `impliesx/4=(y+1)/6=(z+1)/2`.................i and cartesian equation of the line passes through `C(3,9,4)` and `D(-4,4,4)` is `(x-3)/(-4-3)=(y-9)/(4-9)=(z-4)/(4-4)` `implies (x-3)/(-7)=(y-9)/(-5)=(z-4)/0`..................ii If the lines intersect, then shortest distance between both of them should be zero. `:.` Shortest distance between the lines `=(|(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((b_(1)c_(2)-b_(2)c_(1))^(2)+(c_(1)a_(2)-c_(2)a_(1))^(2)+(a_(1)b_(2)-a_(2)b_(1))^(2))` ltbgt `=(|(3-0,9+1,4+1),(4,6,2),(-7,-5,0)|)/(sqrt((6.0+10)^(2)+(-14-0)^(2)+(-20+42)^(2))` `=(|(3,10,5),(4,6,2),(-7,-5,0)|)/(sqrt(100+196+484))` `=(3(0+10)-10(14)+55(-20+42))/(sqrt(780))` `=(30-140+110)/(sqrt(780))=0` So, the given lines intersect. |
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| 344. |
Find the equation of a line passes through the point `hati+hatj+5hatk` and parallel to line joining the points `(2,-4,1)` and `(0,1,3)`. |
| Answer» Correct Answer - `vecr = (hati+hatj+5hatk)+lambda(2hati-5hatj-2hatk)` | |
| 345. |
Find the value of `lambda`. If the points `A(-1,3,2), B(-4,2,-2)` and `C(5,lambda,10)` are collinear. |
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Answer» `(x+1)/(-4-(-1))= (y-3)/(2-3) = (z-2)/(2-2)` `rArr (x+1)/(-3) = (y-3)/(-1) = (z-2)/(1)` `rArr (x+1)/(3) = (y-3)/(1) = (z-2)/(4) "….."(1)` `:. A, B` and C points are collinear. `:.` Point C will lie on `(1)` `(5+1)/(3) = (lambda - 3)/(1) = (10-2)/(4)` `rArr 2 = lambda - 3 = 2` `rArr lambda = 5`. |
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| 346. |
Direction ratios of twolines are `a ,b ,ca n d1//b c ,1//c a ,1//a bdot`Then the lines are ______. |
| Answer» Since `(a)/((1//bc))=(b)/((1//ca))=(c)/((1//ab))`, lines are parallel. | |
| 347. |
Find the vector equation of the line through `A(3,4,-7)a n dB(1,-1,6)dot`Find also, its Cartesian equations. |
| Answer» Correct Answer - `(x-3)/(-2)=(y-4)/(-5)=(z+7)/(13)` | |
| 348. |
Direction ratios of two lines satisfy the relations `2a-b+2c=0` and `ab+bc+ca=0`. Then the angle between two lines isA. `60^(@)`B. `30^(@)`C. `90^(@)`D. `145^(@)` |
| Answer» Correct Answer - C | |
| 349. |
Find the equation of a line passes through the points whose position vectors are `(hati+4hatj+hatk)` and `(2hati-hatj+5hatk)`. |
| Answer» Correct Answer - `vecr=(hati+4hatj+hatk)+lambda(hati-5hatj+4hatk)` | |
| 350. |
Find the vector equation of a lin e passes through the point whose position vector is `(2hati-hatj-hatk)` and parallel to vector `hati+5hatk`. |
| Answer» Correct Answer - `vecr=(2hati-hatj-hatk)+lambda(hati+5hatk)` | |