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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`
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.
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)`.
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.
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.
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))`
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`.
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).
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)`
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`
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))`
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.
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))`
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^(@)`.
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.
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)`
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.
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))`
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.
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
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))`
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^(@)`
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`
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`
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`
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)`
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`.
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.
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)`
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)`.
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`