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51.

Find the cartesian form of the equation of the plane. `vecr=(lambda-mu)hati+(1-mu)hatj+(2lambda+3mu)hatk`

Answer» `vecr=(lambda-mu)hati+(1-mu)hatj+(2lambda+3mu)hatk`
`rArr vecr= hatj+lambda(hati+2hatk)+mu(-hati-hatj+3hatk)`
Comparing with equation `vecr = veca+lambdavecb+muvecc`
`veca = hatj, vecb=hati+2hatk` and `vecc = -hati+hatj+3hatk`
`:. vecn=vecbxxvecc=|{:(hati,hatj,hatk),(1,0,2),(-1,-1,3):}|`
`= hati(0+2)-hatj(3+2)hatk(-1-0)`
`=2hati-5hatj-hatk`
Equation of plane in scalar product form
`vecr.vecn = veca.vecn`
`rArr (xhati+yhatj+zhatk).(2hati-5hatj-hatk)`
`= hatj.(2hati-5hatj-hatk)`
`rArr 2x-5y - z = - 5`.
52.

The direction ratios of a line which is perpendicular to the two vectors ` -2bar(i) +3 bar(j) -bar(k) and 4 bar(i) +bar(j) +3 bar(k)` isA. `1,2,-7`B. `5,1,-7`C. `5,2,-3`D. `6,1,-2`

Answer» Correct Answer - b
53.

If the line `vec(OR)` makes angles ` theta_(1),theta_(2),theta_(3)` with the planes ` XOY, YOZ, ZOX` respectively , then ` cos^(2)theta_(1)+cos^(2)theta_(2)+cos^(2)theta_(3)` is equal toA. `-1`B. 0C. 1D. 2

Answer» Correct Answer - d
54.

Find the intercepts cut off by the plane `2x + y z = 5`.

Answer» Given plane `2x+y-z=5` can be written as :
`(2x)/(5) + (y)/(5) - z/5 = 1`
`rArr (x)/(5//2) + y/5 + (z)/(-5) = 1`
Here, `a = 5/2 , b = 5` and `c = - 5`
Thereform, intercepts are `5/2,5` and `-5`
55.

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Answer» Equation of plane `ZOX` is `y = 0`
Equation of a plane parallel to this plane is ` y = a`.
y-intercept of plane is `3`.
`:. a = 3`
Therefore, required equation of plane is ` y = 3`.
56.

If the line passing through origin makes angles `theta_(1), theta_(2), theta_(3)` with the planes XOY, YOZ and ZOX, then ` sin^(2)theta_(1)+sin^(2)theta_(2)+sin^(2)theta_(3)=`A. 4B. 2C. 3D. 1

Answer» Correct Answer - D
57.

The length of projection ofthe line segment joining the points `(1,0,-1)a n d(-1,2,2)`on the plane `x+3y-5z=6`is equal toa. `2`b. `sqrt((271)/(53))`c. `sqrt((472)/(31))`d. `sqrt((474)/(35))`A. 2B. `sqrt((271)/(53))`C. `sqrt((472)/(31))`D. `sqrt((474)/(35))`

Answer» Correct Answer - d
Let A(1,0,-1), B(-1,2,2) Direction rations of segment AB are `lt2,-2,-3gt`
`becausecostheta=(|2xx1+3(-2)-5(-3)|)/(sqrt(1+9+25)sqrt4+4+9)`
`=(11)/(sqrt17sqrt35)=(11)/sqrt595`
Length of projection `=(AB)sintheta`
`=sqrt((2)^(2)+(2)^(2)+(3)^(2))xxsqrt((474)/(35))` units
58.

A horizontal plane `4x-3y+7z=0`is given. Find a line of greatest slope passesthrough the point `(2,1,1)`in the plane `2x+y-5z=0.`

Answer» The required line through the point `P(2, 1, 1)` in the plane `2x+y-5z=0` and of greatest slope is perpendicular to line of intersection of the planes
`" "2x+y-5z=0 and 4x-3y+7z=0`
Vector along line of intersection of planes is
`" "|{:(hati,,hatj,,hatk),(2,,1,,-5),(4,,-3,,7):}|=-8hati-34hatj-10hatk`
Then equation of line is `(x-2)/(-8)=(y-1)/(-34)=(z-1)/(-10) or (x-2)/(4)=(y-1)/(17)=(z-1)/(5)`
59.

If the distance of thepoint `P(1,-2,1)`from the plane `x+2y-2z=alpha,w h e r ealpha>0,i s5,`then the foot of the perpendicular from `P`to the place isa. `(8/3,4/3,-7/3)`b. `(4/3,-4/3,1/3)`c. `(1/3,2/3,(10)/3)`d. `(2/3,-1/3,-5/3)`A. `((8)/(3), (4)/(3), -(7)/(3))`B. `((4)/(3),-(4)/(3), (1)/(3))`C. `((1)/(3), (2)/(3), (10)/(3))`D. `((2)/(3), -(1)/(3), (5)/(2))`

Answer» Correct Answer - a
Distance of point `(1, -2, 1)` from plane `x+2y-2z=alpha ` is `5rArr alpha = 10`.
Equation of `PQ, (x-1)/(1)= (y+2)/(2)=(z-1)/(-2)=t`
`Q-=(t+1, 2t-2, -2t+1) and PQ=5`
`rArr" "t= (5+alpha)/(9)= (5)/(3) rArr Q-= ((8)/(3), (4)/(3), (-7)/(3))`
60.

If a line drawn from point (1, 2, 1) is perpendicular to the line joining points (1, 4, 6) and (5, 4, 4), then the foot of the perpendicular isA. (2, 4, 5)B. (3, 4, 1)C. (3, 4, 5)D. (3, 0, 5)

Answer» Correct Answer - C
61.

Find the coordinates of thefoot of the perpendicular drawn from point `A(1,0,3)`to the join of points `B(4,7,1)a n d C(3,5,3)dot`

Answer» Let D be the foot of the perpendicular and let is divide BC in the ratio `lamda:1` . Then the coordinates of D are `(3lamda+4)/(lamda+1),(5lamda+7)/(lamda+1)and(3lamda+1)/(lamda+1)`.
Now, `" "vec(AD)botvec(BC)rArrvec(AD)*vec(BC)=0`
`rArr" "(2lamda+3)+2(5lamda+7)+4=0`
`or" "lamda=-(7)/(4)`
Thus, coordinates of D are `(5)/(3), (7)/(3) and (17)/(3)`.
62.

The direction consines of a line which lies in XZ-plane and making an angle of `30^(@)` with positive Z-axis areA. `0,1/2,pm(sqrt(3))/2`B. `(sqrt(3))/2 , 0,pm1/2`C. 1,0,0D. `pm1/2,0,(sqrt(3))/2`

Answer» Correct Answer - d
63.

If a line drawn from point (4, 3, 2) is perpendicular to the line joining points (2, 4, 1) and (4, 5, 3), then the foot of the perpendicular isA. `((28)/(9), 1, (19)/(9))`B. `((28)/(9), (41)/(9), (11)/(9))`C. `((4)/(3), (41)/(9), (19)/(9))`D. `((28)/(9), (41)/(9), (19)/(9))`

Answer» Correct Answer - D
64.

Two lines `(x-x_(i))/(l_(i))=(y-y_(i))/(m_(i))=(z-z_(i))/(n_(i)), (i=1,2)` are perpendicular to each other, if their direction ratios satisfyA. `(l_(1))/(l_(2))=(m_(1))/(m_(2))=(n_(1))/(n_(2))`B. `l_(i)=m_(i)=n_(i)`C. `l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=-1`D. `l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=0`

Answer» Correct Answer - D
65.

If `|bar(u)|=2 and bar(u)` makes angles of `60^(@) and 120^(@)` with OX and OY respectively , then vector `bar(u)=`A. `i+j-sqrt(jk)`B. `i-j+sqrt(2)k`C. `i+j+sqrt(2)k`D. `i+sqrt(2)j-k`

Answer» Correct Answer - b
66.

The direction consines of a line which lies in XZ-plane and making an angle of `30^(@)` with positive Z-axis areA. `pm(1)/(2), 0, (sqrt(3))/(2)`B. `pm (1)/(2), 0, (-sqrt(3))/(2)`C. `pm(sqrt(3))/(2), 0,(1)/(2)`D. `pm (sqrt(3))/(2), 0, (-1)/(2)`

Answer» Correct Answer - A
67.

The direction Cosines of two lines at right angles are `l_(1),m_(1),n_(1)and l_(2),m_(2),n_(2)` Then the direction cosines of a line which is perpendicular to both these lines areA. `l_(2)-l_(2),m_(1)-m_(2),n_(1)-n_(2)`B. `l_(1)+l_(2),m_(1)+m_(2),n_(1)+n_(2)`C. `m_(1)n_(2)-m_(2)n_(1),n_(1)l_(2)-n_(2)l_(1),l_(1)m_(2)-l_(2)m_(1)`D. `l_(1)+kl_(2),m_(1)+km_(2),n_(1)+kn_(2)`

Answer» Correct Answer - c
68.

If OP=8 and OP makes angles `45^(@) and 60^(@)` with OX-axis and OY-axis respectively, then OP is equal toA. `12 (sqrt(2)i+hatjpm hatk)`B. `6(sqrt(2)hati+hatjpm hatk)`C. `3(sqrt(2)hati+hatj pm hatk)`D. `2(sqrt(2)i+hatj pm hatk)`

Answer» Correct Answer - B
69.

If the line makes angles `45^(@), 60^(@), 120^(@)` with X, Y, Z axes respectively, then the direction consines of line areA. `(1)/(sqrt(2)), (-1)/(2), (1)/(2)`B. `(1)/(sqrt(2)), (-1)/(2), (-1)/(2)`C. `(1)/(sqrt(2)),(1)/(2), (1)/(2)`D. `(1)/(sqrt(2)), (1)/(2), (-1)/(2)`

Answer» Correct Answer - D
70.

Of line makes angle `60^(@), 120^(@), 45^(@)` with X, Y, Z axes respectively, then the direction consines of a line areA. `(1)/(2), (-1)/(2), (1)/(sqrt(2))`B. `(1)/(sqrt(2)), (1)/(2), (sqrt(3))/(2)`C. `(1)/(2),(1)/(sqrt(2)), (-1)/(sqrt(2))`D. `(1)/(sqrt(2)), (1)/(2), (-1)/(sqrt(2))`

Answer» Correct Answer - A
71.

OA, OB and OC, with O as the origin, are three mutually perpendicular lines whose lines whose direction cosines are `l_(r),m_(r)andn_(r)(r=1,2 and 3)`. If the projections of OA and OB on the plane z=0 make angles `phi_(1)andphi_(2)`, respectively, with the x-axis, prove that `tan(phi_(1)-phi_(2))=+-n_(3)//n_(1)n_(2)`.

Answer» Let `A_(1)andB_(1)` be the projections of A and B on the plane z=0. Let OA, OB and OC be of the unit length each so that the coordinates of A,B and C are `A(l_(1),m_(1),n_(1)),B(l_(2),m_(2),n_(2))andC(l_(3),m_(3),n_(3))`. The coordinates of `A_(1)andB_(1)`, therefore, are `A_(1)(l_(1),m_(1),0)andB_(1)(l_(2),m_(2),0)`. Since `OA_(1)andOB_(1)` make angles `phi_(1)andphi_(2),` respectively, with the x-axis, the angle between `OA_(1)andOB_(1)" is "phi_(1)~phi_(2)`. Hence,
`cso(phi_(1)~phi_(2))=(l_(1)l_(2)+m_(1)m_(2))/(sqrt(l_(1)^(2)+m_(1)^(2))sqrt(l_(2)^(2)+m_(2)^(2)))" "(i)`
Aslo OA, OB and OC are mutually perpendicular so that
`l_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=0`
and `l_(1)^(2)+m_(1)^(2)+n_(1)^(2)=1`
Eq. (i), therefore, yields
`cos(phi_(1)-phi_(2))=(-n_(1)n_(2))/(sqrt(1-n_(1)^(2))sqrt(1-n_(2)^(2)))`
or `sec^(2)(phi_(1)-phi_(2))=(1-n_(1)^(2)-n_(2)^(2))/(n_(1)^(2)n_(2)^(2))=1+(n_(3)^(2))/(n_(1)^(2)n_(2)^(2))`
`=1+(1-n_(1)^(2)-n_(2)^(2))/(n_(1)^(2)n_(2)^(2))=1+(n_(3)^(2))/(n_(1)^(2)n_(2^(2)))`
or `tan^(2)(phi_(1)-phi_(2))=(n_(3)^(2))/(n_(1)^(2)n_(2)^(2))`
or `tan(phi_(1)-phi_(2))=pm(n_(3))/(n_(1)n_(2)`
72.

Let a plane `a x+b y+c z+1=0,w h e r ea ,b ,c`are parameters, make anangle `60^0`with the line `x=y=z ,45^0`with the line `x=y-z=0`and`theta`with the plane `x=0.`The distance of the plane from point `(2,1,1)`is 3 units. Findthe value of `theta`and the equation ofthe plane.

Answer» Correct Answer - `theta= 60^(@) , 2x+ (2pm sqrt2) y + (2 pm sqrt2) z + 4 =0, 2x+ (2pm sqrt2)y + (2 pm sqrt2)z - 20 =0`
`ax+by+cz+1=0" "(i)`
It makes an `angle60^(@)` with the line x=y=z. So we get
`sin60^(@)=(a+b+c)/(sqrt3suma^(2))`
or `3sqrtsuma^(2)=2(a+b+c)" "(ii)`
Plane (i) makes an angle of `45^(@)` with the line
`x=y-z=0(or(x)/(0)=(y)/(1)=(z)/(1))`
`sin45^(@)=(b+c)/(sqrt2sqrt(suma^(2)))orsqrt(suma^(2))=a+c" "(iii)`
Plane (i) makes an `angletheta` with plane x=0. So we get `costheta=(a)/(sqrt(suma^(2)))" "(iv)`
From (ii) and (iii) we get
`(sqrtsuma^(2))=2a`
or `(a)/(sqrtsuma^(2))=(1)/(2)`
From (iv), `costheta=1//2ortheta=60^(@)`
Distance of plane (i) from the point (2,1,1) is 3 units.
`implies(2a+b+c+1)/(sqrt(suma^(2)))=+-3`
or `+-3sqrt(suma^(2))=2a+b+c+1`
Case I:
`+-3sqrt(suma^(2))=2a+b+c+1" "(v)`
From (ii) and (iv), we get `b+c-1=0" "(vi)`
and from (iii) and (iv), we get
`2a+b+c+1=3(b+c)" "(vii)`
From (vi) and (vii), we get `a=(1)/(2),b =(2pmsqrt2)/(4)andc=(2pmsqrt2)/(4)`
Hence, the set of such planes is
Case II:
`-3sqrtsuma^(2)=2a+b+c+1`
`a=(-1)/(10),b=(-(2pmsqrt2))/(20) andc=(-(2pmsqrt2))/(20)`
Hence, the other set of the planes is
`2x+(2pmsqrt2)y+(2pmsqrt2)z-20=0.`
73.

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`

Answer» given planes are : `(2x+y+2z=8)xx 2 = 4x + 2y + 4z = 16`
`4x+2y + 4z = -5`
distance`= (|16 - (-5)|)/sqrt((4)^2 + (2)^2 + (4)^2)`
`= 21/sqrt36 = 21/6 = 7/2`
Answer
option 2 is correct
74.

The co-ordiantes of the foot of perpendicular from origin to a plane are `(3,-2,1)`. Find the equation of the plane.

Answer» Correct Answer - `3x-2y+z-14=0`
75.

Prove that for all valuesof `lambdaa n dmu`, the planes `(2x)/a+y/b+(2z)/c-1+lambda(x/a-(2y)/b-z/c-2)=0`and `(4x)/a+(3y)/b-5+mu((5y)/b-(4z)/c+3)=0`intersect on thesame line.

Answer» Let the given planes intersect on the line with direction ratios l, m and n. In that case.
`(2+lamda)(l)/(a)+(1-2lamda)(m)/(b)+(2-lamda)(n)/(c)=0" "(i)`
and `(4l)/(a)-(3-5mu)(m)/(b)+4mu.(n)/(c)=0" "(ii)`
Hence, `(l//a)/(6-6mu-3lamda-3lamdamu)=(m//b)/(8-8mu-4lamda=4lamdamu)`
`=(n//c)/(-10+10mu+5lamda+5lamdamu)`
or `(l//a)/(3(2-2mu-lamda-lamdamu))=(m//b)/(4(2-2mu=lamda-lamdamu))`
`=(n//c)/((-5(2-2mu-lamda-lamdamu))`
`(l//a)/(3)=(m//b)/4=(n//c)/(-5)" "("provided " 2-2mu-lamda-lamdamune0)`
which are independent of `lamdaand mu`. Hence, a line with driection ratios (3a, 4b, -5c) lies in both the planes.
For `2-2mu-lamda-lamdamu=0orlamda=(2(l-mu))/(1+mu)`, planes (i)
and (ii) coincide with each other. Hence, the two given familise of planes intersect on the same line.
76.

The planes: `2x y + 4z = 5 a n d 5x 2. 5 y + 10 z = 6`are(A) Perpendicular (B) Parallel(C) intersect y-axis (D) passes through `(0,0,5/4)`

Answer» `P_1=2x-y+4z=5`
`P_2=5x-2.5y+10z=6`
`2/5-1/-2.5=4/10`
these are same ratios
so,`P_1`is parallel to `P_2`
77.

Find the coordinates of the foot of perpendicular drawn from origin to the planes: `2x-3y+4z-6=0`

Answer» Correct Answer - `(12/29,18/29,24/29)`
78.

Find the distance between the planes `2x-y+2z=4 and 6x-3y+6z=2.`

Answer» Correct Answer - `(25)/(3sqrt(14))`
79.

Determine whether the following pair of lines intersect or not. i. `vecr=hati-hatj+lamda(2hati+hatk), vecr=2hati-hatj+mu(hati+hatj-hatk)`

Answer» Here `vec(a_(1))=veci-vecj, vec(a_(2))=2hati -hatj, vec(b_(1))=2hati+hatk and vec(b_(2))=hati+hatj-hatk`
Now `" "[vec(a_(2))-vec(a_(1))vec(b_(1))vec(b_(2))]=|{:(2-1, ,-1+1,,0),(2,,0,,1),(1,,1,,-1):}|`
`" "=|{:(1,,0,,0),(2,,0,,1),(1,,1,,-1):}|`
`" "=-1 ne 0`
Thus, the two given lines do not intersect.
ii. Here `vec(a_(1))=hati+hatj-hatk, vec(a_(2))=4hati-hatk, vec(b_(1))=3hati-hatj and vec(b_(2))=2hati+ 3hatk`
`rArr" "[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`
Thus, the two given lines intersect. Let us obtain the point of intersection of the two given lines.
For some values of `lamda` and `mu`, the two values of `vecr` must coincide. Thus,
`" "hati+hatj-hatk+ lamda(3hati-hatj)=4hati-hatk+mu(2hati+3hatk)`
or `" "(3+2mu-3lamda)hati+(lamda-1)hatj+3muhatk=0`
or `" "3+2mu-3lamda=0, lamda-1=0, 3mu=0`
Solving, we obtain `lamda=1 and mu=0`
Therefore, the point of intersection is `vecr=4hati-hatk`( by putting `mu=0` in the second equation).
80.

The planes: `2x y + 4z = 5 a n d 5x 2. 5 y + 10 z = 6`are(A) Perpendicular (B) Parallel(C) intersect y-axis (D) passes through `(0,0,5/4)`A. perpendicularB. parallelC. intersect y-axisD. passes through `(0,0,5/4)`

Answer» Correct Answer - B
Direction of the given planes are `(2,-1,4)` and `(5,-2.5,10)`
`:. 2/5 = (-1)/(-2.5) = (4)/(10) rArr (4)/(10) = 4/10 = 4/10`
Therefore, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))`
So, the given planes are parallel.
81.

Find the equations of thebisectors of the angles between the planes `2x-y+2z+3=0a n d3x-2y+6z+8=0`and specify the plane which bisects the acuteangle and the plane which bisects the obtuse angle.

Answer» The given planes are `2x-y+2z+3=0 and 3x-2y+6z+8=0` , where `d_(1), d_(2) gt 0 and a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)= 6+2+12gt 0`.
`" "(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)))" "` (obtuse angle bisector)
`" "(a_(1)x+b_(1)y+c_(1)z+d)/(sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))=(a_(2)x+b_(2)y+c_(2)z+d_(2))/(sqrt(a_(2)^(2)+b_(2)^(2)+c_(2)^(2)))" "` (acute angle bisector)
i.e., `" "(2x-y+2z+3)/(sqrt(4+1+4))=pm(3x-2y+6z+8)/(sqrt(9+4+36))`
or `" "(14x-7y+14z+21)=pm(9x-6y+18z+24)`
Taking the positive sign on the right hand side, we get
`" "5x-y-4z-3=0`
Taking the negative sign on the right hand side, we get
`" "23x-13y+32z+45=0" "` (acute angle bisector)
82.

The length of the perpendicular from the origin to the plane passing through the points `veca` and containing the line `vecr = vecb + lambda vecc` isA. `([vecavecbvecc])/(|vecaxxvecb+vecbxxvecc+veccxxveca|)`B. `([vecavecbvecc])/(|vecaxxvecb+vecbxxvecc|)`C. `([vecavecbvecc])/(|vecbxxvecc+veccxxveca|)`D. `([vecavecbvecc])/(|veccxxveca+vecaxxvecb|)`

Answer» Correct Answer - c
The given plane passes through `veca` and is parallel to the vectors `vecb-vecaandvecc`. So it is normal to
`(vecb-veca)xxvecc`. Henec, its equation is `(vecr-veca).((vecb-veca)xxvecc)=0`
or `vecr.(vecbxxvecc+veccxxveca)=[vecavecbvecc]`
The length of the perpendicular from the origin to this plane is
`[vecavecbvecc]/(|vecbxxvecc+veccxxveca|)`
83.

Find the vector of the plane passing through the intersection of the planes `vecr.(2hati+2hatj-3hatk)=7, vecr.(2hati+5hatj+3hatk)=9` and the point (2,1,3)`.

Answer» Equation of given planes are
`vecr.(2hati+2hatj-3hatk)=7` and `vecr.(2hati+5hatj+3hatk) = 9`
The equation of these planes can be written as
` v ecr.(2hati+2hatj-3hatk) - 7 = 0 "………"(1)`
and `vecr.(2hati+5hatj+3hatk) - 9 = 0 "….."(2)`
Equation of a plane through the intersection of plane (1) and (2) is,
`[vecr.(2hati+2hatj-3hatk)-7]+lambda[vecr.(2hati+5hatj+3hatk)-9]=0`
`rArr vecr.[(2hati+2hatj-3hatk)+lambda(2hati+5hatj+3hatk)]=9lambda+7`
`rArr [vecr.(2+2lambda)hati+(2+5lambda)hatj+(3lambda-3)hatk]=9lambda+7"......."(3)`
This intersecting plane passes through the point (2,1,3), whose position vector is `vecr.(2hati+hatj+3hatk)`. Put this value of `vecr` in eqaution (3),
`(2hati+hatj+ 3hatk) .[(2+2lambda) hati+(2+5lambda)hatj+(3lambda-3)hatk] = 6 lambda + 7`
`rArr 2(2+2lambda)+(2+5lambda)+3(3lambda-3)=9lambda+7`
`rArr (4+4lambda)+(2+5lambda)+(9lambda-9)=9lambda+7`
`rArr -3+18lambda=9lambda+7`
`rArr 9lambda = 10 lambda = 10/9`
`:.` Equation of plane
`[vecr.(2hati+2hatj-3hatk)-7]+10/9[vecr.(2hati+5hatj+3hatk)-9]=0`
`rArr vecr .(18hati+18hatj-27hatk+20hati+50hatj+30hatk)-63-90=0`
`rArr vecr.(38hati+68hatj+3hatk) = 153`
which is the required equation of the plane.
84.

Find the shortest distance between the lines gives by `vecr=(8+3lamda)hati-(9+16lamda)hatj+(10+7lamda)hatk` and `vecr=15hati+29hatj+5hatk+mu(3hati+8hatj-5hatk)`.

Answer» We have `vecr=(8+3lamda)hati-(9+16lamda)hatj+(10+7lamda)hatk`
`=8hati-9hatj+10hatk+3lamdahati-16lamdahatj+7lamdahatk`
`=8hati-9hatj+10hatk+lamda(3hati-16hatj+7hatk)`
`implies veca_(1)=8hati-9hatj+10hatk` and `vecb_(1)=3hati-16hatj+7hatk`
Also `vecr=15hati+29hatj+5hatk+mu(3hati+8hatj-5hatk)`
`implies veca_(2)=15hati+29hatj+5hatk` and `vecb_(2)=3hati+8hatj-5hatk`
Now, shortest distance between two lines is given by `|((vecb_(1)xxvecb_(2)).(veca_(2)-veca_(1)))/(|vecb_(1)xxvecb_(2)|)|`
`:. vecb_(1)xxvecb_(2)=|(hati, hatj, hatk),(3,-16,7),(3,8,-5)|`
`=hati(80-56)-hatj(-15-21)+hatk(24+48)`
`=24hati+36hatj+72hatk`
Now `|vecb_(1)xxvecb_(2)|=sqrt((24)^(2)+(36)^(2)+(72)^(2))`
`=12sqrt(2^(2)+3^(2)+6^(2))=84` ltbgt and `(veca_(2)-veca_(1))=(15-8)hati+(29-9)hatj+(5-10)hatk`
`=7hati+38hatj-5hatk`
`:.` Shortest distance `=|((24hati+36hatj+72hatk).(7hati+38hatj-5hatk))/84|`
`=|(168+1368-360)/84|=|1176/84|=14` units
85.

Find the equation of the plane through the points (2,1, 1) and (1, 3, 4) and perpendicular to the plane `x 2y+4z=10.`

Answer» The equation of the plane through `(2,1,-1)` is
`a(x-2)+b(y-1)+c(z+1)=0`
Since this passes through `(-1,3,4)`
`:. a(-1-2)+b(3-1)+c(4+1)=0`
`implies-3a+2b+5c=0`
Since the plane (i) is perpendicular to the plane `x-2y+4z=10`
`:. 1.a-2.b+4.c=0`
`implies a-2b+4c=0`
On solving eqs (ii) and (iii) we get
`a/(8+10)=(-b)/(-17)=c/4=lamda`
`implies a=18lamda,b=17lamda,c=4lamda`.
From eq. (i)
`18lamda(x-2)+17lamda(y-1)+4lamda(z+1)=0`
`implies18x-36+17y-17+4z+4=0`
`implies 18x+17y+4z-49=0`
`. 18x+17y+4z=49`
86.

The equations of the planewhich passes through `(0,0,0)`and which is equally inclined to the planes `x-y+z-3=0a n dx+y=z+4=0`is/area. `y=0`b. `x=0`c. `x+y=0`d. `x+z=0`A. `y=0`B. `x=0`C. `x+y=0`D. `x+z=0`

Answer» Correct Answer - a, c
The required plane is parallel to the bisector of the given planes.
Bisector are `(x-y+z-3)/(sqrt3)= pm (x+y+z+4)/(sqrt3)`
or `2y+7 =0 and 2x+2y+1=0`. Hence, the planes are `y=0 and x+y=0`.
87.

The vector equation of the plane passing through the origin and the line of intersection of the planes `vecr.veca=lamdaandvecr.vecb=mu` isA. `vecr.(lamdaveca-muvecb)=0`B. `vecr.(lamdavecb-muveca)=0`C. `vecr.(lamdaveca+muvecb)=0`D. `vecr.(lamdavecb+muveca)=0`

Answer» Correct Answer - b
The equation of a plane through the line of intersection of the planes `vecr*veca=lamda and vecr*vecb= mu` is
`" "(vecr*veca-lamda)+k(vecr*vecb-mu)=0`
or `" "vecr*(veca+kvecb)= lamda+kmu" "`(i)
This passes through the origin, therefore
`" "vec0(veca+k vecb) = lamda+ muk or k= (-lamda)/(mu)`
Putting the value of k in (i), we get the equation of the required plane as
`" "vecr*(muveca-lamdavecb)= 0 or vecr*(lamda vecb-mu veca)=0`
88.

The equation of the planewhich is equally inclined to the lines `(x-1)/2=y/(-2)=(z+2)/(-1)a n d=(x+3)/8=(y-4)/1=z/(-4)`and passing through the origin is/area. `14 x-5y-7z=0`b. `2x+7y-z=0`c. `3x-4y-z=0`d. `x+2y-5z=0`A. `14x-5y-7z=0`B. `2x+7y-z=0`C. `3x-4y-z=0`D. `x+2y-5z=0`

Answer» Correct Answer - a, b
The plane is equaly inclined to the lines. Hence, it is perpendicular to the angle bisector of the vectors
`2hati-2hatj-hatk and 8hati+hatj-4hatk`.
Vector along the angle bisectors of the vectors are
`" "(2hati-2hatj-hatk)/(3) pm (8hati+hatj-4hatk)/(9)`
or `" "(14hati-5hatj-7hatk)/(9) and (-2hati-7hatj+hatk)/(9)`
Hence, the equations of the planes are
`" "14x-5y-7z=0 or 2x+7y-z=0`
89.

The plane `4x+7y+4z+81=0`is rotated through a right angle about itsline of intersection with the plane `5x+3y+10 z=25.`The equation of the plane in its new positionisa. `x-4y+6z=106`b. `x-8y+13 z=103`c. `x-4y+6z=110`d. `x-8y+13 z=105`A. `x-4y+6z=106`B. `x-8y+13z=103`C. `x-4y+6z=110`D. `x-8y+13z=19=105`

Answer» Correct Answer - a
The equation of the plane through the line of intersection of the planes `4x+7y+4z+81=0 and 5x+3y+10z=25` is
`" "(4x+7y+ 4z + 81)+ lamda(5x+ 3y+ 10z- 25)=0`
or `" "(4+5lamda)x +(7+ 3lamda ) y + (4+ 10 lamda)z+ 81- 25 lamda =0 " "` (i)
which is perpendicular to `4x+ 7y + 4z+ 81=0`
`rArr" "4(4+ 5lamda)+7(7+ 3lamda) + 4(4+ 10 lamda) =0`
or `" " 81 lamda + 81=0`
or `" "lamda=-1`
Hence, the plane is `x-4y+6z= 106`
90.

The x-y plane is rotatedabout its line of intersection with the y-z plane by `45^0`, then the equation of thenew plane is/area. `z+x=0`b. `z-y=0`c. `x+y+z=0`d. `z-x=0`A. `z+x=0`B. `z-y=0`C. `x+y+z=0`D. `z-x=0`

Answer» Correct Answer - a, d
The equation of a plane passing through the line of intersection of the `x-y and y-z` planes is
`" "z+lamdax=0, lamdain R`
This plane makes an angle `45^(@)` with the `x-y` plane `(z=0)`. Thus,
`" "cos45^(@) = (1)/(sqrt(1)sqrt(lamda^(2)+1))`
or `" "lamda= pm 1`
91.

The plane denoted by `P_(1) : 4x+7y+4z+81=0` is rotated through a right angle about its line of intersection with the plane `P_(2) : 5x+3y+ 10 z = 25`. If the plane in its new position is denoted by `P`, and the distance of this plane from the origin is d, then find the value of `[k//2]` (where `[*]` represents greatest integer less than or equal to k).

Answer» Correct Answer - `7`
`" "4x+7y+ 4z+ 81=0" "` (i)
`" "5x+3y+10z=25" "`(ii)
Equation of plane passing through their line of intersection is
`(4x+7y+4z+81)+ lamda(5x+3y+10z-25)=0`
or `" "(4+5lamda)x+ (7+3lamda)y+ (4+10lamda)z+ 81- 25lamda=0" "` (iii)
Plane (iii) `bot` to (i), so
`" "4(4+5lamda)+7(7+3lamda)+ 4(4+10lamda)=0`
`therefore" "lamda=-1`
From (iii), equation of plane is
`" "-x+4y-6z+106=0" "`(iv)
Distance of (iv) from (0, 0, 0)`=(106)/(sqrt(1+16+ 36))=(106)/(sqrt(53))`
92.

The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle `alpha`. Prove that the equation of the plane in its new position is ` ax + by +- (sqrt(a^2 + b^2) tan alpha)z = 0`

Answer» Equation of the plane is `ax+by=0`……i
`:.` Equation of the plane after new position is
`(ax cos alpha)/(sqrt(a^(2)+b^(2)))+(by cos alpha)/(sqrt(b^(2)+a^(2)))+-z sin alpha =0`
`implies (ax)/(sqrt(a^(2)+b^(2)))+(by)/(sqrt(b^(2)+a^(2)))+-z tan alpha =0` [on dividing by `cos alpha`]
`implies ax+by+-z tan alpha sqrt(alpha^(2)+b^(2))=0` [on multiplying with `sqrt(a^(2)+b^(2))`]
Alternate Method
Given planes are `ax+by=0`.................i
and `z=0`.................ii
Therefore the equation of any plane passing through the line of intersection of planes i and ii may be taken as `ax+by+k=0` ....................iii
Then, direction cosines of a normal to the plane iii are `a/(sqrt(a^(2)+b^(2)+k^(2))),b/(sqrt(a^(2)+b^(2)+k^(2))),c/(sqrt(a^(2)+b^(2)+k^(2)))` and direction cosines of the normal to the plane i are `a/(sqrt(a^(2)+b^(2))),b/(sqrt(a^(2)+b^(2))),0`
Since the angle between the plane i and i is `alpha`,
`:. cos alpha=(a.a+b.b+k.0)/(sqrt(a^(2)+b^(2)+k^(2))sqrt(a^(2)+b^(2)))`
`= sqrt((a^(2)+b^(2))/(a^(2)+b^(2)+k^(2)))`
`implies k^(2)cos^(2)alpha=a^(2)(1-cos^(2)alpha)+b^(2)(1-cos^(2)alpha)`
`implies k^(2)=((a^(2)+b^(2)sin^(2)alpha))/(cos^(2) alpha)`
`=k=+-sqrt(a^(2)+b^(2)) tan alpha`
On putting this value in plane iii we get the equation of the planes as
`ax+by+zsqrt(a^(2)+b^(2)) tan alpha=0`
93.

Find Cartesian and vectorequation of the line which passes through the point `(-2,4,-5)`and parallel to the linegiven by `(x+3)/3=(y-4)/5=(z+8)/6`.

Answer» Equation of given line is `(x+3)/(3)=(y-4)/(5)=(z+8)/(6)" "`(i)
Direction ratios of line (i) are 3, 5, 6
Therefore, direction ratios of required line are also 3, 5, 6. ltBrgt Required line passes through the point (-2, 4, -5).
Hence, equation of required line is `(x+2)/(3)=(y-4)/(5)=(z+5)/(6)`
Its vector form is `vecr=(-2hati+4hatj-5hatk)+lamda(3hati+5hatj+6hatk)`
94.

Find the cartesian equation of the line which passes through the point `(-2,4,-5)` and parallel to the line `(x+3)/3 = (y-4)/5 = (z+8)/6`

Answer» The direction ratios of the line `(x+3)/(3) = (y-4)/(5) = (z+8)/(6)` are `(3,5,6)`.
Here the line passes through the point (-2,4,-5) and the direction ratios of the parallel line are `(3,5,6)` . So, the equation of line is,
`(x-(-2))/(3) = (y-4)/(5) = (z-(-5))/(6)`
`rArr (x+2)/(3) = (y-4)/(5) = (z+5)/(6)`
95.

Find the equation of a straight line in the plane `vecr.vecn=d` which is parallel to `vecr.vecn=d("where "vecn.vecb=0)`.A. `vecr=veca+((d-veca.vecn)/(n^(2)))vecn+lamdavecb`B. `vecr=veca+((d-veca.vecn)/(n))vecn+lamdavecb`C. `vecr=veca+((veca.vecn-d)/(n^(2)))vecn+lamdavecb`D. `vecr=veca+((veca.vecn-d)/(n))vecn+lamdavecb`

Answer» Correct Answer - a
Foot of the perpendicular from point `A(veca)` on the plane `vecr*vecn=d` is `veca+ ((d-veca*vecn))/(|vecn|^(2))vecn`
Therefore, equation of the line parallel to `vecr=veca+lamdavecb` in the plane `vecr*vecn =d` is given by
`" "vecr=veca+ ((d-veca*vecn))/(|vecn|^(2))vecn+lamdavecb`
96.

The equation of the line `x+y+z-1=0`, `4x+y-2z+2=0` written in the symmetrical form isA. `(x-1)/(2)=(y+2)/(-1)=(z-2)/(2)`B. `(x+(1//2))/(1)=(y-1)/(-2)=(z-(1//2))/(1)`C. `(x)/(1)=(y)/(-2)=(z-1)/(1)`D. `(x+1)/(1)=(y-2)/(-2)=(z=0)/(1)`

Answer» Correct Answer - b, c, d
`" "x+y+z-1=0`
`" "4x+y-2z+2=0`
Therefore, the line is along the vector
`" "(hati+hatj+hatk)xx(4hati+hatj-2hatk)=3hati-6hatj+3hatk`
Let `z=k`. Then `x=k-1 and y=2-2k`
Therefore, `(k-1, 2-2k, k)` is any point on the line
Hence, `(-1, 2, 0), (0, 0, 1) and (-1//2, 1, 1//2)` are the points on the line.
97.

What is the nature of theintersection of the set of planes `x+a y+(b+c)z+d=0,x+b y+(a+a)z+d=0a n dx+c y+(a+b)z+d=0?`a. they meet at a pointb. the form a triangular prismc. the pass through a lined. they are at equal distance from the originA. They meet at a pointB. They form a triangular prismC. They pass through a lineD. They are at equal distance from the origin

Answer» Correct Answer - c
`|{:(1,,a,,b+c),(1,,b,,c+a),(1,,c,,a+b):}|=|{:(1,,a,,a+b+c),(1,,b,,a+ b+ c),(1,,c,,a+b+c):}|=0`
98.

Equation of a plane passing through the intersection of the planes `vecr.(3hati-hatj+hatk)=1` and `vecr.(hati+4hatj-2hatk)=2` and passing through the point `(hati+2hatj-hatk)` is :A. `vecr.(2hati-7hatj-13hatk)=1`B. `vecr.(2hati+7hatj+13hatk)=1`C. `vecr.(2hati-7hatj-13hatk)=4`D. None of these

Answer» Correct Answer - A
99.

Find the vector equation to the plane through the point `(2,1,-1) ` passing through the line of intersection of the planes `vecr.(hati+3hatj-hatk)=0 ad vecr.(hatj+2hatk)=0`A. `vecr.(hati+9hatj+11hatk) = 0`B. `vecr.(hati+9hatj+11hatk) = 6`C. `vecr.(hati-9hatj-11hatk) 0`D. None of the above

Answer» Correct Answer - A
100.

Find the angle between the following lines and the planes : (i) line `vecr=(hati+2hatj-hatk)+lambda(hati-hatj+hatk)` and planes `vecr.(2hati-hatj+hatk) = 4`. (ii) line `vecr = (2hati+3hatj+9hatk)+lambda(2hati+3hatj+4hatk)` and plane `vecr.(hati+hatj+hatk) =5`. (iii) line `(x+1)/(3) = y/2 = z/4` and plane `2x+y-3z=5`. (iv) line `(x-3)/(9) = (y+4)/(6) = (z+2)/(2)` and plane `3x-y+z=0`.

Answer» Correct Answer - (i) `sin^(-1)((2sqrt(2))/(3))`, (ii) `sin^(-1)((3sqrt(3))/(sqrt(29)))`, (iii) `sin^(-1)((-4)/(sqrt(406)))` , (iv) `sin^(-1)((23)/(11sqrt(11)))`