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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.
| 251. |
The distance of the point `P(3, 8, 2)` from the line `(x-1)/2 = (y-3)/4=(z-2)/3` measured parallel to the plane `3x+2y-2z+17=0` is |
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Answer» plane given is `3x+2y - 2z + 17=0` `3x+2y-2z+c=0-> (3,8,2)` `9+16-4 + c =0` `c=-21` so eqn of plane will be `3x + 2y-2z-21=0` for coordinates lets the points are equal to `lamda` so, `x=2lamda +1` `y= 4 lamda+3` `z= 3 lamda + 2` putting it in eqn we get `3(2 lamda+1) + 2(4 lamda+3) - 2( 3lamda+2) - 21=0` `6 lamda + 3 + 8 lamda+ 6 - 6 lamda - 4 -21=0` `E= sum_(r=1)^n(a+b) omega^(r-1)` `= (a+b) +((2a+b) omega + (3a+b)omega^2 + (4a+b) omega^3 + .... + (na+b)omega^(n-1)` `=b (1+ omega + omega^2 + .....+ omega^(n-1)) + a(1+ 2omega + 3omega^2 + 4 omega^3 + .... + n omega^(n-1)` S`= 1 + 2omega + 3omega^2 + 4omega^3 + ..... + nomega^(n-2)` S`omega= omega + 2omega^2 + 3omega^3 + .... + nomega^n` subtracting these equations we get so,`lamda = 2` so points will be like `x= 5` `y=11` `z=8` now we have to find the distance from point `(3,8,2)` of point (5,11,8) `` `E= sum_(r=1)^n(a+b) omega^(r-1)` `= (a+b) +((2a+b) omega + (3a+b)omega^2 + (4a+b) omega^3 + .... + (na+b)omega^(n-1)` `=b (1+ omega + omega^2 + .....+ omega^(n-1)) + a(1+ 2omega + 3omega^2 + 4 omega^3 + .... + n omega^(n-1)` S`= 1 + 2omega + 3omega^2 + 4omega^3 + ..... + nomega^(n-2)` S`omega= omega + 2omega^2 + 3omega^3 + .... + nomega^n` subtracting these equations we get `d= sqrt(4+9+36) =7` option 1 is correct |
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| 252. |
Find the equation of aplane which is parallel to the plane `x-2y+2z=5`and whose distance fromthepoint `(1,2,3)`is 1. |
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Answer» The equation of the plane parallel to `" "x-2y+2z=5` is `x-2y+2z+k=0" "` (i) Now, according to the equation, `" "(1-4+6+k)/(sqrt(9))=pm1` `" "k+3=pm3 rArr k=0 or -6` Then `x-2y+2z-6=0 or x-2y+2z=6` |
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| 253. |
If a line makes angles of measure `60^(@)` and `45^(@)` with OX and OZ, then the angle made by line with Y-axis isA. `45^(@)`B. `0^(@)`C. `30^(@)`D. `60^(@)` |
| Answer» Correct Answer - D | |
| 254. |
If `cos alpha=(sqrt(3))/(2), cos gamma=(-1)/(2)`, then the direction angle of lines areA. `(pi)/(6), (pi)/(4),(pi)/(3)`B. `(5pi)/(6),(pi)/(2),(2pi)/(3)`C. `(pi)/(6), (pi)/(2),(2pi)/(3)`D. `(pi)/(6),(pi)/(2),(pi)/(3)` |
| Answer» Correct Answer - C | |
| 255. |
Let `l_(1) and l_(2)` be the two skew lines. If P, Q are two distinct points on `l_(1)` and R, S are two distinct points on `l_(2)`, then prove that PR cannot be parallel to QS. |
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Answer» Let equation of the line `l_(1)` be `vecr=veca+lamdavecb` and equation of the line `l_(2)` be `vecr=vecc+muvecd`, where `veca-vecc, vecb and vecd` are non-coplanar. Let the position vectors of points P and Q be `veca+lamda_(1)vecb and vecaa+lamda_(2)vecb`, respectively. Let the position vectors of points R and S be `vecc+mu_(1)vecd and vecc+mu_(1)vecd`, respectively. Then the lines PR and QS are parallel if and only if `vecc-veca+mu_(1)vecd-lamda_(1)vecb=k(vecc-veca+mu_(2)vecd-lamda_(2)vecb)` i.e., `" "(1-k)(vecc-veca)+(mu_(1)-kmu_(2))vecd - (lamda_(1)-klamda_(2))vecb=veco` `therefore" "1-k=0, mu_(1)-kmu_(2)=0, lamda_(1)-klamda_(2)=0` i.e., `" "mu_(1)=mu_(2) and lamda_(1)=lamda_(2)` which is not possible Therefore, PR can not be parallel to QS. |
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| 256. |
If the lines `(x-1)/2=(y+1)/3=(z-1)/4 and (x-3)/1=(y-k)/2=z/1` intersect then the value of k is (A) `3/2` (B) `9/2` (C) `-2/9` (D) `-3/2`A. `3//2`B. `9//2`C. `-2//9`D. `-3//2` |
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Answer» Correct Answer - b `(x-1)/(2)= (y+1)/(3)= (z-1)/(4)=lamda` `rArr" "x=2lamda+1, y=3lamda-1 and z=4lamda+1` `" "(x-3)/(1)= (y-k)/(2)= (z)/(1)= mu` `rArr" "x=3+mu, y=k+2mu and z=mu ` Since the above lines intersect, `" "2lamda+1= 3+mu" " `(i) `" "3lamda-1= 2mu +k" "` (ii) `" "mu = 4lamda +1" "` (iii) Solving (i) and (iii) and putting the value of `lamda and mu` in (ii), `k = 9//2`. |
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| 257. |
If a line makes angles of measure `45^(@)` and `60^(@)` with the positive direction of the Y and Z axex respectively, then the angle made by the line with the positive directions of the X-axis isA. `30^(@)" or " 150^(@)`B. `30^(@)" or " 120^(@)`C. `60^(@)" or " 120^(@)`D. `60^(@)" or " 150^(@)` |
| Answer» Correct Answer - C | |
| 258. |
If a line is inclined at `60^(@)` and `30^(@)` with the X-and Y-axes respectively, then the angle which makes with the Z-axes is |
| Answer» Correct Answer - C | |
| 259. |
The lines which intersectthe skew lines `y=m x ,z=c ; y=-m x ,z=-c`and the x-axis lie on the surfacea. `c z=m x y`b. `x y=c m z`c. `c y=m x z`d. none of theseA. `cz=mxy`B. `xy=cmz`C. `cy=mxz`D. none of these |
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Answer» Correct Answer - c Equations of the planes through `y=mx,z=candy=-mx,z=-c` are respectively, `(y-mx)+lamda_(1)(z-c)=0" "(i)` and `(y+mx)+lamda_(2)(z+c)=0" "(ii)` It meets at x-axis, i.e., y=0=z. `becauselamda_(2)=lamda_(1)` From (i) and (ii), `(y-mx)/(z-c)=(y+mx)/(z+c)` `becausecy=mzx` |
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| 260. |
A line makes angles of measures `(pi)/(6)` and `(pi)/(3)` with X-and Z-axes respectively. Find the angle made by the line with the Y-axis.A. `(pi)/(2)`B. `(2pi)/(3)`C. `(5pi)/(6)`D. `(3pi)/(4)` |
| Answer» Correct Answer - A | |
| 261. |
The centre of the circle given by `vecr.(hati+2hatj+2hatk)=15and|vecr-(hatj+2hatk)=4` isA. (0,1,2)B. (1,3,4)C. (-1,3,4)D. none of these |
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Answer» Correct Answer - b The equation of the line through the center `hatj+2hatj` and normal to the given plane is `vecr=hatj=2hatk+lamda(hati+2hatj+2hatk)" "(i)` This meets the plane for which `[hatj+2hatk+lamda(hati+2hatj+2hatk)].(hati+2hatj+2hatk)=15` or `6+9lamda=15orlamda=1` Putting in (i), we get `vecr=hatj+2hatk+(hati+2hatj+2hatk)=hati+3hatj+4hatk` Hence, centre is (1,3,4). |
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| 262. |
The dr. of normal to the plane through `(1,0,0), (0,1,0)` which makes an angle `pi/4` with plane , `x+y=3` areA. `lt1,sqrt2,1gt`B. `lt1,1,sqrt2gt`C. `lt1,1,2gt`D. `ltsqrt2,1,1gt` |
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Answer» Correct Answer - b Any plane through (1,0,0) is `a(x-1)+by+cz=0" "(i)` It passes through (0,1,0). Therefore, `a(0-1)+b+(1)+c(0)=0or-a+b=0" "(ii)` (i) make an angle of `(pi)/(4)` with `x+y=3`, therefore `cos""(pi)/(4)=(a(1)+b(1)c+(0))/(sqrt(a^(2)+b^(2)+c^(2))sqrt(01+1+0))` or `(1)/(sqrt2)=(a+b)/(sqrt2sqrt(a^(2)+b^(2)+c^(2)))` or `a+b=sqrt(a^(2)+b^(2)+c^(2))` Squaring, we get `a^(2)b^(2)+2ab=a^(2)+b^(2)+c^(2)` or `2ab=c^(2)or2a^(2)=c^(2)" "["using"(ii)]` or `c=sqrt2a` Hence, `a:b:c=a:a:sqrt2a` `=1:1:sqrt2` |
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| 263. |
Find the equation of a plane which passes through thepoint (3, 2, 0) and contains the line `(x-3)/1=(y-6)/5=(z-4)/4dot`A. `x-y+z=1`B. `x+y+z=5`C. `x+2y-z=1`D. `2x-y+z=5` |
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Answer» Correct Answer - a The required plane is `|[x-3,y-6,z-4],[3-3,2-6,0-4],[1,5,4]|=0` or `|[x-3,y-z-2,z-4],[0,0,-4],[1,1,4]|=0` `("Operating"C_(2)toC_(2)-C_(3))` or `4(x-3-y+z+2)=0` or `x-y+z=1` |
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| 264. |
Find the equation of a linewhich passes through the point `(1,1,1)`and intersects the lines `(x-1)/2=(y-2)/3=(z-3)/4a n d(x+2)/1=(y-3)/2=(z+1)/4dot` |
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Answer» Any line passing through the point (1, 1, 1) is `(x-1)/(a)=(y-1)/(b)=(z-1)/(c)" "`(i) This line intersects the line `(x-1)/(2)=(y-2)/(3)=(z-3)/(4)`. If `a:b:c ne 2:3:4 and |{:(1-1,,2-1,,3-1),(a,,b,,c),(2,,3,,4):}|=0` `rArr" "a-2b+c=0` Again, line (i) intersects line `(x-(-2))/(1)=(y-3)/(2)=(z-(-1))/(4)" "`(ii) `a:b:c ne 1:2:4 and |{:(-2-1,,3-1,,-1-1),(a,,b,,c),(1,,2,,4):}|=0` `rArr" "6a+5b-4c=0" "`(iii) From (ii) and (iii) by cross multiplication, we have `(a)/(8i-5)= (b)/(6+4)=(c)/(5+12)` or `" "(a)/(3)=(b)/(10)=(c)/(17)` So, the required line is `(x-1)/(3)=(y-1)/(10)=(z-1)/(17)` |
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| 265. |
Find the equation of plane which is at a distance `(4)/(sqrt(14))` from the origin and is normal to vector `2hati+hatj-3hatk`. |
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Answer» Here `vecn=2hati+hatj-3hatk.` Then `(vecn)/(|vecn|)=(2hati+hatj-3hatk)/(sqrt(2^(2)+1^(2)+(-3)^(2)))=(2hati+hatj-3hatk)/(sqrt(14)` Hence, required equation of plane is `vecr*(1)/(sqrt(14))(2hati+hatj-3hatk)=pm(1)/(sqrt(14))` or `" "vecr*(2hati+hatj-3hatk)=pm1` (vector form) or `" "2x+y-3z=pm 1` (Cartesian form) |
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| 266. |
A plane cuts the co-ordinate axes at A, B and C respectively. If the centroid of `DeltaABC` is `(2,-3,4)`, find the equation of the plane. |
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Answer» Let the equation of plane be `x/a+y/b+z/c = 1"……"(1)` Which cuts x-axis at point A(a,0,0), y-axis at poin t `B(0,b,0)` and z-axis at point `C(0,0,z)` Now, Centroid of `DeltaABC` `= ((a+0+0)/(3),(0+b+0)/(3),(0+0+c)/(3))` `= (a/3,b/3,c/3)` Given that, `(a/3,b/3,c/3) = (2,-3,4)` `rArr a=6, b = - 9, c = 12` Therefore equation of plane. `x/6-y/9 + z/12 = 1` `rArr 6x-4y+3z = 36`. |
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| 267. |
Find the vector and equation of a plane which is at a distance of 10 units from origin and normal vector from origin to this plane is `2hati-hatj+2hatk`. |
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Answer» Here,` vecn=2hati-hatj+2hatk` and `p = 10` `:. hatn = (vecn)/(|vecn|)= (2hati-hatj+2hatk)/(sqrt(2^(2)+(-1)^(2)+2^(2)))=(2hati-hatj+2hatk)/(3)` Equation of plane `vecr.hatn=p` `rArr vecr.((2hati-hatj+2hatk))/(3) = 10` `rArr vecr.(2hati-hatj+2hatk) = 30`. |
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| 268. |
A point P lies on a line whose ends are A (1,2,3) and B(2,10,1). If z-coordinate of P is 7, then point P is |
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Answer» `(x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1)=(z-z_!)/(z_2-z_1)=k` `(x-1)/1=(y-2)/8=(z-3)/(-2)=k` `(x-1)=(y-2)/8=-2=k` `x=1+(-2)=-1` `y=-16+2=-14` option 1 is corrrect. |
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| 269. |
If a point lies on the y-axis then what are its x-coordinate nnd z-coordinate? |
| Answer» If a point lies on the y-axis then its x-coodrinate is 0 its z-coortinate is 0. | |
| 270. |
The vector equation of a plane is `vecr.(3hati+2hatj-6hatk) = 56`. Convert it into normal form. Also find the length of perpendicular from origin and direction cosines of normal to the plane. |
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Answer» `vecr.(3hati+2hatj-6hatk) = 56` Here `vecn=3hati+2hatj-6hatk` `rArr |vecn |=sqrt(9+4+36) = 7` `:. vecr.((3hati+2hatj-6hatk))/(7) = (56)/(7)` `rArr vecr.(3/7hati+2/7hatj-6/7hatk) = 8` Which is the normal form of the plane. Direction cosines of the perpendicular drawn from ` =3/7,2/7,(-6)/(7)` origin and length of perpendicular from origin `= 8` units. |
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| 271. |
If a point lies in xy-plane then what is it z-coordinate?A.B.C.D. |
| Answer» If a point lies in xy-plane then its z-coordinate is 0. | |
| 272. |
Find the vector equation of the line which is parallel to the vector `3hati-2hatj+6hatk` and which passes through the point `(1,-2,3)`. |
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Answer» Let `veca=3hati-2hatj+6hatk` and `vecb=hati-2hatj+3hatk` so, vector equation of the line, which is parallel to the vector `veca=3hati-2hatj+6hatk` and passes through the vector `vecb=hati-2hatj+3hatk` is `vecr=vecb+lamdaveca` `:. vecr=(hati-2hatj+3hatk)+lamda(3hati-2hatj+6hatk)` `implies (xhati+yhatj+zhatk)-(hati-2hatj+3hatk)=lamda(3hati-2hatj+6hatk)` `implies (x-1)hati+(y+2)hatj+(z-3)hatk=lamda(3hati-2hatj+6hatk)` |
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| 273. |
In which plane does the point (0,5,-4) lie ? |
| Answer» Clearly, the point (0,5-4) lies in the yz-plane. | |
| 274. |
Show that the two lines`(x-1)/2=(y-2)/3=(z-3)/4 nd (x-4)/5=(y-1)/z=z` interect. Find also the point of intersection of these lines. |
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Answer» We have `x_(1)=1,y_(1)=2,z_(1)=3` and `a_(1)=2,b_(1)=3,c_(1)=4` Also `x_(2)=4,y_(2)=1,z_(2)=0` and `a_(2)=5,b_(2)=2,c_(2)=1` If two lines intersect then shortest distance between them should be zero. `:.` Shortest distance between two given 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))` `(|(4-1,1-2,0-3),(2,3,4),(5,2,1)|)/(sqrt((3.1-2.4)^(2)+(4.5-1.2)^(2)+(2.2-5.3)^(2))` `=(|(3,-1,-3),(2,3,4),(5,2,1)|)/(sqrt(25+34+121))` `=(3(3-8)+1(2-20)-3(4-15))/(sqrt(470))` `=(-15-18+33)/(sqrt(470))=0/(sqrt(470))=0` Therefore the given two lines are intersecting. For finding their point of intersection for the first line `(x-1)/2=(y-2)/3=(z-3)/4=lamda` `implies x=2lamda+1,y=3lamda+2` and `z=4lamda+3` Since the lines are intersecting. So, let us put these values in the equation of another line. Thus, `(2lamda+1-4)/5=(3lamda+2-1)/2=(4lamda+3)/1` `implies (2lamda-3)/5=(3lamda+1)/2=(4lamda+3)/1` `implies (2lamda-3)/5=(4lamda+3)/1` `implies 2lamda-3=20lamda+15` `implies 18lamda=-18=-1` So the required point of intersection is `x=2(-1)+1=-1` `y=3(-1)+2=-1` `z=4(-1)+3=-1` Thus, the ines intersect at `(-1,-1,-1)`. |
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| 275. |
The acute angle between the vectors `bar(AB)` and `bar(CD)`, where `A-=(1,2-1), B-=(2, 1,1), C-=(2, 1, -2), D-=(3, 2,1)` isA. `"cos"^(-1)sqrt((6)/(11))`B. `"cos"^(-1)sqrt((3)/(11))`C. `cos^(-1)((6)/(sqrt(11)))`D. `cos^(-1)((3)/(sqrt(11)))` |
| Answer» Correct Answer - A | |
| 276. |
Find the values of `lambda` if the following of lines perpendicular : `(1-x)/(3) = (7y-14)/(3lambda)=(z+1)/(2)` and `(7-7x)/(3lambda) = y/1 = (1-z)/(5)` |
| Answer» Correct Answer - `35/6` | |
| 277. |
If `triangleABC` is right angled at B , where A(5,6,4), B(4,4,1) and C(8,2,x) , then find the value of x. |
| Answer» Correct Answer - C | |
| 278. |
If `DeltaABC`, if `A-=(3,2,6), B-=(1,4,5) ` and `C-=(3,5,3)`, then `m angleABC=`A. `90^(@)`B. `60^(@)`C. `45^(@)`D. `30^(@)` |
| Answer» Correct Answer - A | |
| 279. |
The angle between `hati` and line of the intersection of the plane `vecr.(hati+2hatj+3hatk)=0andvecr.(3hati+3hatj+hatk)=0` isA. `cos^(-1)((1)/(3))`B. `cos^(-1)((1)/(sqrt3))`C. `cos^(-1)((2)/(sqrt3))`D. none of these |
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Answer» Correct Answer - d Line of intersection of `vecr.(hati+2hatj+3hatk)=0andvecr.(3hati+3hatj+hatk)=0` will be parellel to `(hati+hatj+hatk)xx(hati+2hatj+3hatk),i.e.,7hati-8hatj+3hatk.` If the required angle is `theta`, then `costheta=(7)/(sqrt(49+64+9))=(7)/(sqrt122)` |
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| 280. |
Show that the following pairs of lines intersect. Also find their point of intersection : (i) `(x-1)/(2) = (y-2)/(3)=(z-3)/(4)` and `(x-4)/(5)=(y-1)/(2)=z` (ii) `(x-4)/(1)=(y+3)/(-4) = (z+1)/(7)` and `(x-1)/(2) = (y+1)/(-3)=(z+10)/(8)` |
| Answer» Correct Answer - (i) `(-1,-1,-1)`, (ii) `(5,-7,6)` | |
| 281. |
The line `(x+6)/5=(y+10)/3=(z+14)/8`is the hypotenuse of an isoscelesright-angled triangle whose opposite vertex is `(7,2,4)dot`Then which of the following in not the sideof the triangle?a. `(x-7)/2=(y-2)/(-3)=(z-4)/6`b. `(x-7)/3=(y-2)/6=(z-4)/2`c. `(x-7)/3=(y-2)/5=(z-4)/(-1)`d. noneof theseA. `(x-7)/(2)=(y-2)/(-3)=(z-4)/(6)`B. `(x-7)/(3)=(y-2)/(6)=(z-4)/(2)`C. `(x-7)/(3)=(y-2)/(5)=(z-4)/(-1)`D. none of these |
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Answer» Correct Answer - c Given one vertex A(7,2,4) and line `(x+6)/(5)=(y+10)/(3)=(z+14)/(8)` General point on above line `B-=(5lamda-6,3lamda-10,8lamda-14)` Direction ratios of line AB are `lt5lamda-13,3lamda-12,8lamda-18gt` Direction ratios of line BC are `lt5,3,8gt` Since angle between AB and BC is `pi//4`, we have `cos""(pi)/(4)=((5lamda-3)+3(3lamda-12)+8(8lamda-18))/(sqrt(5^(2)+3^(2)+8^(2)).sqrt((5lamda-13)^(2)+(3lamda-12)^(2)+(8lamda-18)^(2)))` Squaring and solving we have `lamda=3,2` Hence, equation of lines are `(x-7)/(2)=(y-2)/(-3)=(z-4)/(6)` and `(x-7)/(3)=(y-2)/(6)=(z-4)/(2)` |
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| 282. |
If `Delta ABC` is right angled at A, where `A-=(4, 2, 3), B-=(3, 1, 8)` and `C-=(x,-1,2)`, then x =A. -4B. -2C. 2D. 4 |
| Answer» Correct Answer - C | |
| 283. |
Find the distance between the following points-pairs `(i) (-2, 1, -3)` and `(4, 3, -6)` |
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Answer» `AB=sqrt({[4-(-2)]^(2)+(3-1)^(2)+[-6-(-3)]^(2)})` `=sqrt({6^(2)+2^(2)+(-3)^(2)})=sqrt(49)=7` units. |
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| 284. |
Show that the points `(0,7,10),(-1,6,6) and (-4,9,6)` form a righat angled isosceles triangle. |
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Answer» We have, `AB=sqrt((-1-0)^(2)+(6-7)^(2)+(6-10)^(2))=sqrt(18)=3sqrt(2)` `BC=sqrt((-4+1)^(2)+(9-6)^(2)+(6-6)^(2))=sqrt(18)=3sqrt(2)` `and AC=sqrt((-4-0)^(2)+(9-7)^(2)+(6-7)^(2))=sqrt(36)=6` Clearly, `AB= BC and AB^(2)+BC^(2)=(18+18)=36=AC^(2)` Hence, triangle ABC is an isosceles right-angled triangle. |
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| 285. |
Prove that the points `A(3,-2,4) , B(3,-2,4), B(1,1,1) and C(-1,4,-2)` are collinear. |
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Answer» We have, `AB=sqrt((1-3)^(2)+(1+2)^(2)+(1-4)^(2))=sqrt(4+9+9)=sqrt(22)` `BC=sqrt((1-1)^(2)+(4-1)^(2)+(-2-1)^(2))=sqrt(4+9+9)=sqrt(22)` `and AC=sqrt((1-3)^(2)+(4-2)^(2)+(-2-4)^(2))=sqrt(16+36+36)` `=sqrt(88)=2sqrt(22)` ` :. AB+BC+AC` This show that the given points are collinear. |
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| 286. |
Find the coordinates of the point which divides the join of the points `P(5,4,-2) and Q(-1,-2,4)` in the ratio `2:3` |
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Answer» Let `R(x,y,z)` be the required point, then `x=(2(-1)+3xx5)/(2+3),y=(2(-2)+3xx4)/(2+3),z=(2xx4+3xx2)/(2+3)` `or x=(13)/(5), y= (8)/(5) and z =(14)/(5)` So, the required point is `R((13)/(5),(8)/(5),(14)/(5))` |
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| 287. |
Find the equation of the set of points P, the sum of whose distances from `A (4, 0, 0)`and `B ( 4, 0, 0)`is equal to 10. |
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Answer» Let `P(x,y,z)` be an arbitrary on the given curve. Then, `PA+PB=10` `rArr sqrt((x-4)^(2)+y^(2)+z^(2))+sqrt((x+4)^(2)+y^(2)+z^(2))=10` `rArr sqrt((x-4)^(2)+y^(2)+z^(2))=10-sqrt((x+4)^(2)+y^(2)+z^(2))` `rArr (x+4)^(2)+y^(2)+z^(2)=100+(x-4)^(2)+y^(2)+z^(2)-20sqrt((x-4)^(2)+y^(2)+z^(2))" " [ " on squaring both side of (i)"]` `rArr 16x=10-20 sqrt((x-4)^(2)+y^(2)+z^(2))` `rArr 5sqrt((x-4)^(2)+y^(2)+z^(2))=(25-4x)` `rArr 9x^(2)+25y^(2)+25z^(2)-225=0` Hence, the required equation of the curve is `rArr 9x^(2)+25y^(2)+25z^(2)-225=0` |
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| 288. |
Find the equation of the set of points P which moves so that its distances from the points `A(3,4,-5) and B(-2,1,4)` are equal. |
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Answer» Let `P(x,y,z)` be any point on the give curve, and let `A(3,4,-5) and B(-2,1,4)` be the given points. Then, PA=PB `rArr PA^(2)=PB^(2)` `rArr (x-3)^(2)+(y-4)^(2)+(z+5)^(2)=(x+2)^(2)+(y-1)^(2)+(z-4)^(2)` `rArr 10x+6y-18x-29=0` Hence, the required curve is `10x+18z-29=0` |
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| 289. |
The straight lines whose direction cosines are given by `al + bm + cn = 0`,` fmn + gnl +hlm=0`are perpendicular ifA. `(f)/(a)+(g)/(b)+(h)/(c )=1`B. `(f)/(a)+(g)/(b)+(h)/(c )=-1`C. `(f)/(a)+(g)/(b)+(h)/(c )=0`D. `(a)/(f)+(b)/(g)+(c )/(h)=0` |
| Answer» Correct Answer - C | |
| 290. |
The plane XOZ divides the join of `(1, -1, 5) and (2, 3, 4)`in the ratio of `lambda:1`, then `lambda` isA. 3B. -3C. `(1)/(3)`D. `(-1)/(3)` |
| Answer» Correct Answer - C | |
| 291. |
Find the direction cosines of that line whose direction ratios are as follows : (i) `1,-2,2` , (ii) `2,6,3` (iii) `3,1,-2` |
| Answer» Correct Answer - (i) `1/3,-2/3,-2/3` , (ii) `2/7,6/7,3/7` , (iii) `3/(sqrt(14)),(1)/(sqrt(14)),(-2)/(sqrt(14))` | |
| 292. |
If ` vec r`is a vector of magnitude 21and has direction ratios `2,-3a n d6,`then find ` vec rdot` |
| Answer» Correct Answer - `6hati-9hatj+18hatk` | |
| 293. |
Find the angle between those lines whose direction ratios are as follows : (i) `(2,3,6)` and `(1,2,2)` (ii) `(4,-3,5)` and `(3,4,5)` (iii) `(1,2,1)` and `(4,-3,2)` |
| Answer» Correct Answer - (i) `cos^(-1)(20/21)` , (ii) `60^(@)` , (iii) `90^(@)` | |
| 294. |
Find the angle between the following vectors : (i) `veca = 2hati-6hatj+3hatk` and `vecb = hati+2hatj-2hatk` (ii) `veca=6hati+3hatj-2hatk` and `vecb=4hati-2hatj+9hatk` |
| Answer» Correct Answer - (i) `cos^(-1)(16/21)`, (ii) `90^(@)` | |
| 295. |
The ratio in which the line segment joining the points whose position vectors are `2hati-4hatj-7hatkand-3hati+5hatj-8hatk` is divided by the plane whose equationis `hatr.(hati-2hatj+3hatk)=13`isA. 13:12 internallyB. 12:25 externallyC. 13:25 internallyD. 37:25 internally |
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Answer» Correct Answer - b Let P be the point and it divides the line segment in the ratio `lamda :1 `. Then, `" "vec(OP)=vecr= (-3lamda+2)/(lamda+1)hati+ (5lamda-4)/(lamda+1)hatj+ (-8lamda-7)/(lamda+1)hatk` It satisfies `vecr*(hati-2hatj+3hatk)= 13`. So, `" "(-3lamda+2)/(lamda+1)-2(5lamda-4)/(lamda+1)+3(-8lamda-7)/(lamda+1)=13` or `" "-3lamda+2-2(5lamda-4)+3(-8lamda-7)= 13 (lamda+1)` or `" "-37lamda-11=13lamda+13 or 50 lamda= -24 or lamda= - (12)/(25)` |
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| 296. |
The co-ordinates of the point P on the line `vecr =(hati +hatj + hatk)+lambda (-hati +hatj -hatk)` which is nearest to the origin isA. `((2)/(3)(4)/(3),(2)/(3))`B. `(-(2)/(3)-(4)/(3),(2)/(3))`C. `((2)/(3)(4)/(3),-(2)/(3))`D. none of these |
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Answer» Correct Answer - a Let the point `P` be `(x, y, z)`, then the vector `xhati+yhatj+zhatk` will lie on the line. Thus, `" "(x-1)hati+(y-1)hatj+ (z-1)hatk= -lamdahati+lamdahatj-lamdahatk` `rArr" "x=1-lamda, y=1+lamda and z=1-lamda` Now point P is nearest to the origin `rArr" "D= (1-lamda)^(2) + (1+lamda)^(2) + (1-lamda)^(2)` or `" "(dD)/(dlamda)= -4(1-lamda)+ 2(1+lamda) = 0 rArr lamda = (1)/(3)` Hence, the point is `((2)/(3), (4)/(3), (2)/(3))` |
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| 297. |
If the co-ordinates of the points P and Q are `(1,-2,1)and(2,3,4)` and O is the origin, thenA. `OP=OQ`B. `OP_|_OQ`C. `OP||OQ`D. OP and OQ intersect at an angle of `45^(@)` |
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Answer» Correct Answer - B |
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| 298. |
Find the co-ordinates of the point at a distance of `sqrt(5)`units from the point `(1,2,3)` on the line `(x+2)/(3) = (y+1)/(2) = (z-3)/(2)`. |
| Answer» Correct Answer - `(5/17,9/17,77/17),(1,1,5)` | |
| 299. |
Find the equations of the plane passing through the following points : (i) `A(2,1,0), B(3,-2,-2), C(3,1,7)` (ii) `A(1,1,1), B(1,-1,2), C(-2,-2,2)` (iii) `A(0,-1,0), B(2,1,-1), C(1,1,1)` (iv) `A(1,-2,5), B(0,-5,-1), C(-3,5,0)` (v) `(4,-1,-1), B(2,0,2),C(3,-1,2)` |
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Answer» Correct Answer - (i) `7x+3y-z=17` , (ii) `x-3y-6z+8=0` (iii) `4x-3y+2z=3`, (iv) `3x+y-z +4 =0` (v) `3x+3y+z=8` |
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| 300. |
Find the co-ordinates of the point at a distance of `sqrt(14)` from the mid-point of AB on the line joining the point `A(1,2,3)` and `B(3,6,9)`. |
| Answer» Correct Answer - `(3,6,9),(1,2,3)` | |