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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. |
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 distance of the point (1,1,1) from the plane passing throught the point (-1,-2,-1) and whose normal is perpendicular to both the lines `L_(1)` and `L_(2)`, isA. `2//sqrt(75)` unitB. `7//sqrt(75)` unitsC. `13//sqrt(75)` unitsD. `23//sqrt(75)` units |
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Answer» Correct Answer - C The equation of the plane passing through the point (-1,-2,-1) and whose normal is perpendicular to both the given lines `L_(1)` and `L_(2)` may be written as `(x+1)+7(y+2)-5(z+1)=0impliesx+7y-5z+10=0` The distance of the point (1,1,1) from the plane `=|(1+7-5+10)/(sqrt(1+49+25))|=(13)/(sqrt(75))`units |
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| 2. |
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. `(-hati+7hatj+7hatk)/(sqrt(99))`B. `(-hati-7hatj+5hatk)/(5sqrt(3))`C. `(-hati+7hatj+5hatk)/(5sqrt(3))`D. `(7hati-7hatj-hatk)/(sqrt(99))` |
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Answer» Correct Answer - B The equations of given lines in vector form may be written as `L_(1):vecr=(-hati-2hatj-hatk)+lambda(3hati+hatj+2hatk)` and `" "L_(2):vecr=(2hati-2hatj+3hatk)+mu(hati+2hatj+3hatk)` Since, the vector is perpendicular to both `L_(1)` and `L_(2).` `|{:(hati,hatj, hatk),(3,1,2),(1,2,3):}|=-hati-7hatj+5hatk` `:.` Required unit vector `=((-hati-7hatj+5hatk))/(sqrt((-1)^(2)+(-7)^(2)+(5)^(2)))=(1)/(5sqrt(3))(-hati-7hatj+5hatk)` |
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| 3. |
Consider the line L 1 : x 1 y 2 z 1 312 +++ ==, L2 : x2y2z3 123A. 0 unitB. `17//sqrt(3)` unitsC. `41//sqrt(5)`D. `17//5sqrt(3)` units |
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Answer» Correct Answer - D The shortest distance between `L_(1)` ad `L_(2)` is `|({(2-(-1))hati+(2-2)hatj+(3-(-1))hatk}.(-hati-7hatj+5hatk))/(5sqrt(3))|` `=|((3hati+4hatk).(-hati-7hatj+5hatk))/(5sqrt(3))|` `(17)/(5sqrt(3))` units |
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| 4. |
If the distance between the plane Ax 2y + z = d and the plane containing the lines2 1x=3 2y=4 3zand3 2x=4 3y=5 4zis 6 , then |d| is |
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Answer» Correct Answer - `|d|=6` Equation of the plane containing the lines `(x-2)/(2)=(y-3)/(5)" and "(x-1)/(2)=(y-2)/(3)=(z-3)/(4)` is `a(x-2)+b(gamma-3)+c(z-4)=0" "...(i)` where, `3a+4b+5c=0" "…(ii)` `2a+3b+4c=0" "(iii)` and`a(1-2)+b(2-3)+c(2-3)=0` i.e.`" "a+b+c=0" "...(iv)` From Eqs. (ii) and (iii), `(a)/(1)=(b)/(-2)=(c)/(1),` which satisfy Eq. (iv). Plane through lines is `x-2y+z=0.` Given plane is `Ax-2y+z=d` is `sqrt(6).` `:.` Planes must be parallel, so A=1 and then `(|d|)/(sqrt(6))=sqrt(6)implies|d|=6` |
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| 5. |
The angle betweenthe lines whose direction cosines satisfy the equations `l+m+n=""0`and `l^2=m^2+n^2`is(1) `pi/3`(2) `pi/4`(3) `pi/6`(4) `pi/2`A. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(2)` |
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Answer» Correct Answer - A We know that, angle between two lines is `cos 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_(2)^(2)+b_(2)^(2)+c_(2)^(2)))` `l+m+n=0` `implies" "l=-(m+n)implies(m+n)^(2)=l^(2)` `implies" "m^(2)+n^(2)+2mn=m^(2)+n^(2)" "[becausel^(2)=m^(2)+n^(2)," given"]` `implies" "2mn=0` When `" "m=0` `implies" "l=-n` Hence, (l, m, n) is (1, 0, -1). When `" "n=0," then "l=-m` Hence, (l, m, n) is (1, 0, -1). `:.costheta=(1+0+0)/(sqrt(2)xxsqrt(2))=(1)/(2)impliestheta=(pi)/(3)` |
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| 6. |
The distance of the point `(1,-5,""9)`from the plane `x-y+z=5`measured along the line `x=y=z`is :(1) `3sqrt(10)`(2) `10sqrt(3)`(3) `(10)/(sqrt(3))`(4) `(20)/3`A. `3sqrt(10)`B. `10sqrt(3)`C. `(10)/(sqrt(3))`D. `(20)/(3)` |
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Answer» Correct Answer - B Equation of line passing through the point (1, -5, 9) and parallel to x=y=z is `(x-1)/(1)=(y+5)/(1)=(z-9)/(1)=lambda" "("say")` Thus, any point on this line is of the form `(lambda+1,lambda-5,lambda+9).` Now, if `P(lambda+1,lambda-5,lambda+9)` is the point of intersection of line and plane, then `(lambda+1)-(lambda-5)+lambda+9=5` `implies" "lambda+15=5implieslambda=-10` `:."Coordinates of point P are "(-9, -15, -1).` Hence, required distance `sqrt((1+9)^(2)+(-5+15)^(2)+(9+1)^(2))` `=sqrt(10^(2)+10^(2)+10^(2))` `=10sqrt(3)` |
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| 7. |
Equation of the plane containing the straight line `x/2=y/3=z/4` and perpendicular to the plane containing the straight lines `x/2=y/4=z/2` and `x/4=y/2=z/3` isA. `5x+2y-4z=0`B. `x+2y-2z=0`C. `ax+2y-3z=0`D. `x-2y+z=0` |
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Answer» Correct Answer - D Let `P_(1)` be the plane containing the lines `(x)/(3)=(y)/(4)=(z)/(2)" and "(x)/(4)=(y)/(2)=(z)/(3).` For these two lies, direction vectors are `b_(1)=3hati+4hatj+2hatk" and "b_(2)=4hati+2hatj+3hatk.` A vector along the normal to the plane `P_(1)` is given by `n_(1)=b_(1)xxb_(2)=|{:(hati,hatj,hatk),(3,4,2),(4,2,3):}|` `hati(12-4)-hatj(9-8)+hatk(6-16)=8hati-hatj-10hatk` Let `P_(2)` be the plane containing the line `(x)/(2)=(y)/(3)=(z)/(4)` and perpendicular to plane` P_(1).` For the line `(x)/(2)=(y)/(3)=(z)/(4)`, the direction vector is `b=2hati+3hatj+4hatk` and it passes through the point with position vector `a=0hati+0hatj+0hatk.` `becauseP_(2)` is perpendicular to `P_(1)`, therefore `n_(1)` and b lies along the plane. Also, `P_(2)` also passes through the point with position vector a. `:.` Equation of plane `P_(2)` is given by `(r-a).(n_(1)xxb)=0implies|{:(x-0,y-0,z-0),(8,-1,-10),(2," "3," "4):}|=0` `impliesx(-4+30)-y(32+20)+z(24-2)=0` `implies26x-52y+26z=0` `implies" "x-2y+z=0` |
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| 8. |
If `L_1` is the line of intersection of the planes `2x-2y+3x-2=0``x-y+z+1=0` and `L_2` is the line of the intersection of the planes `x+2y-z-3=0` `3x-y+2z-1=0`then the distance of the origin from the plane containing the lines `L_1` and `L_2` isA. `(1)/(4sqrt(2))`B. `(1)/(3sqrt(2))`C. `(1)/(2sqrt(2))`D. `(1)/(sqrt(2))` |
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Answer» Correct Answer - B `L_(1)` is the line of intersection of the plane `2x-2y+3z-2=0` and `x-y+z+1=0` and `L_(2)` is the line of intersection of the plane `x+2y-z-3=0` and `3x-y+2z-1=0` Since `L_(1)` is parallel to `|{:(hati,hatj,hatk),(2,-2,3),(1,-1,1):}|=hati+hatj` `L_(2)` is parallel to `|{:(hati,hatj,hatk),(1,2,-1),(3,-1,2):}|=3hati-5hatj-7hatk` Also, `L_(2)` passes through `((5)/(7),(8)/(7),0).` [put z=0 in last two planes] So, equation of plane is `|{:(x-(5)/(7),y-(8)/(7),z),(1,1,0),(3,-5,-7):}|=0implies7x-7y+8z+3=0` Now, perpendicular distance from origin is `|(3)/(sqrt(7^(2)+7^(2)+8^(2)))|=(3)/(sqrt(162))=(1)/(3sqrt(2))` |
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| 9. |
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines `r=(hati+hatj)+lambda(hati+2hatj-hatk)" and "r=(hati+hatj)+mu(-hati+hatj-2hatk)` isA. 3B. `(1)/(3)`C. `sqrt(3)`D. `(1)/(sqrt(3))` |
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Answer» Correct Answer - C Length of the perpendicular drawn from point `(x_(1), y_(1), z_(1))` to the plane `ax+by+cz+d=0`is `d_(1)=(|ax_(1)+by_(1)+cz_(1)+d|)/(sqrt(a^(2)+b^(2)+c ^(2)))` Given line vectors `r=(hati+hatj)+lambda(hati+2hatj-hatk)" and "" "...(i)` `r=(hati+hatj)+mu(-hati+hatj-2hatk)" "...(ii)` Now, a vector perpendicular to the given vectors (i) and (ii) is `n=|{:(" "hati,hatj," "hatk),(" "1,2,-1),(-1,1,-2):}|` `=hati(-4+1)-hatj(-2-1)+hatk(1+2)` `=-3hati+3hatj+3hatk` `:.`The equation of plane containing given vecotrs (i) and (ii) is `-3(x-1)+3(y-1)+3(z-0)=0` `implies" "-3x+3y+3z=0` `implies" "x-y-z=0" "...(iii)` Now, the length of perpendicular drawn from the point (2, 1, 4) to the plane `x-y-z=0`, is `d_(1)=(|2-1-4|)/(sqrt(1+1+1))` `=(3)/(sqrt(3))=sqrt(3)` |
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| 10. |
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) 27A. `2sqrt(14)`B. 8C. `3sqrt(21)`D. 13 |
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Answer» Correct Answer - D Given equation of line is `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)=lambda" "["say"]…(i)` and equation of plane is `x-y+z=16" "…(ii)` Any point on the line (i) is, `(3lambda+2,4lambda-1,12lambda+2)` Let this point of intersection of the line and plane. `:.(3lambda+2)-(4lambda-1)+(12lambda+2)=16` `implies" "11lambda+5=16` `implies" "11lambda=11` `implies" "lambda=1` So, the point of intersection is (5, 3, 14) Now, distance between the points (1, 0, 2) and (5, 3, 14) `sqrt((5-1)^(2)+(3-0)^(2)+(14-2)^(2))` `=sqrt(16+9+144)=sqrt(169)=13` |
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| 11. |
The plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` and parallel to Y-axis also passes through the pointA. (3, 3, -1)B. (-3, 1, 1)C. (3, 2, 1)D. (-3, 0, -1) |
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Answer» Correct Answer - C Equation of poane through the intersection of two planes `P_(1)` and `P_(2)` is given by `P_(1)+lambdaP_(2)=0` The plane through the intersection of the planes `x+y+z-1=0` and `2x+3y-z+4=0` is given by `(x+y+z-1)+lambda(2x+3y-z+4)=0,` where `lambdainR` `implies(1+2lambda)x+(1+3lambda)y+(1-lambda)z+(4lambda-1)=0,` where `lambdainR" "...(i)` Since, this plane is parallel to Y-axis, therefore its normal is perpendicular to Y-axis, therefore its normal is perpendicular to Y-axis. `implies" "{(1+2lambda)hati+(1+3lambda)hatj+(1-lambda)hatk}.hatj=0` `implies" "1+3lambda=0implies" "lambda=-(1)/(3)` Now, required equation of plane is `(1-(2)/(3))x+(1-(3)/(3))y+(1+(1)/(3))z+(-(4)/(3)-1)=0` [substituting `lambda=(-1)/(3)` in Eq. (i)] `impliesx+4z-7=0` Here, only (3, 2, 1) satisfy the above equation. |
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| 12. |
The magnitude of the projection of the vector `2hati+3hatj+hatk` on the vector perpendicular to the plane containing the vectors `hati+hatj+hatk" and "hati+2hatj+3hatk`, isA. `sqrt(3)/(2)`B. `sqrt(6)`C. `3sqrt(6)`D. `sqrt((3)/(2))` |
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Answer» Correct Answer - D The normal vector to the plane containing the vecotors `(hati+hatj+hatk)" and "(hati+2hatj+3hatk)` is `n=(hati+hatj+hatk)xx(hati+2hatj+3hatk)` `=|{:(hati,hatj,hatk),(1,1,1),(1,2,3):}|` `=hati(3-2)-hatj(3-1)+hatk(2-1)=hati-2hatj+hatk` Now, magnitude of the projection of vector `2hati+3hatj+hatk` on normal vector n is `(|(2hati+3hatj+hatk).n|)/(|n|)=(|(2hati+3hatj+hatk).(hati-2hatj+hatk)|)/(sqrt(1+74+1))` `=(|2-6+1|)/(sqrt(6))=(3)/(sqrt(6))=sqrt((3)/(2))`units |
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| 13. |
The driection ratios of normal to the plane through the points (0, -1, 0) and (0, 0, 1) and making an angle `pi//4` with the plane `y-z+5=0` areA. 2, -1, 1B. `sqrt(2), 1, -1`C. `2, sqrt(2), -sqrt(2)`D. `2sqrt(3), 1, -1` |
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Answer» Correct Answer - B::C Let the equation of plane be `a(x-0)+b(y+1)+c(z-0)=0` [`because` Equation of plane passing through a point `(x_(1),y_(1),z_(1))` is given by `a(x-x_(1))+b(y-y_(1))+c(z-z_(1))=0`] `impliesax+by+cz+b=0" "...(i)` Since, it also passes through (0, 0, 1) therefore, we get `c+b=0" "(ii)` Now, as angle between the planes `ax+by+cz+b=c` and `" "y-z+5=0" is "(pi)/(4).` `:.cos((pi)/(4))=(|n_(1).n_(2)|)/(|n_(1)||n_(2)|)`, where `n_(1)=ahati+bhatj+chatk` and `n _(2)=0hati+hatj+hatk` `implies" "(1)/(sqrt(2))=(|(ahati+bhatj+chatk).(0hati+hatj-hatk)|)/(sqrt(a^(2)+b^(2)+c^(2))sqrt(0+1+1))` `=(|b-c|)/(sqrt(a^(2)+b^(2)+c^(2))sqrt(2))` `impliesa^(2)+b^(2)+c^(2)=|b-c|^(2)=(b-c)^(2)=b^(2)+c^(2)-2bc` `implies" "a^(2)=-2bc` `implies" "a^(2)=2b^(2)" "["Using Eq. (ii)"]` `implies" "a=+-sqrt(2)b` `implies"Direction ratios (a, b, c)"=(+-sqrt(2), 1, -1)` So, options (b) and (c) are correct because `2, sqrt(2), -sqrt(2)` and `sqrt(2), 1, -1`. are multiple of each other. |
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| 14. |
On which of the following lines lies the point of intersection of the line, `(x-4)/(2)=(y-5)/(2)=(z-3)/(1)` and the plane, `x+y+z=2`?A. `(x-4)/(1)=(y-5)/(1)=(z-5)/(-1)`B. `(x+3)/(3)=(4-y)/(3)=(z+1)/(-2)`C. `(x-2)/(2)=(y-3)/(2)=(z+3)/(3)`D. `(x-1)/(1)=(y-3)/(2)=(z+4)/(-5)` |
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Answer» Correct Answer - D Given equation of line is `(x-4)/(2)=(y-5)/(2)=(z-3)/(1)=r ("let")" "…(i)` `impliesx=2r+4,y=2r+5" and "z=r+3` `:."General point on the line (i) is "` `P(2r+4,2r+5,r+3)` So, the point of intersection of line (i) and plane `x+y+z=2` will be of the form `P(2r+4,2r+5,r+3)` for some `rinR.` `implies(2r+4)+(2r+5)+(r+3)=2` [`because` the point will lie on the plane] `implies5r=-10impliesr=-2` So, the point of intersection is P(0, 1, 1) [putting `r=-2` in `(2r+4,2r+5,r+3)]` Now, on checking the options, we get `(x-1)/(1)=(y-3)/(2)=(z+4)/(-5)` contain the point (0,1,1) |
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| 15. |
The equation of a plane passing through the line of intersection of the planes x+2y+3z = 2 and x y + z = 3 and at a distance 2 3 from the point (3, 1, 1) is (A) 5x 11y + z = 17 (B) 2x y 3 2 1 (C) x + y + z = 3 (D) x 2y 1 2A. `5x-11y+z=17`B. `sqrt(2)x+y=3sqrt(2)-1`C. `x+y+z=sqrt(3)`D. `x-sqrt(y)=1-sqrt(2)` |
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Answer» Correct Answer - A Key Idea (i) Equation on plane through intersection of two planes, i.e. `(a_(1)x+b_(1)y+c_(1)z+d_(1))+lambda` `(a_(2)x+b_(2)y+c _(2)z+d_(2))=0` (ii) Distance of a point `(x_(1),y_(1),z_(1))` from `ax+by+cz+d=0` `=(|ax_(1)+by_(1)+cz_(1)+d|)/(sqrt(a^(2)+b^(2)+c^(2)))` Equation of plane passing through intersection of two planes `x+2y+3z=2" and "x-y+z=3` is `(x+2y+3z-2)+lambda(x-y+z-3)=0` `implies(1+lambda)x+(2-lambda)y+(3+lambda)z-(2+3lambda)=0` whose distance from (3, 1, -1) is `(2)/(sqrt(3))` `implies(|3(1+lambda)+1.(2-lambda)-1(3+lambda)-(2+3lambda)|)/(sqrt((1+lambda)^(2)+(2-lambda)^(2)+(3+lambda)^(2)))=(2)/(sqrt(3))` `implies(|-2lambda|)/(sqrt(3lambda^(2)+4lambda+14))=(2)/(sqrt(2))` `implies" "3lambda^(2)=3lambda^(2)+4lambda+14` `implies" "lambda=-(7)/(2)` `:.(1-(7)/(2))x+(2+(7)/(2))y+(3-(7)/(2))z-(2-(21)/(2))=0` `implies" "-(5x)/(2)+(11)/(2)y-(1)/(2)z+(17)/(2)=0` or`" "5x-11y+z-17=0` |
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| 16. |
If `Q(0, -1, -3)` is the image of the point P in the plane `3x-y+4z=2` and R is the point (3, -1, -2), then the area (in sq units) of `DeltaPQR` isA. `(sqrt(91))/(2)`B. `2sqrt(13)`C. `(sqrt(91))/(4)`D. `(sqrt(65))/(2)` |
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Answer» Correct Answer - A Given , equation of plance `3x-y+4z=2" "(i)` and the point `Q(0,-1,-3)` is the image of point P in the plane (i), so point P is also image of point Q w.r.t. plane (i). Let the coordinates of point P is `(x_(1), y_(1), z_(1)),` then `(x_(1)-0)/(3)=(y_(1)+1)/(-1)=(z_(1)+3)/(4)` `=-2([3(0)-1(-1)+4(-3)-2])/(3^(2)+(-1)^(2)+4^(2))` "["`:.` image of the point `(x_(1), y_(1), z_(1))` in the plane `ax+by+cz+d=0` is (x, y, z), where `(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))"]"` `implies(x_(1)-0)/(3)=(y_(1)+1)/(-1)=(z_(1)+3)/(4)` `=2((1-12-2))/(26)=(26)/(26)=1` `implies" "P(x_(1), y_(1), z_(1))=(3, -2, 1)` Now, area of `DeltaPQR`, where point `R(3, -1, -2)` `=(1)/(2)|vec(PQ)xxvec(PR)|=(1)/(2)|"("-3hati+hatj+4hatk")"xx"("0hati+hatj-3hatk")"|` `=(1)/(2)||{:(hati,hatj," "hatk),(-3,1,-4),(0,1,-3):}||=(1)/(2)|hati-9hatj-3hatk|` `=(1)/(2)sqrt(1+81+9)=(sqrt(91))/(2)` sq units |
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| 17. |
The equation of the plane passing through the point 1,1,1) andperpendicular to the planes `2x+y-2z=5a n d3x-6y-2z=7,`is`14 x+2y+15 z=3``14 x+2y-15 z=1``14 x+2y+15 z=31``14 x-2y+15 z=27`A. `14x+2y-15z=1`B. `-14x+2y+15z=3`C. `14x-2y+15z=27`D. `14x+2y+15z=31` |
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Answer» Correct Answer - D Let the equation of plane e `ax+by+cz=1.` Then `a+b+c=1` `2a+b-2c=0` `3a-6b-2c=0impliesa=7b,c=(15b)/(2)` `b=(2)/(31),a=(14)/(31),c=(15)/(31)` `:.14x+2y+15z=31` |
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| 18. |
(i) Find the equation ofthe plane passing through the points`(2,1,0),(5,0,1)a n d(4,11)dot`(ii) If `P`s the point `(2,1,6),`then the find the point `Q`such that `P Q`is perpendicular to the plane in (i) and themidpoint of `P Q`lies on it. |
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Answer» Correct Answer - (i) `x+y-2z=3" "(ii) Q(6,5,-2)` (i) Equation of plane passing throuth (2,1,0) is `a(x-2)+b(y-1)+c(z-0)=0` It also passes through (5,0,1) and (4,1,1). `implies3a-b+c=0" and "2a-0b+c=0` On solving, we get `(a)/(-1)=(b)/(-1)=( c )/(2)` `:.` Equation of plane is `-(x-2)-(y-1)+2(z-0)=0` `-(x-2)-y+1+2z=0` `implies" "x+y-2z=3` (ii) Let the coordinates of Q be `(alpha,beta,gamma).` Equation of line `PQimplies(x-2)/(1)=(y-1)/(1)=(z-6)/(-2)` Since, mid-point of P and Q `((alpha+2)/(2),(beta+1)/(2),(gamma+6)/(-2)),` which lies in line PQ. `implies" "((a-2)/(2)-2)/(2)=((beta+1)/(2)-1)/(1)=((gamma+6)/(2)-6)/(-2)` `(1((alpha+2)/(2)-2)+1((beta+1)/(2)-1)-2((gamma+6)/(2)-6))/(1.1+1.1+(-2)(-2))=2` `[because((alpha+2)/(2))-1((beta+1)/(2))-2((gamma+6)/(2))=3]` `impliesalpha=6,beta=5,gamma=-2impliesQ(6,5,-2)` |
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| 19. |
If the line `(x-1)/(2)=(y+1)/(3)=(z-2)/(4)` meets the plane, `x+2y+3z=15` at a point P, then the distance of P from the origin isA. `7//2`B. `9//2`C. `sqrt(5)//2`D. `2sqrt(5)` |
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Answer» Correct Answer - B Equation of given plane is `x+2y+3z=15" "...(i)` and line is, `(x-1)/(2)=(y+1)/(3)=(z-2)/(4)=r("let")" "...(ii)` So, the coordinates of any point on line (ii) is `P(1+2r,-1+3r,2+4r)`. `because` Point P is intersecting point of plane (i) and line (ii) `:.(1+2r)+2(-1+3r)+3(2+4r)=15` `implies1+2r-2+6r+6+12r=15implies20r=10` `impliesr=(1)/(2)` `:.` Coordinates of `P=(1+1,-1+(3)/(2),2+2)=(2,(1)/(2),4)` Now, distance of the point P from the origin `=sqrt(4+(1)/(4)+16)=sqrt(20+(1)/(4))=sqrt((81)/(4))=(9)/(2)` units |
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| 20. |
Find the equation of the plane containing the lines 2x-y+z-3=0,3x+y+z=5 and a t a distance of `1/sqrt6` from the point (2,1,-1). |
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Answer» Correct Answer - `2x-y+z-3=0;62x+29y+19z-105=0` Equation of plane containing the lines `2x-y+z-3=0" and "3x+y+z=5` is `(2x-y+z-3)+lambda(3x+y+z-5)=0` `implies(2+3lambda)x+(lambda-1)y+(lambda+1)z-3-5lambda=0` Since, distance of plane from (2,1,-1) to above plane is `1sqrt(6).` `:.|(6lambda+4+lambda-1-lambda-1-3-5lambda)/(sqrt((3lambda+2)^(2)+(lambda-1)^(2)+(lambda+1)^(2)))|=(1)/(sqrt(6))` `implies" "6(lambda-1)^(2)=11lambda^(2)+12lambda+6` `implies" "lambda=0,-(24)/(5)` `:.` Equations of planes are `2x-y+z-3=0" and "62x+29y+19z-105=01` |
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| 21. |
A plane passing through the points (0, -1, 0) and (0, 0, 1) and making an angle `(pi)/(4)` with the plane `y-z+5=0`, also passes through the pointA. `(sqrt(2), 1, 4)`B. `(-sqrt(2), 1, -4)`C. `(-sqrt(2), -1, -4)`D. `(sqrt(2), -1, 4)` |
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Answer» Correct Answer - A Let the equation of plane is `ax+by+cz=d" "…(i)` Since plane (i) passes through the points `(0, -1, 0)` and (0, 0, 1), then -b=d and c = d `:.` Equation of plane becomes `ax-dy+dz=d" "…(ii)` The plane (ii) makes an angle of `(pi)/(4)` with the plane `y-z+5=0.` `"cos"(pi)/(4)=|(-d-d)/(sqrt(a^(2)+d^(2)+d^(2))sqrt(1+1))|` [`because` The angle between the two plances `a_(1)x+b_(1)y+c_(1)z+d=0" and "a_(2)x+b_(2)y+c_(2)z+d=0`is `costheta=|(a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2))sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))|`] `implies(1)/(sqrt(2))=(|2-d|)/(sqrt(a^(2)+2d^(2))sqrt(2))impliessqrt(a^(2)+2d^(2))=|-d|` `impliesa^(2)+2d^(2)=4d^(2)" "["squaring both sides"]` `impliesa^(2)-2d^(2)impliesa=+-sqrt(2)d` So, the Eq. (ii) becomes `+-sqrt(2)x-y+z=1" "...(i)` Now, from options `(sqrt(2), 1, 4)` satisfy the plane `-sqrt(2)x-y+z=1` |
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| 22. |
A plane passes through (1,-2,1) and is perpendicualr to two planes` 2x-2y+z=0" and "x-y+2z=4,` then the distance of the plane from the point (1,2,2) is |
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Answer» Correct Answer - D Let the equation of plane be `a(x-1)+b(y+2)+c(z-1)=0` which is perpendicular to `2x-2y+z=0` and `x-y+2z=4.` `implies" "2a-2b+c=0" and "a-b+2c=0` `implies" "(a)/(-3)=(b)/(-3)=(c)/(0)implies(a)/(1)=(b)/(1)=(c)/(0).` So, the equation of plane is `x-1+y+2=0` or`" "x+y+1=0` Its distance from the point (1,2,2) is `(|1+2+1|)/(sqrt(2))=2sqrt(2)` |
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| 23. |
Let `vec u` be a vector coplanar with the vectors `vec a = 2 hat i + 3 hat j - hat k` and `vec b= hat j+hatk` If `vec u` is perpendicular to `vec a` and `vec u.vecb=24` then `|vecu|^2` is equal toA. 336B. 315C. 256D. 84 |
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Answer» Correct Answer - A If any vector x is coplanar with the vector y and z, then `x=lambday+muz` Here, u is coplanar with a and b. `:." "u=lambdaa+mub` Dot product with a, we get `u.a=lambda(a.a)+mu(b.a)implies0=14lambda+2mu" "…(i)` `[becausea=2hati+3hatj-hatk,b=hatj+hatk,u.a=0]` Dot product with b, we get `u.b=lambda(a.b)+mu(b.b)` `24=2lambda+2mu" "...(ii)[becauseu.b=24]` Solving Eqs. (i) and (ii), we get `lambda=-2,mu=14` Dot product with u, we get `|u|^(2)=lambda(u.a)+mu(u.b)` `|u|^(2)=-2(0)+14(24)implies|u|^(2)=336` |
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| 24. |
If the plane `2x-y+2z+3=0` has the distances `(1)/(3)` and `(2)/(3)` units from the planes `4x-2y+4z+lambda=0` and `2x-y+2z+mu=0`, respectively, then the maximum value of `lambda+mu` is equal toA. 13B. 15C. 5D. 9 |
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Answer» Correct Answer - A Equation of given planes are `2x-y+2z+3=0" "…(i)` `4x-2y+4z+lambda=0" "...(ii)` and `" "2x-y+2x+mu=0" "...(iii)` `because`Distance between two parallel planes `ax+by+cz+d_(1)=0` and`" "ax+by+cz+d_(2)=0` is distance`=(|d_(1)-d_(2)|)/(sqrt(a^(2)+b^(2)+c^(2)))` `:.`Distance between planes (i) and (ii) is `(|lambda-2(3)|)/(sqrt(16+4+16))=(1)/(3)" "["given"]` `implies|lambda-6|=2implies" "lambda-6=+-2implieslambda=8" or "4` and distance between planes (i) and (iii) is `(|mu-3|)/(sqrt(4+1+4))=(2)/(3)" "["given"]` `implies" "|mu-3|=2` `implies" "mu-3=+-2impliesmu=5` or 1 So, maximum value of `(lambda+mu)` at `lambda=8` and `mu=5` it is equal to 13. |
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| 25. |
Let `P`be a pointin the first octant, whose image `Q`in theplane `x+y=3`(that is,the line segment `P Q`isperpendicular to the plane `x+y=3`and themid-point of `P Q`lies in theplane `x+y=3)`lies on the z-axis. Let the distance of `P`from the x-axis be 5. If `R`is theimage of `P`in thexy-plane, then the length of `P R`is _______. |
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Answer» Correct Answer - 8 Let `P(alpha,beta,gamma)` and R is image of P in the xy-plane. `:.R(alpha,beta,-gamma)` Also, Q is the imgae of P in the plane `x+y=3` `:.(x-alpha)/(1)=(y-beta)/(1)=(z-gamma)/(0)=(-2(alpha+beta-3))/(2)` `x=3-beta,y=3-alpha,z=gamma` Since, Q is lies on Z-axis `:." "beta=3,alpha=3,z=gamma` `:.P(3,3,gamma)` Given, distance of P from X-axis be 5. `:." "5=sqrt(3^(2)-gamma^(2))` `25-9=gamma^(2)` `impliesgamma=+-4` Then, `" "PR=|2gamma|=|2xx4|=8` |
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| 26. |
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. any valueB. exactly one valueC. exactly two valuesD. exactly three values |
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Answer» Correct Answer - C Condition for two lines are coplanar. `|{:(x_(1)-x_(2),y_(1)-y_(2),z_(1)-z_(2)),(" "l_(1)," "m_(1)," "n_(1)),(" "l_(2)," "m_(2)," "n_(2)):}|=0` where, `(x_(1), y_(a), z_(1))` and `(x_(2), y_(2), z_(2))` are the points lie on lines (i) and (ii) respectively and `ltl_(1), m_(1), n_(1)gt` and `ltl_(2), m_(2), n_(2)gt` are the dirction cosines of the lines (i) and (ii), respectively. `:." "|{:(2-1,3-4,4-5),(" "1," "1," "-k),(" "k," "2," "1):}|=0` `implies" "|{:(1,-1," "-1),(1," "1," "-k),(k," "2," "1):}|=0` `implies1(1+2k)+(1+k^(2))-(2-k)=0` `implies" "k^(2)+2k+k=0` `implies" "k^(2)+3k=0` `implies" "k=0,-3` If 0 appears in the denominator, then the correct way of representing the equation of straight line is `(x-2)/(1)=(y-3)/(1),z=4` |
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| 27. |
if the lines `(x-1)/2=(y-1)/3=(z-1)/4 and (x-3)/2=(y-k)/1=z/1` intersect then the value of `k` is (a) `1/3` (b) `2/3` (c) `-1/3` (d) `1`A. `(3)/(2)`B. `(9)/(2)`C. `-(2)/(9)`D. `-(3)/(2)` |
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Answer» Correct Answer - B Since, the lines intersect, therefore they must have a point in common, i.e. `(x-1)/(2)=(y+1)/(3)=(z-1)/(4)=lambeda` and`" "(x-3)/(1)=(y-k)/(2)=(z)/(1)=mu` `implies" "x=2lambda+1,y=3lambda-1` `z=4lambda+1` and`" "x=mu+3,y=2mu+k,z=mu` are same. `implies" "2lambda+1=mu+3` `3lambda-1=2mu+k` `4lambda+1=mu` On solving Ist and IIIrd terms, we get, `lambda=-(3)/(2)" and "mu=-5` `:." "k=3lambda-2mu-1` `implies" "k=3(-(3)/(2))-2(-5)-1=(9)/(2)` `:." "k=(9)/(2)` |
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| 28. |
If the straight linesx 1 y 1 z 2 k 2andx 1 y 1 z 5 2 kare coplanar, then the plane (s)containing these two lines is (are) (A) y + 2z = 1 (B) y + z = 1 (C) y z = 1 (D) y 2z = 1 55A. `y+2z=-1`B. `y+z=-1`C. `y-z=-1`D. `y-2z=-1` |
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Answer» Correct Answer - B::C PLAN If the straight lines are coplanar. They the should lie in same plane. Description of Situation If straight lines are coplanar. `implies" "|{:(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)):}|=0` Since,`" "(x-1)/(2)=(y+1)/(K)=(z)/(2)` and`" "(x+1)/(5)=(y+1)/(2)=(z)/(k)` are coplanar. `implies" "|{:(2,0,0),(2,K,2),(5,2,K):}|=0impliesK^(2)=4impliesK=+-2` `:." "n_(1)=b_(1)xxd_(1)=6j-6k," for "k=2` `:." "n_(2)=b_(2)xxd_(2)=14j-14k," for "k=-2` So, equation of planes are `(r-a).n_(1)=0` `implies" "y-z=-1" and "(r-a).n_(2)=0` `implies" "y+z=-1` |
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| 29. |
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) 2A. 26B. 18C. 5D. 2 |
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Answer» Correct Answer - D Since, the line `(x-3)/(2)=(y+2)/(-1)=(z+4)/(3)` line in the plane `lx+my-z=9,` therefore we have `2l-m-3=0` [`because` normal will be perpendicular to the line] `implies" "2l-m=3" "...(i)` and `" "3l-2m+4=9` [`because` point (3, -2, -4) lies on the plane] `implies" "3l-2m=5" "...(ii)` On solving Eqs. (i) and (ii), we get `l=1" and "m=-1` `:." "l^(2)+m^(2)=2` |
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| 30. |
The length of the perpendicular from the point (2, -1, 4) on the straight line, `(x+3)/(10)=(y-2)/(-7)=(z)/(1)` isA. greater than 3 but less than 4B. less than 2C. greater than 2 but less than 3D. greater than 4 |
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Answer» Correct Answer - A Equation of given line is `(x+3)/(10)=(y-2)/(7)=(z)/(1)=r("let")" "...(i)` Coordinates of a point on line (i) is `A(10r-3,-7r+2,r)` Now, let the line joining the points `P(2, -1, 4)` and `A(10-r-3, -7r+2, r)` is perpendicular to line (i). Then, `PA.(10hati-7hatj+hatk)=0` [`because` vector along line (i) is `(10hati-7hatj+hatk)`] `implies[(10r-5)hati+(-7r+3)hatj+(r-4)hatk].[10hati-7hatj+hatk]=0` `implies10(10r-5)-7(3-7r)+(r-4)=0` `implies100r-50-21+49r+r-4=0` `implies150r=75impliesr=(1)/(2)` So, the foot of perpendicular is `A(2,-(3)/(2),(1)/(2))` [put `r=(1)/(2)` in the coordinatcs of point A] Now, perpendicualr distance of point `P(2, -1, 4)` from the line (i) is `PA=sqrt((2-2)^(2)+(-(3)/(2)+1)^(2)+((1)/(2)-4)^(2))` `=sqrt((1)/(4)+(49)/(4))=sqrt((50)/(4))=(5)/(sqrt(2))` which lies in (3, 4). |
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| 31. |
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)/(-1)=(z-2)/(5)` |
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Answer» Correct Answer - A Here, plane line and its image are parallel to each other. So, find any point on the normal to the plane from which the image line will be passed and then find equation of image line. Here, plane and line are parallel to each other. Equation of normal to the plane throught the point (1, 3, 4) is `(x-1)/(2)=(y-3)/(-1)=(z-4)/(1)=k" "["say"]` Any point in this normal is (2k+1,-k+3,4+k).` Then, ((2k+1+1)/(2),(3-k+3)/(2),(4+k+4)/(2))` lies on plane. `implies" "(k+1)-((6-k)/(2))+((8+k)/(2))+3=0impliesk=-2` Hence, point through which this image pass is `(2k+1,3-k,4+k)` i.e.`[2(-2)+1,3+2,4-2]=(-3,5,2)` Hence, equation of image line is `(x+3)/(2)=(y-5)/(1)=(z-2)/(-5)` |
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| 32. |
The value of `k`such that `(x-4)/1=(y-2)/1=(z-k)/2`lies in the plane `2x-4y=z=7`isa. 7 b. -7 c. no real value d. 4A. 7B. -7C. No real valueD. 4 |
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Answer» Correct Answer - A Given equation of straight line `(x-4)/(1)=(y-2)/(1)=(z-k)/(2)` Since, the line lies in the plane `2x-4y+z=7.` Hence, point (4,2,k) must satisfy the plane. `implies" "8-8+k=7impliesk=7` |
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| 33. |
Let `P_(1):2x+y-z=3" and "P_(2):x+2y+z=2` be two planes. Then, which of the following statement(s) is (are) TRUE?A. The line of intersection of `P_(1)` and `P_(2)` has direction ratios 1,2,-1B. The line `(3x-4)/(9)=(1-3y)/(9)=(z)/(3)` is perpendicular to the line of intersection of `P_(1)` and `P_(2)`C. The acute angle between `P_(1)` and `P_(2)` is `60^(@)`D. If `P_(3)` is the plane passing through the point (4,2,-2) and perpendicular to the line of intersection of `P_(1)` and `P_(2)`, then the distance of the point (2,1,1) from the plane `P_(3)" is "(2)/(sqrt(3))` |
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Answer» Correct Answer - C::D We have, `P_(1):2x+y-z=3` and `" "P_(2):x+2y+z=2` Here, `" "vecn_(1)=2hati+hatj-hatk` and`" "vecn_(2)=hati+2hatj+hatk` (a) Direction ratio of the line of intersection of `P_(1)` and `P_(2)` is `thetavecn_(1)xxvecn_(2)` `i.e.|{:(hati,hatj,hatk),(2,1,-1),(1,2,1):}|=(1+2)hati-(2+1)hatj+(4-1)hatk` `=3(hati-hatj+hatk)` Hence, statement a is false. (b) We have, `(3x-4)/(9)=(1-3y)/(9)=(z)/(3)` `implies" "(x-(4)/(3))/(3)=((y-(1)/(3)))/(-3)=(z)/(3)` This line is parallel to the line of intersection of `P_(1)` and `P_(2)`. Hence, statement (b) is false. (c) Let acute angle between `P_(1)` and `P_(2)` be `theta`. We know that, `costheta=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)=((2hati+hatj-hatk).(hati+2hatj+hatk))/(|2hati+hatj-hatk||hati+2hatj+hatk|)` `(2+2-1)/(sqrt(6)xxsqrt(6))=(1)/(2)` `theta=60^(@)` Hence, statement (c) is true. (d) Equation of plane passing through the point (4,2,-2) and perpendicular to the line of intersection of `P_(1)` and `P_(2)` is `3(x-4)-3(y-2)+3(z+2)=0` `implies" "3x-3y+3z-12+6+6=0` `implies" "x-y+z=0` Now, distance of the point (2, 1, 1) from the plane `x-y+z=0` is `D=|(2-1+1)/(sqrt(1+1+1))|=(2)/(sqrt(3))` Hence, statement (d) is true. |
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| 34. |
Find the equation of the plane through the line of intersection of the planes `x+y+z=1 and 2x+3y+4z=5` which is perpendicular to the plane `x-y+z=0`A. `r.(hati-hatk)-2=0`B. `rxx(hati+hatk)+2=0`C. `rxx(hati-hatk)+2=0`D. `r.(hati-hatk)+2=0` |
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Answer» Correct Answer - D Since, equation of planes passes through the line of intersection of the planes `x+y+z=1` and `2x+3y+4z=5,` is `(x+y+z-1)+lambda(2x+3y+4z-5)=0` `implies" "(1+2lambda)x+(1+3lambda)y+(1+4lambda)z-(1+5lambda)=0" "...(i)` `because` The plane (i) is perpendicular to the plane `x-y+z=0`. `:.(1+2lambda)-(1+3lambda)+(1+4lambda)=0` [`because` if plane `a_(1)x+b_(1)y+c_(1)z+d_(1)=0` is perpendicular to plane `a_(2)x+b_(2)y+c_(2)z+d_(2)=0," then"a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`] `implies" "3lambda+1=0` `implies" "lambda=-(1)/(3)" "...(ii)` So, the equation of required plane, is `(1-(2)/(3))x+(1-(3)/(3))y+(1-(4)/(3))z-(1-(5)/(3))=0` `implies(1)/(3)x-(1)/(3)z+(2)/(3)=0impliesx-z+2=0` Now, vector form is `r.(hati-hatk)+2=0` |
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| 35. |
Perpendiculars are drawn from points on the line `(x+2)/2=(y+1)/(-1)=z/3` to the plane `x + y + z=3` The feet of perpendiculars lie on the line (a) `x/5=(y-1)/8=(z-2)/(-13)` (b) `x/2=(y-1)/3=(z-2)/(-5)` (c) `x/4=(y-1)/3=(z-2)/(-7)` (d) `x/2=(y-1)/(-7)=(z-2)/5`A. `(x)/(5)=(y-1)/(8)=(z-2)/(-13)`B. `(x)/(2)=(y-1)/(3)=(z-2)/(-5)`C. `(x)/(4)=(y-1)/(3)=(z-2)/(-7)`D. `(x)/(2)=(y-1)/(-7)=(z-2)/(5)` |
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Answer» Correct Answer - D To find the foot of perpendiculars and find its locus. Fromula used Footof perpendicular from `(x_(1),y_(1),z_(1))`to `ax+by+cz+d=0`be`(x_(2),y_(2),z_(2))`, then `(x_(2)-x_(1))/(a)=(y_(2)-y_(1))/(b)=(z_(2)-z_(1))/(c)=(-(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))` Any point on `(x+2)/(2)=(y+1)/(1)=(z)/(3)=lambda` `implies" "x=2lambda-2,y=-lambda-1,z=3lambda` Let foot of perpendicular from `(2lambda-2,-lambda-1,3lambda)` to `x+y+z=3` be `(x_(2),y_(2),z_(2)).` `:.(x_(2)-(2lambda-2))/(1)=(y_(2)-(-lambda-1))/(1)=(z_(2)-(3lambda))/(1)` `=-((2lambda-2-lambda-1+3lambda-3))/(1+1+1)` `implies" "x_(2)-2lambda+2=y_(2)+lambda+1=z_(2)-3lambda=2-(4lambda)/(3)` `:." "x_(2)=(2lambda)/(3),y_(2)=1-(7lambda)/(3),z_(2)=2+(5lambda)/(3)` `implies" "lambda=(x_(2)-0)/(2//3)=(y_(2)-1)/(-7//3)=(z_(2)-2)/(5//3)` Hence, foot of perpendicular lie on `(x)/(2//3)=(y-1)/(-7//3)=(z-2)/(5//3)implies(x)/(2)=(y-1)/(-7)=(z-2)/(5)` |
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| 36. |
Two lines `(x-3)/1 = (y+1)/3 = (z-6)/(-1)` and `(x+5)/7 = (y-2)/(-6) = (z-3)/4` intersect in point R. The reflection of R in the xy-plane has coordinates: (a) `(2,-4,-7)` (b) `(2,4,7)` (c) `(2,-4,7)` (d) `(-2,4,7)`A. (2, -4, -7)B. (2, -4, 7)C. (-2,4,7)D. (2, 4, 7) |
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Answer» Correct Answer - A Let `(x-3)/(1)=(y+1)/(3)=(z-6)/(-1)=a` (say). Then any point on this line is of the from `P(a+3, 3a-1,-a+6)` Similarly, any point on the line. `(x+5)/(7)=(y-2)/(-6)=(z-3)/(4)=b`(say), is of the from `Q(7b-5,-6b+2,4b+3)` Now, if the lines are intersect, then P=Q for some a and b. `implies" "a+3=7b-5` `3a-1=-6b+2` and`" "-a+6=4b+3` `impliesa-7b=-8,a+2b=1" and "a+2b=3` On solving `a-7b=-8" and "a+2b=1`, we get b=1 and a=1, which also satisfy `a+4b=3` `:.P=Q-=(2, -4,7)" for "a=-1"and"b=1` Thus, coordinates of point R are (2, -4, 7) and reflection of R in xy-plane is (2, -4, -7) |
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| 37. |
A plane which bisects the angle between the two given planes `2x-y+2z-4=0" and "x+2y+2z-2=0`, passes through the pointA. (1, -4, 1)B. (1, 4, -1)C. (2, 4, 1)D. (2, -4, 1) |
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Answer» Correct Answer - D Equation of planes bisecting the angles between the planes `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," and"` `(a_(1)x+b_(1)y+c_(1)z+d_(1))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2)))=+-(a_(2)x+b_(2)y+c_(2)z+d_(2))/(sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))` Equation of given planes are `2x-y+2z-4=0" "...(i)` and `" "z+2y+2z-2=0" "...(ii)` Now, equation of planes bisecting the angles between the planes (i) and (ii) are `(2x-y+2z-4)/(sqrt(4-1-4))=+-(x+2y+2z-2)/(sqrt(1+4+4))` `implies" "2x-y+2z-4=+-(x+2y+2z-2)` On taking (+ve) sign, we get a plane `x-3y=2" "...(iii)` On taking(-ve)sing, we get a plane `3x+y+4z=6" "...(iv)` Now from the given options, the point (2, -4, 1) satisfy the plane of angle bisector `3x+y+4z=6` |
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| 38. |
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 Given planes are `2x+y+2z-8=0` and `2x+y+2z+(5)/(2)=0` `:."Distance between two planes"` `(|c_(1)-c_(2)|)/(sqrt(a^(2)+b^(2)+c^(2)))=|(-8-(5)/(2))/(sqrt(2^(2)+1^(2)+2^(2)))|=(21//2)/(3)=(7)/(2)` |
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| 39. |
The equation of a plane containing the line of intersection of the planes `2x-y-4=0` and `y+2z-4=0` and passing through the point (1, 1, 0) isA. `x-3y-2z=-2`B. `2x-z=2`C. `x-y-z=0`D. `x+3y+z=4` |
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Answer» Correct Answer - C Equations of given planes are `2x-y-4=0" "…(i)` and`" "y+2z-4=0" "…(ii)` Now, equation of family of planes passes through the line of intersection of given planes (i) and (ii) is `(2x-y-4)+lambda(y+2z-4)=0" "...(iii)` According to the question, Plane (iii) passes through the point, (1, 1, 0), so `(2-1-4)+lambda(1+0-4)=0` `implies" "-3-3lambda=0` `implies" "lambda=-1` Now, equation of required plane can be obtained by putting`lambda=-1` in the equation of plane (iii). `implies(2x-y-4)-1(y+2z-4)=0` `implies" "2x-y-4-y-2z+4=0` `implies" "2x-2y-2z=0` `impliesx-y-z=0` |
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| 40. |
The perpendicular distance from the origin to the plane containing the two lines, `(x+2)/3=(y-2)/5=(z+5)/7` and `(x-1)/1=(y-4)/4=(z+4)/7` is: (a) `11sqrt6` (b)`11/(sqrt6)` (c) 11 (d) `6sqrt(11)`A. `11sqrt(6)`B. `(11)/(sqrt(6))`C. 11D. `6sqrt(11)` |
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Answer» Correct Answer - B Let the equation of plane, containing the two lines `(x+2)/(3)=(y-2)/(5)=(z+5)/(7)" and "(x-1)/(1)=(y-4)/(4)=(z+4)/(7)` is `a(x+2)+b(y-2)+c(z+5)=0" "...(i)` `because` Plane (i) contain lines, so `3a+5b+7c=0" "...(ii)` and`" "a+4b+7c=0" "...(iii)` From Eqs. (ii) and (iii) , we get `(a)/(35-28)=(b)/(7-21)=(c)/(12-5)` `implies" "(a)/(7)=(b)/(-14)=(c)/(7)implies(a)/(1)=(b)/(-2)=(c)/(1)` So, equation of plane will be `1(x+2)-2(y-2)+1(z+5)=0` `implies" "x-2y+z+11=0" "...(iv)` Now, perpendicular distance from origin to the plane is `=(11)/(sqrt(1+4+1))=(11)/(sqrt(6))` [`because` perpendicular distance from origin to the plane `ax+by+cz+d=0," is "(|d|)/(sqrt(a^(2)+b^(2)+c^(2)))`] |
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| 41. |
The equation of the line passing through `(-4, 3, 1),` parallel to the plane `x +2y-z-5=0` and intersecting the line `(x+1)/-3=(y-3)/2=(z-2)/-1`A. `(x+4)/(3)=(y-3)/(-1)=(z-1)/(1)`B. `(x+4)/(-1)=(y-3)/(1)=(z-1)/(1)`C. `(x+4)/(1)=(y-3)/(1)=(z-1)/(3)`D. `(x-4)/(2)=(y+3)/(1)=(z+1)/(4)` |
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Answer» Correct Answer - A Any point on the line`(x+1)/(-3)=(y-3)/(2)=(z-2)/(-1)` is of the form `(-3lambda-1,2lambda+3,-lambda+2)` `["take"(x+1)/(-3)=(y-3)/(2)=(z-2)/(-1)=lambdaimpliesx=-3lambda-1,` `y=2lambda+3" and "z=-lambda+2`] So, the coordinates of point of intersection of two lines will be `(-3lambda-1,2lambda+3,-lambda+2)` for some `lambdainR` Let the point `A=(-3lambda-1,2lambda+3,-lambda+2)` and `B-=(-4, 3, 1)` Then, `AB=OB-OA` `=(-4hati+3hatj+hatk)-{(-3lambda-1)hati+(2lambda+3)hatj+(-lambda+2)hatk}` `=(3lambda-3)hati-2lambdahatj+(lambda-1)hatk` Now, as the line is parallel to the given plane and so AB will be parallel to the given plane and so AB will be perpendicular to the normal of plane. `impliesAB.lambda=0," where" n=hati+2hatj-hatk` is normal to the plane. `implies((3llambda-3)hati-2lambdahatj+(lambda-1)hatk).(hati+2hatj-hatk)=0` `implies3(lambda-1)-4lambda+(-1)(lambda-1)=0` `[because` If `a=a_(1)hatja_(2)hatj+a_(3)hatk" and "b=b_(1)hati+b_(2)hatj+b_(3)hatk,` then `a.b=a_(1)b_(1)+a_(2)b_(2)+a_(3)b_(3)`] `implies3lambda-3-4lambda-lambda+1=0` `implies" "-2lambda=2implieslambda=-1` Now, the required equation is the equation of line joining A(2, 1 , 3) and B(-4, 3, 1), which is `(x-(-4))/(2-(-4))=(y-3)/(1-3)=(z-1)/(3-1)` [`because` Equation of line joining `(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)-a_(1))`] `implies" "(c+4)/(6)=(y-3)/(-2)=(z-1)/(2)` or`" "(x+4)/(3)=(y-3)/(-1)=(z-1)/(1)" "["multiplying by "2]` |
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| 42. |
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=0` |
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Answer» Correct Answer - B::D Here, `P_(3):(x+z-1)+lambday=0` i.e.`" "P_(3):x+lambday+z-1=0` whose distance from (0,1,0) is 1. `:." "(|0+lambda+0-1|)/(sqrt(1+lambda^(2)+1))=1` `implies" "|lambda-1|=sqrt(lambda^(2)+2)` `implies" "lambda^(2)-2lambda+1=lambda^(2)+2implieslambda=-(1)/(2)` `:."Equation of "P_(3)" is "2x-y+2z-2=0.` `because` Distance from `(alpha,beta,gamma)` is 2. `:." "(|2alpha-beta+2gamma-2|)/(sqrt(4+1+4))=2` `implies" "2alpha-beta+2gamma-2=+-6` `implies2alpha-beta+2gamma=8" and "2alpha-beta+2gamma=-4` |
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| 43. |
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))`unitsB. `(10)/(sqrt(83))` unitsC. `(5)/(sqrt(83))`unitsD. `(10)/(sqrt(74))`units |
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Answer» Correct Answer - B Given, equations of lines are `(x-1)/(1)=(y+2)/(-2)=(z-4)/(3)` and`" "(x-2)/(2)=(y+1)/(-1)=(z+7)/(-1)` Let `n_(1)=hati-2hatj+3hatk` and `n_(2)=2hati-hatj-hatk` `:.` Any vector n perpendicular to both `n_(1),n_(2)` is given by `n=n_(1)xxn_(2)` `implies" "n=|{:(hati,hatj,hatk),(1,-2,3),(2,-1,-2):}|=5hati+7hatj+3hatk` `:.` Eauation of a plane passing through (1,-1,-1) and perpendicular to n is given by `5(x-1)+7(y+1)+3(z+1)=0` `implies" "5x+7y+3z+5=0` `:.` Required distance`=|(5+21-21+5)/(sqrt(5^(2)+7^(2)+3^(2)))|=(10)/(sqrt(83))` units |
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| 44. |
Let A be a point on the line `vec(r)=(1-3mu)hati+(mu-1)hatj+(2+5mu)hatk` and `B(3,2,6)` be a point in the space. Then the value of `mu` for which the vector `vec(AB)` is parallel to the plane `x-4y+3z=1` is: (a) `1/8` (b) `1/2` (c) `1/4` (d) `-1/4`A. `(1)/(4)`B. `-(1)/(4)`C. `(1)/(8)`D. `(1)/(2)` |
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Answer» Correct Answer - A Given equation of line is `r=(1-3mu)hati+(mu-1)hatj+(2+5mu)hatk` Clearly, any point on the above line is of the form `(1-3mu,mu-1,2+5mu)` Let A be `(-3mu+1,mu-1,5mu+2)` for some `muinR.` Then, `AB(3-(-3mu+1))hati+(2-(mu-1))hatj` `+(6-(5mu+2))hatk[becauseAB=OB-OA]` `=(3mu+2)hati+(3-mu)hatj+(4-5mu)hatk" "...(i)` Normal vector (n) of the plane `x-4y+3z=1` is `n=hati-4hatj+3hatk" "...(ii)` `because`AB is parallel to the plane. `:.` n is perpendicular to the AB. `impliesAB.n=0` `implies[(3mu+2)hati+(3-mu)hatj+(4-5mu)hatk].[hati-4hatj+3hatk]=0` [From Eqs. (i) and (ii)] `implies(3mu+2)-4(3-mu)+3(4-5mu)=0` `implies" "-8mu+2=0` `implies" "mu=(1)/(4)` |
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| 45. |
The plane passing through the point (4, -1, 2) and perallel to the lines `(x+2)/(3)=(y-2)/(-1)=(z+1)/(2)` and `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)` also passes through the pointA. (-1, -1, -1)B. (1, 1, -1)C. (1, 1, 1)D. (-1, -1, 1) |
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Answer» Correct Answer - C Let a be the position vector of the given point (4,-1,2). Then, `a=4hati-hatj+2hatk` The direction vector of the lines `(x+2)/(3)=(y-2)/(-1)=(z+1)/(2)` and `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)` are respectively `b_(1)=3hati-hatj+2hatk` and`" "b_(2)=hati+2hatj+3hatk` Now, as the plane is parallel to both `b_(1)` and `b_(2)` [`because` plane is parallel to the given lines] So, normal vector (n) of the plane is perpendicular to both `b_(1)` and `b_(2)`. `implies" "n=b_(1)xxb_(2)`and Required equation of plane is `(r-a).n=0` `implies" "(r-a).(b_(1)xxb_(2))=0` `implies|{:(x-4,y+1,z-2),(3,-1," "2),(1,2," "3):}|=0` `[{:(becauser-a=(xhati+yhatj+zhatk)-(4hati-hatj+2hatk)),(=(x-4)hati+(y+1)hatj+(z-2)hatk),("and we know that"","[a b c]=a.(bxxc)),(|{:(a_(1),b_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|)]` `implies(x-4)(-3-4)-(y+1)(9-2)+(z-2)(6+1)=0` `implies" "-7(x-4)-7(y+1)+7(z-2)=0` `implies" "(x-4)+(y+1)-(z-2)=0` `implies" "x+y-z-1=0` (1, 1, 1) is the only point that satisfies. |
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