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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 distance between the line `x=2+t,y=1+t,z=-1/2-t/2` and the plane `vecr.(hati+2hatj+6hatk)=10`, isA. `1/6`B. `1/(sqrt(41))`C. `1/7`D. `9/(sqrt(41))` |
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Answer» Correct Answer - D The equation of the lines is `(x-2)/1=(y-1)/1=(z+1/2)/(-1//2)=t` Clearly, it is parallel to the plane `vecr.(hati+2hatj+6hatk)=10`. So required distance is equal to the length of perpendicular form `(2,1,-1/2)` to the plane `vecr.(hati+2hatj-6hatk)-10=0` and is given by `d=|((2hati+hatj-1/2hatk).(hati+2hatj+6hatk)-10)/(sqrt(1+4+36))|=9/(sqrt(41))` |
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| 2. |
A line with directional cosines prontional to `2, 1,2` meets each of the lines `x = y + a = z and x+ a = 2y = 2z`. The coordinates of each of the points deintersection are given byA. `(2a,3a,a),(2a,a,a)`B. `(3a,2a,3a),(a,a,a)`C. `(3a,2a,3a),(a,a,2a)`D. `(3a,3a,3a),(a,a,a)` |
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Answer» Correct Answer - B We observe that the points `P(3a,2a,3a)` and `Q(a,a,a)` satisfy the equations of the lines `x=y+a=z` and `x+a=2y=2z` respectively. Also, direction ratios of `PQ` are proportional to `3a-a,2a-a,3a-a` i.e.`2,1,2`. Hence option b is correct. |
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| 3. |
The shortest distance between the lines `2x + y + z - 1 = 0 = 3x + y + 2z - 2` and `x = y = z`, isA. `1/(sqrt(2))`B. `sqrt(2)`C. `3/(sqrt(2))`D. `(sqrt(3))/2` |
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Answer» Correct Answer - A The shortest distance between the given lines is equal to the length of the perpendicular from any point on `(x-0)/y=(y-0)/1=(z-0)/1` to the plane through the line `2x+y=z-1=10, 3x+y+2z-2=0` and parallel to `(x-0)/1=(y-0)/1=(z-0)/1` The equation of any plane through the line `2x+y+z-1=10, 3x+y+2z-2=0` is `(2x+y+z-1)+lamda(3x+y+2z-2)=0` or `x(3lamda+2)+y(lamda+1)+z(2lamda+1)-2lamda-1=0`...........i If it is parallel to `(x-0)/1=(y-0)/1=(z-0)/1`, then `3lamda+2+lamda+1+2lamda+1=0implieslamda=(-2)/3` Putting `lamda=(-2)/3` in (i) we obtain `-y+z-1=0` The distance of this plane from any point on `(x-0)/1=(y-0)/1=(z-0)/1` is `d=|(-0+0-1)/(sqrt(1+1))|=1/(sqrt(2))` |
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| 4. |
Find the vector equation of the plane in which the lines `vecr=hati+hatj+lambda(hati+2hatj-hatk)` and `vecr=(hati+hatj)+mu(-hati+hatj-2hatk)` lie.A. ` vecr.(2hati+hatj-3hatk)=-4`B. `vecrxx(-hati+hatj+hatk)=vec0`C. `vecr.(-hati+hatj+hatk)=0`D. none of these |
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Answer» Correct Answer - C The lines are parallel to the vector `vecb_(1)=hati+2hatj-hatk` and `vecb_(2)=-hati+hatj-hatk`. Therefore, the plane is normal to the vector `vecn=vecb_(1)xxvecb_(2)=|(hati,hatj,hatk),(1,2,-1),(-1,1,-2)|=-3hati+3hatj+3hatk` The required plane passes through `(hati+hatj)` and is normal to the vector `vecn`. Therefore its equation is `vecr.vecn=veca.vecn` `impliesvecr.(-3hati+3hatj+3hatk)=(hati+hatj).(-3hati+hatj+3hatk)` `impliesvecr.(-3hati+3hatj+3hatk)=3+3impliesvecr.(-hati+hatj+hatk)=0` |
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| 5. |
Find the vector equation of the plane in which the lines `vecr=hati+hatj+lambda(hati+2hatj-hatk)` and `vecr=(hati+hatj)+mu(-hati+hatj-2hatk)` lie.A. `vecr.(hati+hatj+hatk)=0`B. `vecr.(-hati+hatj+hatk)=0`C. `vecr.(-hati+hatj+hatk)=1`D. `vecr.(hati+hatk-hatk)=0` |
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Answer» Correct Answer - B The two given lines pass through the point having position vector `veca=hati+hatj` and are parallel to the vector `vecb_(1)=hati+2hatj-hatk` and `vecb_(2)=-hati+hatj-2hatk` respectively. Therefore, the plane containing the given lines also passes through the point with position vector `veca=hati+hatj`. Since the plane contains the lines which aare parallel to the vectors `vecb_(1)` and `vecb_(2)` respectively. Therefore, the plane is normal to the vector `vecn` given by `vecn=vecb_(1)xxvecb_(2)=|(hati,hatj,hatk),(1,2,-1),(-1,1,-2)|=-3hati+3hatj+3hatk` Thus, the vector equation of the required planes is `vecr.vecn=veca.vecn` `implies vecr.(-3hati+3hatj+3hatk)=(hati+hatj).(-3hati+3hatj+3hatk)` `impliesvecr.(-hati+hatj+hatk)=0` |
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| 6. |
The angle between the lines `vecr=(hati+hatj+hatk)+lamda(hati+hatj+2hatk)` and `vecr=(hati+hatj+hatk)+mu{(-sqrt(3)-1)hati+(sqrt(3)-1)hatj+4hatk}` isA. `(pi)/6`B. `(pi)/4`C. `(pi)/3`D. `(2pi)/3` |
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Answer» Correct Answer - C Given lines are parallel to the vectors `vecb_(1)=hati+hatj+2hatk` and `vecb_(2)=(-sqrt(3)-1)hati+(sqrt(3)-1)hatj+4hatk` Let `theta` the angel between the given lines. Then, `cos theta=(vecb_(1).vecb_(2))/(|vecb_(1)||vecb_(2)|)` `impliescos theta=(-sqrt(3)-1+sqrt(3)-1+8)/(sqrt(1+1+4)sqrt((-sqrt(3)-1)^(2)+(sqrt(3)-1)^(2)+16))=1/2` `implies theta=(pi)/3` |
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| 7. |
The equation of the plane contaiing the lines `vecr=veca_(1)+lamda vecb` and `vecr=veca_(2)+muvecb` isA. `[(vecr, veca, vecb)]=0`B. `[(vecr,veca,vecb)]=veca.vecb`C. `[(veca,vecb,veca)]=veca.vecb`D. none of these |
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Answer» Correct Answer - A The required plane passes through the points having position vectors `veca` and `vecb`. It is perpendicular to `vecaxxvecb`. So theequation of the plane is `(vecr-veca).(vecaxxvecb)=0impliesvecr.(vecaxxvecb)=veca.(vecaxxvecb)implies[(vecr,veca,vecb)]=0` |
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| 8. |
The shortest distance between the lines `vecr=(hati-hatj)+lamda(hati+2hatj-3hatk)` and `vecr=(hati-hatj+2hatk)+mu(2hati+4hatj-5hatk)` isA. `6`B. `6/(sqrt(5))`C. `3/(sqrt(5))`D. `2/(sqrt(5))` |
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Answer» Correct Answer - B We know that the shortest distance between the lines `vecr=veca_(1)+lamdavecb_(1)` and `vecr=veca_(2)+muvecb_(2)` is given by `d=|((veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2)))/(|vecb_(1)xxvecb_(2)|)|` Hence `veca_(1)=4hati-hatj,veca_(2)=hati-hatj+2hatk`, `vecb_(1)=hati+2hatj-3hatk` and `vecb_(2)=2hati+4hatj-5hatk` `:.veca_(2)-veca_(1)=-3hati+0hatj+2hatk` and `vecb_(1)xxvecb_(2)=|(hati,hatj,hatk),(1,2,-3),(2,4,-5)|=2hati-hatj+0hatk` `implies(veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2))=(-3hati+0hatj+2hatk).(2hati-hatj+0hatk)=-6` and `|vecb_(1)xxvecb_(2)|=sqrt(4+1+0)=sqrt(5)` `:.d=|((veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2)))/(|vecb_(1)xxvecb_(2)|)|=|(-6)/(sqrt(5))|=6/(sqrt(5))` |
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| 9. |
the lines `(x-2)/1` = `(y-3)/1` = `(z-4)/-k` and `(x-1)/k` = `(y-4)/1` = `(z-5)/1` are coplanar if k=?A. `k=3` or `-3`B. `k=0` or `-1`C. `k=1` or `-1`D. `k=0` or `-3` |
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Answer» Correct Answer - D We know that the lines `(x-x_(1))/(l_(1))=(y-y_(1))/(m_(1))=(z-z_(1))/(n_(1))` and `(x-x_(2))/(l_(2))=(y-y_(2))/(m_(2))=(z-z_(2))/(n_(2))` are coplanar iff `|(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2))|=0` so the given lines will be coplanar iff `|(1-2,4-3,5-4),(1,1,-k),(k,2,1)|=0impliesk^(2)+3k=0impliesk=0,-3` |
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| 10. |
Two lines `L_1x=5, y/(3-alpha)=z/(-2)a n dL_2: x=alphay/(-1)=z/(2-alpha)`are coplanar. Then `alpha`can take value (s)a. `1`b. `2`c. `3`d. `4`A. 1,4,5B. 1,2,5C. 3,4,5D. 2,4,5 |
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Answer» Correct Answer - A The equations of given lines are: `L_(1):(x-5)/0=y/(3-alpha)=z/(-2),L_(2):(x-alpha)/0=y/(-1)=z/(2-alpha)` These lines are coplanar. Therefore `|(5-alpha, 0-0,0-0),(0,3-alpha,-2),(0,-1,2-alpha)|=0` `implies(5-alpha){(3-alpha)(2-alpha)-2}=0` `=((5-alpha)(6-5alpha+alpha^(2)-2)=0` `implies(5-alpha)(alpha^(2)-5alpha+4)=0` `implies(alpha-1)(alpha-4)(alpha-5)=0impliesalpha=1,4,5` |
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| 11. |
If the lines `(x-1)/2=(y+1)/3=(z-1)/4a n d(x-3)/1=(y-k)/2=z/1`intersect, then find thevalue of `kdot`A. `3/2`B. `9/2`C. `-2/9`D. `-3/2` |
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Answer» Correct Answer - B We know that the lies `(x-x_(1))/(l_(1))=(y-y_(1))/(m_(1))=(z-z_(1))/(n_(1))` and `(x-x_(2))/(l_(2))=(y-y_(2))/(m_(2))=(z-z_(2))/(n_(2))` Intersect if `|(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2))|=0` So given lines will intersect, if `|(3-1,k+1,0-1),(2,3,4),(l,2,1)|=0` `implies 2(3-8)-(k+1)(2-4)-(4-3)=0impliesk=9/2` |
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| 12. |
If the lines `(x-4)/1=(y-2)/1=(z-lamda)/3` and `x/1=(y+2)/2=z/4` intersect each other, then `lamda` lies in the intervalA. (9,11)B. (-5,-3)C. (13,15)D. (11,13) |
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Answer» Correct Answer - D Given lines will intersect if `|(4-0,2-(-2),lamda-0),(1,1,3),(1,2,4)|=0` `implies4(4-6)-4(4-3)+lamda(2-1)-0` `implieslamda=12implieslamda epsilon(11,13)` |
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| 13. |
The equations of two straight lines are `(x-1)/2=(y+3)/1=(z-2)/(-3)` and `(x-2)/1=(y-1)/(-3)=(z+3)/2` Statement 1: The given lines are coplanar. Statement 2: The equations `2r-s=1` `r+3s=4` `3r+2s=5` are consistent.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - A The coordinates of arbitrary positions of the given lines are `(2r+1,r-3,-3r+2)` and `(s+2,-3s+1,2s-3)` respectively. Given lines will intersect (be coplanar) if `2r+1=s+2,r-3=-3s+1` and `-3r+2=2s-3` are conistent i.e `2r-s=1,r+3s=4` and `3r+2s=5` are consistent. Clearly, values of `r` and `s` obtain from any two equations satisfy the third equation. So, these equations are consistent. Hence, both the statements are true and statement -2 is a correct explanation for statement -1. |
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| 14. |
Find the image of the point `(1,3,4)`in the plane `2x-y+z+3=0.`A. `(3,5,-2)`B. `(-3,5,2)`C. `(3,-5,2)`D. `(3,5,2)` |
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Answer» Correct Answer - B We know that the image of the point `(x_(1),y_(1),z_(1))` in the plane `ax+by+cz+d=0` is given by `(x-x_(1))/a=(y-y_(1))/b=(z-z_(1))/c=(-2(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))` So the image of the point `P(1,3,4)` in the plane `2x-y+z+3=0` is given by `(x-2)/2=(y-3)/(-1)=(z-4)/1=(-2(2-3+4+3))/(4+1+1)` `implies(x-1)/2=(y-3)/(-1)=(z-4)/1=-2impliesx=-3,y=5,z=2` Hence the required pointis `(-3,5,2)`. |
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| 15. |
The image of theline `(x-1)/3=(y-3)/1=(z-4)/(-5)`in the plane `2x-y+z+3=0`is the line(1) `(x+3)/3=(y-5)/1=(z-2)/(-5)`(2) `(x+3)/(-3)=(y-5)/(-1)=(z+2)/5`(3) `(x-3)/3=(y+5)/1=(z-2)/(-5)`(3) `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`A. `(x-3)/3=(y+5)/1=(z-2)/(-5)`B. `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`C. `(x+3)/3=(y-5)/1=(z-2)/(-5)`D. `(x+3)/(-3)=(y-5)/(-3)=(y-5)/(-1)=(z+2)/5` |
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Answer» Correct Answer - C Clearly, given line is parallel to the given plane. So, its image will also be parallel to the plane and so its direction ratios are proportional to 3,1,-5. Given line passesthrough (1,3,4). Let `(x_(1),y_(1),z_(1)` be its image in the given plane. Then `x_(1),y_(1),z_(1)` are given by `(x_(1)-1)/2=(y_(1)-3)/(-1)=(z_(1)-4)/1=-((2xx1-3+4+3))/(2^(2)+(-1)^(2)+1^(2))` `implies(x_(1)-1)/2=(y_(1)-3)/(-1)=(z_(1)-4)/1=-2` `impliesx_(1)=-3,y_(1)=5,z_(1)=2` So image of (1,3,4) in the given planeis (-3,5,2). Hence the equations of the required line are `(x+3)/3=(y-5)/1=(z-2)/(-5)`. |
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| 16. |
Ifthe straight lines `(x-1)/k=(y-2)/2=(z-3)/3`and `(x-2)/3=(y-3)/k=(z-1)/2`intersect at a point, then the integer k isequal to(1)`-5`(2) 5(3) 2(4) `-2`A. `2`B. `-2`C. `-5`D. `5` |
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Answer» Correct Answer - C If given lines intersect, then they are coplanar. `:.|(1-2,2-3,3-1),(k,2,3),(3,k,2)|=0` `implies-(4-3k)+(2k-9)+2(k^(2)=6)=0` `implies2k^(2)+5k-25=0implies(k+5)(2k-5)=0impliesk=-5,5//2`. |
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| 17. |
then image of the point `(-1,3,4)` in the plane `x-2y=0`A. `(-17/3,-19/3,4)`B. `(15,11,4)`C. `(-17/3,-19/3,1)`D. none of these |
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Answer» Correct Answer - D The image `(alpha,beta,gamma)` of `(-1,3,4)` in the plane `x-2y+0z=0` is given by `(alpha+1)/1=(beta-3)/(-2)=(gamma-4)/0=(-2(-1-6+0))/(1+4+0)` `implies(alpha+1)/1=(beta-3)/(-2)=(gamma-4)/0=14/5impliesalpha=9/4,beta=(-13)/5,gamma=4` Hence, he image is `(6/5,-13/5,4)`. |
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| 18. |
The line passing through the points `(5,1,a)` and `(3,b,1)` crosses the yz-plane at the point `(0,17/2,(-13)/2)`.ThenA. `a=6,b=4`B. `a=8,b=2`C. `a=2,b=8`D. `a=4,b=6` |
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Answer» Correct Answer - A The equation of the line passing through `(5,1,a)` and `(3,b,1)` is `(x-5)/(-2)=(y-1)/(b-1)=(z-a)/(1-a)` The coordinates of any point on this line are `(-2r+5,(b-1)r+1,(1-a)r+a)` If it lies on yz-plane. Then, `-2r+5=0impliesr=5/2` Thus, the line cuts yz-plane at `(0,(5b-3)/2,(5-3a)/2)` But the coordinates of this point are given as `(0,17/2,(-13)/2)` `:.(5b-3)/2=17/2` and `(5-3a)/2=(-13)/2impliesb=4` and `a=6` ALITER The equation of the line passing through `(5,1,a)` and `(3,b,1)` is `(x-5)/(-2)=(y-1)/(b-1)=(z-a)/(1-a)` It cuts yz-plane at `(0,17/2,(-13)/2)`. This means that this point lies on the line. `:. 5/2=(17/1-1)/(b-1)=((-13)//2-a)/(1-a)` `impliesb-1=3`an `-13-a=5(1-a)impliesb=4` and `a=6` |
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| 19. |
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`1/(sqrt(3))``1/2`1`1/(sqrt(2))`A. `1`B. `1/(sqrt(2))`C. `1/(sqrt(3))`D. `1/2` |
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Answer» Correct Answer - C Vectors normals to the given planes are `vecn_(1)=2hati+3hatj+k` and `vecn_(2)=hati+3hatj+2hatk`. So the line `L` is parallel `vecn_(1)=vecn_(1)xxvecn_(2)=|(hati,hatj,hatk),(2,3,1),(1,3,2)|=3hati-3hatj+3hatk` `implies cos alpha=(vecn.hati)/(|vecn||hati|)=3/(3sqrt(3))=1/(sqrt(3))` |
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| 20. |
The equation of the plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4 = 0` and parallel to x-axis isA. `y-z+6=0`B. `3y-z+6=0`C. `y+3z+6=0`D. `3y-2z+6=0` |
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Answer» Correct Answer - A The equation of the plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` is `(x+y+z-1)+lamda(2x+3y-z+4)=0` or `(2lamda+1)x+(3lamda+1)y+(1-lamda)z+4lamda-1)=0`………..i It is parallel to `X` -axis i.e. `x/1=y/0=z/0` `:.1(2lamda+1)+0xx(3lamda+1)+0(1-lamda)=0implieslamda=-1/2` Substituting `lamda=1/2` in (i) we get `y-3z+6=0` as the equation of the required plane. |
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| 21. |
If the equation of the plane through the line of interesection of `vecr.(2hati-3hatj+hatk)=1` and `vecr.(hati-hatj)+4=0` and perpendicular to `vecr.(2hati+hatj+hatk)+8=0` is `vecr.(5hati-2hatj-12hatk)=lamda` Then `lamda=`A. 47B. -47C. 37D. -37 |
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Answer» Correct Answer - A The equation of any plane through the line of intersection of the given planes is `[vecr.(2hati-3hatj+4hatk)-1]+lamda[vecr.(hati-hatj)+4]=0` `impliesvecr.[(2+lamda)hati-(3+lamda)hatj+4hatk]=1-4lamda`…………..i If i is perpendicular to `vecr.(2hati-hatj+hatk)+8=0` then `[(2+lamda)hati-(3+lamda)hatj+4hatk].(2hati-hatj+hatk)=0` `implies2(2+lamda)+(3+lamda)+4=0implieslamda=-11//3` Putting `lamda=-11//3` in i we obtain the equation of the required plane as `vecr.(-5hati+2hatj+12hatk)=47` Hence `lamda=47`. |
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| 22. |
Find the equation of the planewhich is perpendicular to the plane `5x+3y+6 z+8=0`and which contains the line of intersectionof the planes `x+2y+3z- 4 = 0 a n d 2x+y - z+5 = 0`A. `15x+15y-20z+4=0`B. `51x+15y-50z+173=0`C. `3x-5y+7=0`D. `3x+5y-5z+9=0` |
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Answer» Correct Answer - B The equation of a plane through the line of intersection of the planes. `x+2y+3z-4=0` and `2x+y-z+5=0` is given by `(x+2y+3z-4)+lamda(2x+y-z+5)=0` `impliesx(1+2lamda)+y(2+lamda)+z(3-lamda)-4+5lamda=0` ………………..i This is perpendicular to the plane `5x+3y+6z+8=0` `:.5(1+2lamda)+3(2+lamda)+6(3-lamda)=0` `implies7 lamda+29=0implieslamda=-29//7` Putting `lamda=-29//7` in i we obtain the equation of the required plane as `-51x-15y+50z-173=0implies51x+15y-50z+173=0` |
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| 23. |
Find the equation of a plane containing the line of intersectionof the planes `x+y+z-6=0a n d2x+3y+4z+5=0`passing through `(1,1,1)`.A. `20x+23y+26z-69=0`B. `20x+26y+23z-69=0`C. `x+y+z-3=0`D. `2x+3y+4z-9=0` |
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Answer» Correct Answer - A The equation of the plane through the line of intersection of the given planes is `(x+y+z-6)+lamda(2x+3y+4z+5)=0` ………..i If it pases through (1,1,1) we have `-3+14lamda=0implieslamda=3//14` Putting `lamda=3//14` in i we obtain the equation of the required plane as `20x+23y+26z-69=0` |
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| 24. |
Read the following passage and answer the questions. Consider the lines `L_(1):(x+1)/(3)=(y+2)/(1)=(z+1)/(2),L_(2):(x-2)/(1)=(y+2)/(2)=(z-3)/(3)` The unit vector perpendicualr to both `L_(1)` and `L_(2)` isA. `1/(sqrt(99))(-hati+7hatj+7hatk)`B. `1/(5sqrt(5))(-hati-7hatj+5hatk)`C. `1/(5sqrt(3))(-hati+7hatj+5hatk)`D. `1/(sqrt(99))(7hati-7hatj-hatk)` |
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Answer» Correct Answer - B Lines `L_(1)` and `L_(2)` are parallel to the vectors `vecb_(1)=3hati+hatj+2hatk` and `vecb_(2)=hati+2hatj+3hatk` respectively. Therefore, a unit vector perpendicular to both `L_(1)` and `L_(2)` is `hatn=(vecb_(1)xxvecb_(2))/(|vecb_(1)xxvecb_(2)|)` Now `vecb_(1)xxvecb_(2)=|(hati,hatj,hatk),(3,1,2),(1,2,3)|=-hati-7hatj+6hatk` `:.hatn=1/(5sqrt(3))(-hati-7hatj+5hatk)` |
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| 25. |
The distance of the point `(1,3,-7)`from the plane passing through the point `(1,-1,-1),`having normal perpendicular to both the lines`(x-1)/1=(y+2)/(-2)=(z-4)/3a n d(x-2)/2=(y+1)/(-1)=(z+7)/(-1)i s:``5/(sqrt(83))`(2) `(10)/(sqrt(74))`(3) `(20)/(sqrt(74))`(4) `(10)/(sqrt(83))`A. `20/(sqrt(74))`B. `10/(sqrt(83))`C. `10/(sqrt(74))`D. `5/(sqrt(83))` |
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Answer» Correct Answer - B Let the equation of the plane through the point (1,-1,-1) be `a(x-1)+b(y+1)+c(z+1)=0`……………..i The direction ratios of the normal to the plane proportional to a,b,c. The normal is perpendicular to the lines `(x-1)/1=(y+2)/(-2)=(z-4)/3` and `(x-2)/2=(y+1)/(-1)=(z+7)/(-1)` `:.a-2b+3c=0` and `2a-b-c=0` Using cross multiplication, we obtain `5(x-1)+7(y+1)+3(z+1)-0` or `5x+7y+3z+5=0` ................ii The distance `d` of the point (1,3,-7) from plane ii is `d=|(5+21-21+5)/(sqrt(25+49+9))|=10/(sqrt(83))` |
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| 26. |
Find the distance between the parallel planes `vecr.(2hati-3hatj+6hatk) = 5` and `vecr.(6hati-9hatj+18hatk) + 20 = 0`.A. `2/3`B. `5/3`C. `4/3`D. `1/3` |
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Answer» Correct Answer - B The equations of two planes are `vecr.(2hati-3hatj+6hatk)=5` and `ver.(2hati-3hatj+6hatk)=-20/3` `:.` Distance between the planes `=(|5-(-20/3)|)/(|2hati-3hatj+6hatk|)=5/3` |
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| 27. |
In the space the equation `by+cz+d=0` represents a plane perpendicular to the planeA. YOZB. ZOXC. XOYD. `Z=k` |
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Answer» Correct Answer - A The equation of YOZ-plane is `x=0` Clearly, planes `by+cz+d=0` and `x=0` are perpendicular as `1xx0+0xxb+0xxc=0` |
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| 28. |
Find the equation of the plane through the point (1,4,-2) and parallel to the plane `-2x+y-3z=7`.A. `2x-y+3z=8`B. `2x-y+3z+8=0`C. `2x-y+3z=4`D. none of these |
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Answer» Correct Answer - B Let the equation of a plane parallel to the plane `-2x+y-3z=7` be `-2x+y-3z+k=0`……………i This passes through (1,4,-2) `:.(-2)(1)+4-3(-2)+k=0=k=-8` Putting `k=-8` in i we obtain `-2x+y-3z-8=0implies2x-y+3z+8=0` |
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| 29. |
Distance betweentwo parallel planes `2x""+""y""+""2z""=""8`and `4x""+""2y""+""4z""+""5""=""0`is(1) `5/2`(2) `7/2`(3) `9/2`(4) `3/2`A. `3/2`B. `5/2`C. `7/2`D. `9/2` |
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Answer» Correct Answer - C Equations of the given planes are `2x+y+2z-8=0` and `2x+y+2z+5/2=0`.Let `d` be the distance between these planes, Then `d=|(-8-5/2)/(sqrt(2^(2)+1^(2)+2^(2)))|=21/6=7/2` |
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| 30. |
Find the distance between the parallel planes `2x-y+2z+3=0a n d4x-2y+4z+5=0.`A. `1//3`B. `2//6`C. `2//3`D. none of these |
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Answer» Correct Answer - B The equations of two planes are `2x-y+2z+3=0` and `2x-y+2z+5/2=0` `:.` Required distance `=(|3-5/2|)/(sqrt(4+1+4))=1/6` |
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| 31. |
Let `P(3,2,6)`be a point in space and `Q`be a point on line ` vec r=( hat i- hat j+2 hat k)+mu(-3 hat i+ hat j+5 hat k)dot`Then the value of `mu`for which the vector ` vec P Q`is parallel to the plane `x-4y+3z=1`isa. 1/4 b. -1/4 c. 1/8 d. -1/8A. `1/4`B. `-1/4`C. `1/8`D. `-1/8` |
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Answer» Correct Answer - A Let the position vector of `Q` be `(hati-hatj+2hatk)+mu(-3hati+hatj+5hatk)` `=(-3mu+1)hati+(mu-1)hatj+(5mu2)hatk`. Then `vec(PQ)=(-3mu-2)hati+(mu-3)hatj+(5mu-4)hatk` It is given that `vec(PQ)` is parallel to the plane `x-4y+3z=1` whose normal is `vecn=hati-4hatj+3hatk`. `:.vec(PQ).vecn(n)=0` `implies(-3mu-2)=4(mu-3)+3(5mu-4)=0impliesmu=1/4` |
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| 32. |
The distance of the plane through (1,1,1) and perpendicular to the line `(x-1)/3=(y-1)/0=(z-1)/4` from the origin isA. `3/4`B. `4/3`C. `7/5`D. `1` |
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Answer» Correct Answer - C The given line is normal to the plane. So direction ratios of the normal to the plane are proportional to 3,0,4. The plane pases through (1,1,1). So its equation is `3(x-1)+0(y-1)+4(z-1)=0` or `3x+4z-7=0` Its distance `d` from the oriign is given by `d=(|-7|)/(sqrt(3^(2)+0^(2)+4^(2)))=7/5` |
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| 33. |
The equation of the plane containing the line `vecr=hati+hatj+lamda(2hati+hatj+4hatk)` isA. `vecr.(-hati-2hatj+hatk)=3`B. `vecr.(hati+2hatj-hatk)=0`C. `vecr.(hati+2hatj-hatk)=3`D. none of these |
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Answer» Correct Answer - C Clearly, the position vector of any point on the given line satisfies the equation of the plane `vecr.(hati+2hatj-hatk)=3` Hence the given line lies in the plane given in option c |
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| 34. |
The acute angle between the planes `2x-y+z=6` and `x+y+2z=3` isA. `45^(@)`B. `60^(@)`C. `30^(@)`D. `75^(@)` |
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Answer» Correct Answer - B Vectors normals to the given planes are `vecn_(1)=2hati-hatj+hatk` and `vecn_(2)=hati+hatj+2hatk` Let `theta` be the acute angle between the planes. Then, `cos theta=(|vecn_(1).vecn_(2)|)/(|vecn_(1)||vecn_(2)|)=(|2-1+z|)/(sqrt(6)sqrt(6))=1/2impliestheta=60^(@)` |
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| 35. |
The equation of the plane passing through the point `(1,1,1)` and perpendicular to the planes `2x+y-2z=5` and `3x-6y-2z=7` isA. `14x+2y-15z=1`B. `14x-2y+15z=27`C. `14x+2y+15z=31`D. `-14x+2y+15z=3` |
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Answer» Correct Answer - C The equation of a plane through (1,1,) is `a(x-1)+b(y-1)+c(z-1)=0`…………i It is perpendicular to the planes `2x+y-2z=5` and `3x-6y-2z=7` `:.2a+b-2c=0` and `3a-6b-b-2c=0` `implies a/(-14)=b/(-2)=c/(-15)` Substituting these values of a,b,c in i we obtain `14x+2y+15z=31` as the equation of the required plane. |
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| 36. |
The distance of the point `(1,-2,4)` from the plane passing through the point `(1,2,2)` perpendicular to the planes `x-y+2z=3` and `2x-2y+z+12=0` isA. `1/(sqrt(2))`B. `2`C. `sqrt(2)`D. `2sqrt(2)` |
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Answer» Correct Answer - D Let `vecn` be a vector normal to the required plane. Then, its equation is `(vecr-(hati+2hatj+2hatk)).vecn=0`…………i This plane is perpendicular to the plane `x-y+2z=3` and `2x-2y+z+12=0` whose normals are `vecn_(1)=hati-hatj+2hatk` and `vecn_(2)=2hati-2hatj+hatk` `:.vecn=vecn_(1)xxvecn_(2)=|(hati,hatj,hatk),(1,-1,2),(2,-2,1)|=3hati+3hatj+0hatk` Substituting `vecn=3hati+3hatj` in i we obtain `vecr.(3hati+3hatj)=9` or `vecr.(hati+hatj)=3`............ii The distance of this plane from `veca=hati-2hatj+4hatk` is given by `d=|((hati-2hatj+4hatk).(hati+hatj)-3)/(sqrt(1+1))|=2sqrt(2)` |
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| 37. |
The value of `m` for which the line `(x-1)/2=(y-1)/3=(z-1)/m` is perpendicular to normal to the plane `vecr.(2hati+3hatj+4hatk)=0` isA. `-13/4`B. `-17/4`C. `4`D. none of these |
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Answer» Correct Answer - A Given line is parallel to the vector `vecb=2hati+3hatj+mhatk` and given plane is normal to `vecn=2hati+3hatj+hatk`. If `vecb` and `vecn` are perpendicular then `vecb.vecn=0implies4+9+4m=0impliesm=-13/4` |
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| 38. |
Find the angle between the planes `x+y+2z=9a d n2x-y+z=15.`A. `(pi)/6`B. `(pi)/3`C. `(pi)/4`D. none of these |
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Answer» Correct Answer - B Let `theta` be the angle between the planes `x+y+2z=9` and `2x-y+z=15`. Then `cos theta=((1)(2)+(1)(-1)+(2)(1))/(sqrt(1^(2)+1^(2)+2^(2))sqrt(2^(2)+(-1)^(2)+1^(2)))=1/2impliestheta=(pi)/3` |
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| 39. |
If the planes `2x-y+lamdaz-5=0` an `x+4y+2z-7=0` are perpendicular then `lamda=`A. 1B. -1C. 2D. -2 |
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Answer» Correct Answer - A Direction ratios of normals to given planes are proportional to `2,-1,lamda` and `1,4,2` respectively. If the planes are perpendicular then `2-4+2lamda=0implieslamda=1` |
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| 40. |
The vector equation of a plane passing through a point having position vector `2hati+3hatj-4hatk` and perpendicular to the vector `2hati-hatj+2hatk`, isA. `vecr.(2hati-hatj+2hatk)=7`B. `vecr.(2hati-hatj+2hatk)=-7`C. `vecr.(2hati-hatj+2hatk=4`D. none of these |
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Answer» Correct Answer - B We know that the vector equation of plane passing through a point `veca` and normal to `vecn` is `(vecr-veca).vecn=0` or `vecr.vecn=veca.vecn` Here `veca=2hati+3hatj-4hatk` and `vecn=2hati-hatj+2hatk` So the equation of the required plane is `vecr.(2hati-hatj+2hatk)=(2hati+3hatj-4hatk).(2hati-hatj+2hatk)` `impliesvecr.(2hati-hatj+2hatk)=-7` |
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| 41. |
The angle between the planes `vecr.(2hati-hatj+hatk)=6` and `vecr.(hati+hatj+2hatk)=5` isA. `(pi)/3`B. `(2pi)/3`C. `(pi)/6`D. `(5pi)/6` |
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Answer» Correct Answer - A We know that the angle between the planes `vecr.vecn_(1)=d_(1)` and`vecr.vecn_(2)=d_(2)` is given by `cos theta=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)` Here `vecn_(1)=2hati-hatj+hatk` and `vecn_(2)=hati+hatj+2hatk` `:.cos theta=((2hati-hatj+hatk).(hati+hatj+2hatk))/(|2hati-hatj+hatk||hati+hatj+2hatk|)=1/2` `impliestheta=pi//3` If `a_(1)x+b_(1)y+c_(1)z+d_(1)=0` and `a_(2)x+b_(2)y+c_(2)z+d_(2)=0` are Cartesian equations of two planes, then vectors normal to them are `vecn_(1)=a_(1)hati+b_(1)hatj+c_(1)hatk` and `vecn_(2)=a_(2)hati+b_(2)hatj+c_(2)hatk` respectively. Therefore, the angle `theta` between the planes is given by `cos thet=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)` `impliescos theta=(a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2))sqrt(a_(1)^(2)+b_(2)^(2)+c_(2)^(2)))` If the planes are perpendicular then `vecn_(1)` and `vecn_(2)` are perpendicular. `:.vecn_(1).vecn_(2)=0impliesa_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0` If the planes are parallel then `vecn_(1)` and `vecn_(2)` are parallel. `:.(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))` |
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| 42. |
If the planes `vecr.(2hati-hatj+2hatk)=4` and `vecr.(3hati+2hatj+lamda hatk)=3` are perpendicular then `lamda=`A. `2`B. `-2`C. `3`D. `-3` |
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Answer» Correct Answer - B Vectors normals to the given planes ae `vecn_(1)=2hati-hatj+2hatk` and `vecn_(2)=3hati+2hatj+lamdahatk` If the planes are perpendicular then `vecn_(1).vecn_(2)=0implies6-2+2lamda=0implieslamda=-2` |
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| 43. |
The vector equation of the plane passing through the points `hati+hatj-2hatk,2hati-hatj+hatk` and `hati+hatj+hatk`, isA. `vecr.(9hati+3hatj-hatk)=-14`B. `vecr.(9hati+3hatj-hatk)=14`C. `vecr.(3hati+9hatj-hatk)=14`D. none of these |
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Answer» Correct Answer - B Let A,B,C be the points with position vectors `hati+hatj-2hatk,2hati-hatj+hatk` and `hati+2hatj+hatk` respectively. Then, `vec(AB)=hati-2hatj+3hatk` and `vec(AB)=-hati+3hatj+0hatk` The vector normal to the plane containing points A,B and C is `vecn=vec(AB)xxvec(AC)=|(hati,hatj,hatk),(1,-2,3),(-1,3,0)|=-9hati-3hatj+hatk` The required plane passes through the point having position vector `veca=hati+hatj-2hatk` and is normal to the vector `vecn=-9hati-3hatj+hatk`. So its vector equation is `impliesvecr.vecn=veca.vecnimpliesveci.(9hati+3hatj-hatk)=14` |
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| 44. |
The equation of plane passing through the point `hati+hatj+hatk` and paralel to the plane `vecr.(2hati-hatj+2hatk)=5` isA. `vecr.(2hati-hatj+2hatk)=5`B. `vecr.(2hati-hatj+2hatk)=-3`C. `vecr.(2hati-hatj+2hatk)=3`D. none of these |
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Answer» Correct Answer - C The equation of a plane parallel to the plane `vecr.(2hati-hatj+2hatk)=5` is `vecr.(2hati-hatj+2hatk)=d`…………….i Since it passes through `hati+hatj+hatk`. `:.(hati+hatj+hatk).(2hati-hatj+2hatk)=dimplies2-1+2=d=d=3`. Putting `d=3` in i we obtain `vecr.(2hati-hatj+2hatk)=3` This is the equation of the required plane. |
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| 45. |
The vector equation of the plane which is at a distance of 8 units from the origin which is normal to the vector `2hati+hatj+2hatk` isA. `vecr.(2hati+hatj+2hatk)=8`B. `vecr.(2hati+hatj+2hatk)=24`C. `vecr.(2hati+hatj+2hatk)=4`D. none of these |
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Answer» Correct Answer - B Here `d=8` and `vecn=2hati+hatj+2hatk` `:. hatn=(vecn)/(|vecn|)=(2hati+hatj+2hatk)/(sqrt(4+1+4))=2/3hati+1/3hatj+2/3hatk` Hence the required equation of the plane is `vecr.(2/3hati+1/3hatj+2/3hatk)=8impliesvecr.(2hati+hatj+2hatk)=24` |
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| 46. |
Consider the plane `pi:x+y-2z=3` and two points `P(2,1,6)` and `Q(6,5,-2)`. Statement 1: PQ is parallel to the normal to the plane. Statement 2: `Q` is the image of point `P` in he plane `pi`A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B A vector normal to the plane `x+y-2z=3` is `vecn=hati+hatj-2hatk` and `vec(PQ)=4hati+4hatk-8hatk` Clearly `vec(PQ)=4vecn`. Therefore PQ is parallel to the normal to the plane `pi`. So statement -1 is true. Let `R` be the mid point of PQ. Then, coordinates of `R` are (4,3,2). Clearly, it lies onthe plane `pi`. So `Q` is the image of `P`. Hence statement -2 is true. |
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| 47. |
In `R^(3)`, consider the planes `P_(1):y=0` and `P_(2),x+z=1.` Let `P_(3)` be a plane, different from `P_(1)` and `P_(2)` which passes through the intersection of `P_(1)` and `P_(2)`, If the distance of the point (0,1,0) from `P_(3)` is 1 and the distance of a point `(alpha,beta,gamma)` from `P_(3)` is 2, then which of the following relation(s) is/are true?A. `2alpha+beta+2gamma+2=0`B. `2alpha-beta+2gamma+4=0`C. `2alpha+beta-2gamma-10=0`D. `2alpha-beta+2gamma-8=8` |
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Answer» Correct Answer - B::D The plane passing through the intersection of planes `P_(1):y=0` and `P_(2):x+z=1` is `P_(3):(x+z-1)+lamday=0` or `x+lamday+z-1=0` ………..i It is given that the distances of the point `(0,1,0)` and `(alpha, beta, gamma)` from i are 1 and 2 respectively. `:.|(0+lamda+0-1)/(sqrt(1+lamda^(2)+1))|=1` and `|(alpha+lamda beta+gamma-1)/(sqrt(1+lamda^(2)+1))|=2` `implies(|lamda-1|)/(sqrt(lamda^(2)+2))=1` and `(|alpha+lamda beta+gamma-1|)/(sqrt(lamda^(2)+2))=2` `implies(lamda-1)^(2)=lamda^(2)+2` and `|alpha+lamda beta+gamma-1|=2sqrt(lamda^(2)+2)` `implieslamda=-1/2` and `|alpha+lamda beta+gamma-1|-2sqrt(lamda^(2)+2)` `implies|2alpha-beta+2gamma-2|=6` `implies2alpha-beta+2gamma=+-2alpha-beta+2gamma-8=0` or `2alpha-beta+2gamma+4=0` |
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| 48. |
The equation of the plane which bisects the line joining `(2, 3, 4)` and `(6,7,8)`A. `x+y+z+15=0`B. `x-y-z-15=0`C. `x-y+z-15=0`D. `x+y+z-15=0` |
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Answer» Correct Answer - D Let the two points be `P(2,3,4)` and `Q(6,7,8)` and let `R` be their mid-point. The required plane passes through the mid point `R` and is normal to `PQ`. So, its equation is `[(vecr.(4hati+5hatj+6hatk)].(4hati+4hatj+4hatk)=0` `impliesvecr.(hati+hatj+hatk)-(4+5+6)=0impliesx+y+z-15=0` |
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| 49. |
If `P=(0,1,0) and Q=(0,0,1)` then the projection of `PQ` on the plane `x+y+z=3` isA. 2B. 3C. `sqrt(2)`D. `sqrt(3)` |
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Answer» Correct Answer - C The projection of `PQ` on the given plane is `PQ cos theta` where `theta` is the angle between `PQ` and the plane. Let `vecn` be a vector normal to the plane. We have `vec(PQ)=-hatj+hatk` and `vecn=hati+hati+hatk` `:.sin theta=(vec(PQ).vecn)/(|vec(PQ)||vecn|)=0impliesvec(PQ)` is parallel to the plane. Hence projection of `vec(PQ)` on the given plane `=|vec(PQ)|cos theta=|vec(PQ)|=sqrt(1+1)=sqrt(2)` |
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| 50. |
If the foot of the perpendicular from `O(0,0,0)` to a plane is `P(1,2,2)`. Then the equation of the plane isA. `-x+2y+8z-9=0`B. `x+2y+2z-9=0`C. `x+y+z-5=0`D. `x+2y-3z+1=0` |
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Answer» Correct Answer - B The plane passes through `P(1,2,2)` and is normal to `vec(OP)=hati+2hatj+2hatk`. So its vector equation is `vecr.vec(OP)=(hati+2hatj+2hatk).vec(OP)` [ Using `vecr.vecn=veca.vecn`] `impliesvecr.(hati+2hatj+2hatk)=(hati+2hatj+2hatk).(hati+2hatj+2hatk)` `impliesvecr.(hati+2hatj+hatk)=9impliesx+2y+2z=9` |
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