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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.
101. |
Let `P=0`be the equation of a plane passing through the line ofintersection of the planes `2x-y=0a n d3z-y=0`and perpendicular to the plane `4x+5y-3z=8.`Then the points which lie on the plane `P=0`is/area. `(0,9,17)`b. `(1//7,21//9)`c. `(1,3,-4)`d. `(1//2,1,1//3)`A. (0,9,17)B. `(1//7,2,1//9)`C. (1,3,-4)D. `(1//2,1,1//3)` |
Answer» Correct Answer - a, d The equation of the plane through the intersection of the planes `2x-y=0 and 3z-y=0` is `" "2x-y+lamda(3z-y)=0" "`(i) or `" "2x-y(lamda+1)+3ladmaz=0` Plane (i) is perpendicular to `4x+5y-3z=8`. Therefore, `" "4xx2-5(lamda+1)-9lamda=0` or `" "8-5lamda-5-9lamda=0` or `" "3-14lamda=0` or `" "lamda= 3//4` `therefore" "2x-y+ (3)/(14) (3z-y)=0` `" "28x-17y+9z=0` |
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102. |
Find the equation of the plane passing through the intersection of theplanes `2x-3y+z-4=0a n dx-y+z+1=0`and perpendicular to the plane `x+2y-3z+6=0.` |
Answer» Correct Answer - `9x-8y+7z-21=0` | |
103. |
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` |
Answer» Correct Answer - b, d Clearly, plane `P_(3)` is `P_(2)+ lamdaP_(1)=0`. `rArr" "x+lamday+z-1=0` Distance of this plane from point `(0, 1, 0)` is 1. `rArr " "(0+lamda+0-1)/(sqrt(1+lamda^(2)+1))= pm 1` `therefore " "lamda= - (1)/(2)` Thus, equation of `P_(3)` is `2x-y+2z-2=0`. Distance of this plane from point `(alpha, beta, gamma)` is 2. `rArr" "|(2alpha-beta+2gamma-2)/(3)|=2` `rArr" "2alpha-beta+2gamma= 2pm 6` Thus, option (b), (d) are correct. |
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104. |
Consider three planes `P_(1):x-y+z=1` `P_(2):x+y-z=-1` and `" "P_(3):x-3y+3z=2` Let `L_(1),L_(2),L_(3)` be the lines of intersection of the planes `P_(2)` and `P_(3),P_(3)` and `P_(1),P_(1)` and `P_(2)` respectively. Statement I Atleast two of the lines `L_(1),L_(2)` and `L_(3)` are non-parallel. Statement II The three planes do not have a common point.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
Answer» Correct Answer - d The direction cosines of each of the lines `L_(1), L_(2), L_(3)` are proportional to `(0, 1, 1)`. |
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105. |
Find the equation of the plane through the intersection of the planes `3x-4y+5z=10a n d2x+2y-3z=4`and parallel to the line `x=2y=3zdot` |
Answer» Correct Answer - `x-20y+27z=14` | |
106. |
Find the equation of a plane passing through the intersection of the planes `x-3y+2z-5 - 0` and `2x+y+3z-1 = 0` and passes through the point `(1,-2,3)`. |
Answer» Correct Answer - `x+7y+13=0` | |
107. |
Consider the planes `3x-6y-2z=15a n d2x+y-2z=5.`Statement 1:The parametricequations of the line intersection of the given planes are `x=3+14 t ,y=2t ,z=15 tdot`Statement 2: The vector `14 hat i+2 hat j+15 hat k`is parallel to theline of intersection of the given planes.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
Answer» Correct Answer - d The line of intersection of the given plane is `3x-6y-2z-15=0= 2x+y-2z-5=0` For `z=0`, we obtain `x=3 and y=-1`. `therefore" "` Line passes through `(3, -1, 0)` Also, the line is parallel to the cross product of normal to given planes, that is `" "|{:(hati,,hatj,,hatk),(3,,-6,,-2),(2,,1,,-2):}|= 14hati+2hatj+15hatk` The equation of line is `(x-3)/(14)= (y+1)/(2)= (z)/(15) = t`, whose parametic form is `" "x=3+14t, y=-1+2t, z=15t` Therefore, Statement 1 is false. However, Statement 2 is true. |
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108. |
Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati+hatj+hatk)=1 and vecr.(2hati+3hatj-hatk)+4=0` and parallel to x-axis. |
Answer» Equation of a plane through the intersection of planes is `[vecr.(hati+hatj+hatk)-1)]+lambda(vecr.(2hati+3hatj-hatk)+4]=0` `rArr vecr.[(2lambda+1)hati+(3lambda+1)hatj + (1-lambda)hatk]+(4lambda-1)=0"……"(1)` The direction ratio of the normal to the plane are `(2lambda+1),(3lambda+1)` and `(1-lambda)`. Required plane is parallel to X-axis, so the normal to the plane will be ratios of X-axis are `1,0` and `0`. `:. 1(2lamda+1)+0(3lambda+1)+0(1-lambda) = 0` `rArr 2 lambda +1=0 rArr lambda = - 1/2` put the value of `lambda` in equation (1), `vecr.(-(1)/(2)hatj+3/2hatk)+(-3)=0 rArr vecr.(hatj-3hatk) + 6 =0` Therefore, cartessian equation of plane is `y - 3z + 6 = 0` Which is the equation of the required plane. |
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109. |
Find the angle between the folowing planes :- (i) `vecr.(2hati-3hatj+4hatk) = 5` and `vecr.(-hati+hatj) = 1` (ii) `vecr.(2hati+3hatj-6hatk) = 1` and `vecr.(hati-2hatj+2hatk) + 3 = 0` (iii) `x+y-2z=3` and `2x-2y+z+1 = 0` (iv) `x+y-z=8` and `-x+2y+z-1 = 0` |
Answer» Correct Answer - (i) `cos^(-1)((-5)/(sqrt(58)))` , (ii) `cos^(-1)((-16)/(21))` , (iii) `cos^(-1)((-2)/(3sqrt(6)))` , (iv) `(pi/2)` | |
110. |
Find the angle between the line `vecr=(hati+2hatj-hatk)+lamda(hati-hatj+hatk)` and the plane `ver.(2hati-hatj+hatk)=4` |
Answer» We know that if `theta` is the angle between the lines `vecr=veca+lamdavecb and vecr*vecn=p`, then `sintheta=|(vecb*vecn)/(|vecb||vecn|)|` Therefore, if `theta` is the angle between `vecr=hati+2hatj-hatk+lamda(hati-hatj+hatk) and vecr*(2hati-hatj+hatk)=4,` then `" "sin theta=|((hati-hatj+hatk)*(2hati-hatj+hatk))/(|hati-hatj+hatk||2hati-hatj+hatk|)|` `" "=(2+1+1)/(sqrt(1+1+1)sqrt(4+1+1))` `" "=(4)/(sqrt(3)sqrt(6))=(4)/(3sqrt2)` or `" "theta=sin^(-1)((4)/(3sqrt2))` |
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111. |
Find the vector equation of a plane passing through the intersection of the planes `vecr.(hati+hatj+hatk) = 6` and `vecr. (2hati+3hatj+4hatk) - 5 = 0` and through the point `(2,2,1)`. |
Answer» Let the equation of the plane through the intersection of given planes is `[vecr.(hati+hatj+hatk)-6]` `+lambda[vecr.(2hati+3hatj+4hatk)-5] = 0` `rArr vecr.[(1+2lambda)hati+(1+3lambda)hatj` `+(1+4lambda)hatk]=6+5lambda` It passes through the point `(2,2,1)` i.e., `(2hati+2hatj+hatk)` `:. (hati+2hatj+hatk),[(1+2lambda)hati+(1+3lambda)hatj` `+(1+4lambda)hatk] = 6+5lambda` `rArr 2(1+2lambda)+2(1+3lambda)+(1+4lambda)=6+5lambda` `rArr 2 + 4 lambda + 2 + 6 lambda+1 + 4lambda= 6 +5lambda` `rArr 9lambda = 1` `rArr lambda = 1/9` Therefore the eqation of the plane is `vecr.[(1+2/9)hati+(1+3/9)hatj+(1+4/9)hatk] = 6+5/9` `rArr vecr.(11hati+12hatj+13hatk) = 59`. |
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112. |
Prove that the plane `vecr*(hati+2hatj-hatk)=3` contains the line `vecr=hati+hatj+lamda(2hati+hatj+4hatk).` |
Answer» To show that `vecr=hati+hatj+lamda(2hati+hatj+4hatk)" "`(i) lies in the plane `vecr* (hati+2hatj-hatk)=3`,`" "`(ii) we must show that each point of (i) lies in (ii). In order words we must show that `vecr` in (i) satisfies (ii) for every value of `lamda`. We have `" "[hati+hatj+lamda(2hati+hatj+4hatk)]*(hati+2hatj-hatk)` `" "=(hati+hatj)*(hati+2hatj-hatk)+lamda(2hati+hatj+4hatk)*(hati+2hatj-hatk)` `" "=(1)(1)+(1)(2)+lamda[(2)(1)+(1)(2)+4(-1)]=3+lamda(0)=3` Hence, line (i) lies in plane (ii). |
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113. |
Find the distance of the point `(21,0)`from the plane `2x+y+2z+5=0.` |
Answer» Correct Answer - `10/3` | |
114. |
Find the perpendicular distance from the point `(2hati+hatj-hatk)` to the plane`vecr.(i-2hatj+4hatk) = 3`. |
Answer» Correct Answer - `(7)/(sqrt(21))` | |
115. |
Find the perpendicular distance from the point `(2hati-hatj+4hatk)` to the plane `vecr.(3hati-4hatj+12hatk) = 1`. |
Answer» Correct Answer - `47/13` | |
116. |
A vector of magnitude 6 units makes equal makes angles from `OX-,OY-` and `OZ`-axes. Find the vector. |
Answer» Let vector is `vecr`. This makes equal angles from `OX-,OY` and `OZ-`axes. `:. l = m = n` Now from `l^(2) = m^(2) + n^(2) = 1` `l^(2) + l^(2) + l^(2) = 1 rArr 3l^(2)= 1` `rArr l = +- (1)/(sqrt(3))` `:. vecr=|vecr|.(lhati+lhat j+lhatk)` `= 6(+-(1)/(sqrt(3))hati+-(1)/(sqrt(3))hatj+-(1)/(sqrt(3))hatk)` `= 2sqrt(3) (+- hati+- hatj+-hatk)` |
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117. |
Find the angle between the lines `vecr=3hati-2hatj+6hatk+(lamda(2hati+hatj+2hatk))` and `vecr=(2hatj-5hatk)+(mu(6hati+3hatj+2hatk))`. |
Answer» We have `vecr=3hati-2hatj|6hatk+lamda(2hati+hatj+2hatk)` and `vecr=(2hatj-5hatk)+mu(6hati+3hatj+2hatk)` where `veca_(1)=3hati-2hatj+6hatk,vecb_(1)=2hati+hatj+2hatk` and `veca_(2)=2hatj-5hatk, vecb_(2)=6hati+3hatj+2hatk` If `theta` is angle between the lines, then `costheta=(|vec(b_(1)).vecb_(2)|)/(|vec(b_(1))|.|vec(b_(2))|)` `=(|(2hati+hatj+2hatk).(6hati+3hatj+2hatk)|)/(|2hati+hatj+2hatk||6hati+3hatj+2hatk|)` `=(|12+3+4|)/(sqrt(9)sqrt(49))=19/21` `:. theta=cos^(-1)19/21` |
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118. |
Find the angle between the following pair of lines : i. `vecr=2hati-5hatj+hatk+lamda(3hati+2hatj+6hatk) and vecr=7hati-6hatk+mu(hati+2hatj+2hatk)` ii. `(x)/(2)=(y)/(2)=(z)/(1) and (x-5)/(4)=(y-2)/(1)=(z-3)/(8)` |
Answer» i. The given lines are parallel to the vectors `vec(b_(1))=3hati+2hatj+6hatk and vec(b_(2))=hati+2hatj+2hatk`, respectively. If `theta` is the angle between the given pair of lines, then `" "costheta=(vec(b_(1))*vec(b_(2)))/(|vec(b_(1))||vec(b_(2))|)=((3)(1)+(2)(2)+(6)(2))/(sqrt(3^(2)+2^(2)+6^(2))sqrt(1^(2)+2^(2)+2^(2)))=(19)/(7xx3)` `therefore" "theta=cos^(-1)((19)/(21))` ii. The given lines are parallel to the vectors `vec(b_(1))=2hati+2hatj+hatk and vec(b_(2))=4hati+hatj+8hatk`, respectively. If `theta` is the angle between the given pair of lines, then `" "costheta=(vec(b_(1))*vec(b_(2)))/(|vec(b_(1))||vec(b_(2))|)=((2)(4)+(2)(1)+(1)(8))/(sqrt(2^(2)+2^(2)+1^(2))sqrt(4^(2)+1^(2)+8^(2)))=(18)/(3xx9)=(2)/(3)` `therefore " "theta=cos^(-1) ((2)/(3))` |
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119. |
Statement 1 : Lines `vecr=hati+hatj-hatk+lamda(3hati-hatj) and vecr=4hati-hatk+ mu (2hati+ 3hatk) intersect. Statement 2 : If `vecbxxvecd=vec0`, then lines `vecr=veca+lamdavecb and vecr= vecc+lamdavecd` do not intersect.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
Answer» Correct Answer - c For the given lines, let `vec(a_1) =hati+hatj-hatk, vec(a_2)= 4hati-hatk, vec(b_1) = 3hati-hatj and vec(b_2) = 2hati+3hatk`. Therefore, `" "[vec(a_(2))-vec(a_(1))vec(b_1)vec(b_2)]=|{:(4-1,,0-1,,-1+1),(3,,-1,,0),(2,,0,,3):}|` `" "= |{:(3,,-1,,0),(3,,-1,,0),(2,,0,,3):}|=0` Hence, the lines are coplanar. Also vector `vec(b_1) adn vec(b_2)` along which the lines are directed are not collinear. Hence, the lines intersect. When `vecbxxvecd=vec0` , vectors and `vecr=vecc+lamdavecd` are parallel and do not intersect. But this statement is not the correct explanation for Statement 1. |
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120. |
Let `L_(1)` be the line `vecr_(1)=2hati+hatj-hatk+lamda(hati+2hatk)` and let `L_(2)` be the line `vecr_(2)=3hati+hatj-hatk+mu(hati+hatj+hatk)`. Let `pi` be the plane which contains the line `L_(1)` and is parallel to `L_(2)`. The distance of the plane `pi` from the origin isA. `sqrt(2//7)`B. `1//7`C. `sqrt6`D. none |
Answer» Correct Answer - a Equation of the plane containing `L_(1),A(x-2)+B(y-1)+C(z+1)=0` where `A+2C=0,A+B-C=0` `impliesA=-2C-,B=3C,C=C` `implies"Plane is"-2(x-2)+3(y-1)+z+1=0` or `2x-3y-z-2=0` Hence, `p=|(-2)/(sqrt14)|=sqrt((2)/(7))` |
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121. |
Direction ratios of the line which is perpendicular to the lines with direction ratios (-1,2,2) and (0,2,1) areA. 1, 1, 2B. 2, -1, 2C. `-2, 1, 2`D. 2, 1, -2 |
Answer» Correct Answer - B | |
122. |
Find the direction cosines of the line which is perpendicular to the lines with direction ratios `4, 1, 3 and 2, -3, 1`.A. `pm (5)/(5sqrt(3)), pm (1)/(5sqrt(3)), pm (7)/(5sqrt(3))`B. `pm (1)/(sqrt(3)), pm (1)/(5sqrt(3)), pm (-7)/(5sqrt(3))`C. `pm (5)/(sqrt(3)), pm (1)/(sqrt(3)), pm (7)/(sqrt(3))`D. `pm (5)/(sqrt(3)), pm (1)/(sqrt(3)), pm (-7)/(sqrt(3))` |
Answer» Correct Answer - B | |
123. |
Find the angle between the lines whose direction ratios are a, b, c and `b c , c a , a b`. |
Answer» The direction ratios of given lines are `(a,b,c)` and `(b-c,c-a,a-b)` Let `theta` be the acute angle between the two lines, then `cos theta =|(a(b-c)+b(c-a)+c(a-b))/(sqrt(a^(2)+b^(2)+c^(2))sqrt((b-c)^(2)+(c-a)^(2)+(a-b)^(2)))|` `= |(ab-ac+bc-ab+ca-bc)/(sqrt(a^(2)+b^(2)+c^(2))+sqrt((b-c)^(2)+(c-a)^(2)+(a-b)^(2)))|=0` `rArr cos theta = 0 rArr theta = (pi)/(2) = 90^(@)`. Therefore, angle between two lines is `90^(@)`. |
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124. |
Show that the line joining the origin to the point`(2,1,1)`is perpendicularto the line determined by the points `(3,5,-1)`and `(4,3,-1)dot` |
Answer» The direction ratios of the line joining the points `(0,0,0)` and `(2,1,1)` are `[ (2-0),(1-0),(1-0) ]` i.e., `(2,1,1)`. `:. a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)` `=2xx1+1xx(-2)+1xx0 = 0` Therefore, the given lines are mutually perpendicular. |
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125. |
The shortest distancebetween the lines `(x-3)/3=(y-8)/(-1)=(z-3)/1a n d(x+3)/(-3)=(y+7)/2=(z-6)/4`isa. `sqrt(30)`b. `2sqrt(30)`c. `5sqrt(30)`d. `3sqrt(30)`A. `sqrt30`B. `2sqrt30`C. `5sqrt30`D. `3sqrt30` |
Answer» Correct Answer - d Given lines are `vecr = 3hati + 8hatj + 3hatk + l(3hati - hatj +hatk)` and `vecr = - 3hati - 7hatj + 6hatk + m(-3hati + 2hatj + 4hatk)` Required shortest distance `=(|(6hati + 15hatj- 3hatk) .((3hati - hatj +hatk)xx(-3hati + 2hatj + 4hatk))|)/(|(3hati - hatj + hatk)xx(-3hati + 2hatj +4hatk)|)` `=(|(6 hati + 15hatj - 3hatk).(-3hati - 15hatj + 3hatk)|)/(|-6 hati - 15hatj + 3hatk|)` `=(36+225+9)/(sqrt(36+225+9)) = (270)/(sqrt(270)) = sqrt(270) = 3sqrt(3)` |
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126. |
The shortest distancebetween the lines `x/(-3)=(y-1)/1=(z+1)/(-1)a n d(x-2)/1=(y-3)/2=((z+(13//7))/(-1))`is zero.Statement 2: The givenlines are perpendicular.A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.B. Both the Statements are true, but Statement 2 is not the correct explanation for Statement 1.C. Statement 1 is true and Statement 2 is false.D. Statement 1 is false and Statement 2 is true. |
Answer» Correct Answer - d Direction ratios of the given lines are `(-3, 1, -1) and (1, 2, -1)`. Hence, the lines are perpendicular as `(-3)(1)+ (1)(2)+ (-1)(-1)=0`. Also lines are coplanar as `" "|{:(0-2,,1-3,,-1+(13//7)),(-3,,1,,-1),(1,,2,,-1):}|=0` But Statement 2 is not enough reason for the shortest distance to be zero, as two skew lines can also be perpendicular. |
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127. |
Find the shortest distance between the following pair of line: `vecr=(1-t)hati+(t-2)hatj+(3-2t)hatk and vecr=(s+1)hati+(2s-1)hatj-(2s+1)hatk.` |
Answer» Given equation of lines can be written as : `vecr = (hati-2hatj+3hatk)+t(-hati+hatj-2hatk)` and `vecr = (hati-hatj-hatk)+s(hati+2hatj-2hatk)` Comparing these equation with `vecr = veca_(1)+tvecb_(1)` and `vecr = veca_(2)+ svecb_(2)`, `veca_(1) = hati-2hatj+3hatk, vecb_(1)=-hati+hatj-2hatk` and `veca_(2) = hati-hatj-hatk, vecb_(2)=hati+2hatj-2hatk` Now, ` veca_(2)-veca_(1) = ( hati-hatj-hatk) - (hati-2hatj+3hatk)= hatj-4hatk` and `vecb_(1)xxvecb_(2)=|{:(hati,hatj,hatk),(-1,1,-2),(1,2,2):}|` `= (-2+4)hati-(2+2)hatj+(-2-1)hatk` `= 2hati-4hatj-3hatk` `rArr |vecb_(1)xxvecb_(2)|=sqrt((2)^(2)+(-4)^(2)+(-3)^(2))` `= sqrt(4+16+9) = sqrt(29)` `:.` Required shortest distance `d=|{:(vecb_(1)xxvecb_(2)).(veca_(2)-veca_(1)))/(|vecb_(1)xxvecb_(2)|)|` `=(|(2hati-4hatj-3hatk).(hatj-4hatk)|)/(sqrt(29))` `= (|-4+12|)/(sqrt(29)) = (8)/(sqrt(29))` |
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128. |
Find the Cartesian equations of the following planes whose vector equations are: `vecr.[(s-2t)hati+93-t)hatj+(2s+t)hatk]=15` |
Answer» (a) Given vector equation of plane is, `vecr.(hati+hatj-hatk) = 2"……"(1)` Put `hatr = xhati+yhatj+zhatk` in equation (1), `(xhati+yhatj+zhatk).(hati+hatj-hatk) = 2` `rArr x+y-z = 2` Which is the required cartesian equation of the plane. (b) Given vector equation of the plane is , `vecr.(2hati+3hatj-4hatk) = 1 "....."(2)` put `vecr - x hati+ yhatj+ zhatk` in equation (2), `(xhati+yhatj+zhatk) (2hati+3hatj-4hatk) = 1` `2x+3y-4z = 1` which is the required cartesian equation of the plane. (c) Given equation of plane is : `vecr.{(s-2t)hati+(3-t)hatj+(2s+t)hatk} = 15 "......."(3)` Put `vecr = xhati+yhatj+zhatk.{(s-2t)hati+(3-t)hatj+(2s+t)hatk} = 15` `rArr x(s-2t)+y(3-t)+z(2s+t)-15=0` Which is the required cartesian equation of the plane. |
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129. |
Find the shortest distance between the following lines:`(x+1)/7=(y+1)/(-6)=(z+1)/1;(3-x)/(-1)=(y-5)/(-2)=(z-7)/1` |
Answer» Given lines are ltbr `(x+1)/(7) = (y+1)/(-6) = (z+1)/(1)` `(x1)/(7) = (y+1)/(-6)=(z+1)/(1)` and `(x-3)/(1) = (y-5)/(-2) = (z-7)/(1)` The direction ratios of first line are `(7,-6,1)` and it passes through the point `(-1,-1,-1)`. Therefore, the vector equation of the given line is `rArr vecr_(1) = -hati-hatj-hatk+lambda(7hati-6hatj+hatk)` Similarly the vector equati on of second line is : `vecr_(2)=3hati+5hatj+7hatk+mu(hati-2hatj+hatk)` which are in the form `vecr_(1) = veca_(1)+lambdavecb_(1)` and `vecr_(2)= veca_(2)+muvecb_(2)` where, `veca_(1) = -hati-hatj-hatk, vecb_(1) = vecb_(1) =7hati-6hatj+hatk` and `veca_(2) = 3hati+5hatj+7hatk,vecb_(2)=hati-2hatj+hatk` Now, `veca_(2)-veca_(1) = (3hati+5hatj+7hatk) - (-hati-hatj-hatk)` `= 4hati+6hatk+8hatk`, and `vecb_(1)xxvecb_(2) = 4hati+6hatj+8hatk`, and `vecb_(1)+vecb_(2) = |{:(hati,hatj,hatk),(7,-6,1),(1,-2,1):}|` `= hati(-6+2)-hatj(7-1)+hatk(-14+6)` `= -4hati-6hatj-8hatk` `|vecb_(1)xxvecb_(2)| = sqrt((-4)^(2)+(-6)^(2)+(-8)^(2))` `=sqrt(166+36+64) = sqrt(116) = 2sqrt(29)` `:.` Shortest distance betweent the given lines `d=|((vecb_(1)xxvecb_(2)).(veca_(2)-veca_(1)))/(|vecb_(1)xxvecb_(2)|)|` `=(|(-4hati-6hatj-8hatk).(4hati+6hatj+8hatk)|)/(2sqrt(29))` `= (|-16-36-64|)/(2sqrt(29))` `= (116)/(2sqrt(29)) = (58)/(sqrt(29)) = 2sqrt(29)` unit |
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130. |
Find the shortest distance between the following pair of line: `vecr=hati+2hatj+hatk+lamd(hati-hatj+hatk) and vecr=2hati-hatj-hatk+mu(2hati+hatj+2hatk)` |
Answer» Here, `veca_(1) = hati+2hatj+hatk, vecb_(1)= hati-hatj+hatk` `veca_(2)=2hati-hatj-hatk, vecb_(2)=2hati+hatj+2hatk` `:. veca_(2)-a_(1)=hati-3hatj-2hatk` `vecb_(1) xx vecb_(2) = |{:(hati,hatj,hatk),(1,-1,1),(2,1,2):}| = - 3 hati+3hatk` Shortest distance between two lines `= |{:((veca_(2)-veca_(1)).(vecb_(1)xxb_(2)))/(|vecb_(1)xxvecb_(2)|):}|` `= |{:((hati-3hatj-2hatk).(-3hati+3hatk))/(|-3hati+3hatk|):}|` `= |{:(-3-6)/(sqrt(9+9)):}|=(9)/(3sqrt(12)) = (3)/(sqrt(2))` |
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131. |
Find the vector and Cartesian equation of the plane that passes through the point (1,4,6) and the normal vector to the plane is `hati-2hatj+hatk`. |
Answer» Correct Answer - `vecr.(hati-2hatj+hatk)+1=0, x-2y+z+1=0` | |
132. |
Convert the equation of the plane `vecr = (hati-hatj)+lambda(-hati+hatj+2hatk)+mu(hati+2hatj+hatk)` into scalar product form. |
Answer» Correct Answer - `vecr.(hati-hatj-hatk)=2` | |
133. |
Find the Cartesian equtionof the followig plane: `vecr=(lamd-2mu)hati+(3-mu)hatj+(2lamda+mu)hatk. |
Answer» Correct Answer - `2x-5y-z+15=0` | |
134. |
If the vector `2 hati-3hatj+7 hatk` makes angles `alpha, beta, gamma` with the co-ordiante axes respectively, then the direction cosine of vector areA. `(-2)/(sqrt(62)), (3)/(sqrt(62)), (-7)/(sqrt(62))`B. `(2)/(sqrt(62)), (-3)/(sqrt(62)), (7)/(sqrt(62))`C. `(-2)/(sqrt(31)), (3)/(sqrt(31)), (-7)/(sqrt(31))`D. `(2)/(sqrt(31)), (-3)/(sqrt(31)), (7)/(sqrt(31))` |
Answer» Correct Answer - B | |
135. |
A line makes angles `alpha, beta, gamma` with X, Y, Z axes respectively. If `alpha=beta` and `gamma=45^(@)`, then `alpha+beta+gamma=`A. `165^(@)`B. `180^(@)`C. `135^(@)`D. `120^(@)` |
Answer» Correct Answer - A | |
136. |
A line makes angles `alpha, beta, gamma` with X, Y, Z axes respectively. If `alpha=beta` and `gamma=45^(@)`, then `alpha=`A. `0^(@)`B. `30^(@)`C. `60^(@)`D. `90^(@)` |
Answer» Correct Answer - C | |
137. |
If the line of the vector `bar (r ) = lambda I + 2j - k ` makes angles `alpha,beta,gamma` with co - ordinate axes ,then :A. 2B. 1C. `1+lambda^(2)`D. `1-lambda^(2)` |
Answer» Correct Answer - a | |
138. |
If a line makes angles `(alpha)/(2), (beta)/(2), (gamma)/(2)` with co-ordinate axes, then `cos alpha+cos beta+cos gamma=` |
Answer» Correct Answer - B | |
139. |
A line makes angles `alpha,betaa n dgamma`with the coordinate axes. If `alpha+beta=90^0,`then find `gammadot` |
Answer» Here `cos^(2)alpha+cos^(2)(90-alpha)+cos^(2)gamma=1` or `" "cos^(2)alpha+sin^(2)alpha+cos^(2)gamma=1` or`" "cos^(2)gamma+1=1 or gamma=90^(@)` |
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140. |
If a line makes angles `alpha,beta,gamma` with co-ordinate axes, then `cos2alpha+cos 2 beta+cos2gamma=` |
Answer» Correct Answer - B | |
141. |
If a line makes angles `alpha,betaa n dgamma`with threew-dimensionalcoordinate axes, respectively, then find the value of `cos2alpha+cos2beta+cos2gammadot` |
Answer» `cos2alpha+cos2beta+cos2gamma` `" "=2cos^(2)alpha-1+2cos^(2)beta-1+2cos^(2) gamma-1` `" "=2(cos^(2)alpha+cos^(2)beta+cos^(2)gamma)-3` ltBrgt `" "=-1` |
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142. |
If O be the origin and thecoordinates of P be`(1," "2," "" "3)`, then find the equationof the plane passing through P and perpendicular to OP. |
Answer» Since `P(1, 2, -3)` is the foot of the perpendicular from the origin to the plane , OP is normal to the plane. Thus, the direction ratios of normal to the plane are 1, 2 and -3. Now, since the plane passes through ( 1, 2, -3), its equation is given by `" "1(x-1)+2(y-2)-3(z+3)=0` or `" "x+2y-3z-14=0` |
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143. |
Find the equation of theplane perpendicular to the line `(x-1)/2=(y-3)/(-1)=(z-4)/2`and passing through the origin. |
Answer» Any plane passing through the origin is `a(x-0) +b (y-0)+c(z-0)=0` This is perpendicular to the given line. Therefore, the normal to the plane is parallel to the give line. Thus, `" "(a)/(2)=(b)/(-1)=(c)/(2)` Therefore, the required plane is `" "2(x-0)-1(y-0)+2(z-0)=0` or `" " 2x-y+2z=0` |
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144. |
Find the equation of theplane passing through `A(2,2,-1),B(3,4,``2)a n dC(7,0,6)dot`Also find a unit vectorperpendicular to this plane. |
Answer» Here `(x_(1), y_(1), z_(1))-= (2, 2,-1), (x_(2), y_(2), z_(2))-= (3, 4, 2) and (x_(3), y_(3), z_(3))-= (7, 0, 6)` Then the equation of the plane is `" "|{:(x-x_(1),,y-y_(1),,z-z_(1)),(x_(2)-x_(1),,y_(2)-y_(1),,z_(2)-z_(1)),(x_(3)-x_(1),,y_(3)-y_(1),,z_(3)-z_(1)):}|=0or |{:(x-2,,y-2,,z-(-1)),(3-2,,4-2,,2-(-1)),(7-2,,0-2,,6-(-1)):}|=0` or `" "5x+2y-3z=17` A normal vector to this plane is `vecd=5hati+2hatj-3hatk` Therefore, a unit vector normal to (i) is given by `" "hatn=(vecd)/(|vecd|)=(5hati+2hatj-3hatk)/(sqrt(25+4+9))=(1)/(sqrt(38))(5hati+2hatj-3hatk)` |
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145. |
Find the equation of theplane passing through the line `(x-1)/5=(y+2)/6=(z-3)/4`and point `(4,3,7)dot` |
Answer» Any plane through `(x-1)/(5) = (y+2)/(6)=(z-3)/(4)` is `" "A(x-1)+B(y+2)+C(z-3)=0" "`(i) where `5A+6B+4C=0" "` (ii) Also, the plane passes through `(4, 3, 7)`. Therefore, `" "3A+5B+4C=0" "` (iii) By (ii) and (iii), `(A)/(4)=(B)/(-8)= (C)/(7)` Therefore, the plane is `" "4(x-1)-8(y+2)+7(z-3)=0` or `" "4x-8y+7z= 41`. |
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146. |
Show that the lines `(x-1)/(1) = (y)/(-5) = z/3` and `(x+1)/(7) = (y)/(2) - (z-3)/(1)` are perpendicular. |
Answer» Here `a_(1) = 1, b_(1) = -5, c_(1) = 3` `a_(2) = 7, b_(2) = 2, c_(2) = 1` Now, `a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=(1)(7)+(-5)(2)+(3)(1)` `= 7-10+3` ` = 0` Therefore given lines are perpendicular. |
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147. |
Show that the lines `(x-4)/(1) = (y+3)/(-4) = (z+1)/(7)` and `(x-1)/(2) = (y+1)/(-3) = (z+10)/(8)` intersect. Also find the co-ordinates of their point of intersection. |
Answer» Let `(x-4)/(1) = (y+3)/(-4) = (z+1)/(7) = lambda` Co-ordinates of any point on A this line. `A(lambda |4, 4 lambda, 4lambda" "3, 7lambda-1)` Again Let, `(x-1)/(2) = (y+1)/(-3) = (z+10)/(8) = mu` Co-ordinates of any point B on this line. `B(2mu+1,-3mu-1,8mu-10)` If these lines intersect, one point will be common. If A and B coincide then `lambda + 4 = 2 mu + 1 rArr lambda = 2mu-3"......"(1)` `-4lambda-3=-3mu-1 rArr 4 lambda=3mu-2"........."(2)` `7lambda-1=8mu-10-10 rArr 7lambda=8mu-9"........."(3)` From eq. (1) and (2) `lambda = 1 " " mu = 2` Eq. (3) satisfies with these values. Therefore given lines intersect. Point of intersection `= A(lambda+4,-4lambda-3,7lambda-1)` `= A (5,-7,6)` |
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148. |
Find the co-ordinates of a point where the line `(x-1)/(-2) = (y-2)/(3) = (z+5)/(-4)`, meets the plane `2x+4-z=3`. |
Answer» Let `(x-1)/(-2) = (y-2)/(3) = (z+5)/(-4) = lambda` Co-ordinates of any point A on this line `A(-2lambda+1,3lambda+2,-4lambda-5)` This point lies on the plane `2x+4y-z=3` `2(-2lambda+1)+4(3lambda+2)-(-4lambda-5)=3` `rArr -4lambda+2+12lambda+8+4lambda+5=3` `rArr 12lambda = -12` `rArr lambda = - 1` `:.` Co-ordinates of point `A = (3,-1,-1)`. |
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149. |
Show that the points `A(-1,4,-3), B(-3,2,1),C(3,2,-5)` and `D(-3,8,-5)` are coplanar. Also find the equation of the plane passing through these points. |
Answer» Correct Answer - `x+y+z=0` | |
150. |
Obtain the equation of the sphere through the four points (4,-1,2); (0,-2,3); (1,-5,-1) and (2,0, 1). |
Answer» Let the `center(a,b,c)` and `radius=r` So, Equation of sphere be:`(x-a)^2+ (y-b)^2+ (z-c)^2=r^2` It passes through the points `A(4.-1.2), B(0,-2,3), C(1,-5,-1) and D(2,0,1)` By point A,`(4-a)^2+ (-1-b)^2+ (2-c)^2 =r^2` By point B,`(0-a)^2+ (-2-b)^2 + (3-c)^2=r^2` By point C,`(1-a)^2 + (-5-b)^2+ (-1-c)^2=r^2` By point D,`(2-a)^2+ (0-b)^2+ (1-c)^2=r^2` On solving We get,`a=2, b=-3, c=1, r=3` Hence, Equation of sphere is=`(x-2)^2+ (y+3)^2 + (z-1)^2=9` |
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