InterviewSolution
Saved Bookmarks
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.
| 151. |
Find the shortest distancebetween the z-axis and the line, `x+y+2z-3=0,2x+3y+4z-4=0.` |
|
Answer» let `p_1` and `p_2` be the points on the given planes having coordinates (1,1,2) & (2,3,4) respectively R = `|(i,j,k),(1,1,2),(2,3,4)|` `=i(-2)-j(0) + k(1) = -2i+k` `(-2,0,1)` if z=0 `x+y-3 = 0 & 2x+3y-4 = 0` we get `2(3-y) + 3y-4 = 0` then, `y=-2 & x = 5` so,intersection point `(5,-2,0)` eqn of line is `(x-5)/-2 = (y+2)/0 = z/1` let direction ratios of this line is `(a_1,b_1,c_1)` eqn of z axis is `x/0 = y/0 = z/0` let direction ratios of this line is `(a_2,b_2,c_2)` S.D = `||(x_2-x_1,y_2-y_1,z_2-z_1),(a_1,b_1,c_1),(a_2,b_2,c_2)||/||(i,j,k),(a_1,b_1,c_1),(a_2,b_2,c_2)||` `= ||(-5,2,0),(-2,0,1),(0,0,1)||/(||(i,j,k),(-2,0,1),(0,0,1)||)` `|(-4)|/(|j(2)|)= 2` |
|
| 152. |
If the angle `theta` between the line `(x+1)/1=(y-1)/2=(z-2)/2` and the plane `2x-y+sqrt(pz)+4=0` is such that `sintheta=1/3`, then the values of p is (A) 0 (B) `1/3` (C) `2/3` (D) `5/3`A. `(-3)/(5)`B. `(5)/(3)`C. `(-4)/(3)`D. `(3)/(4)` |
|
Answer» Correct Answer - b The line is `(x+1)/(1)=(y-1)/(2)=(z-2)/(2)` and the plane is `2x-y+sqrtlamdaz+4=0`. If `theta` is the angle between the line and the plane, then `90^(@)-theta` is the angle between the line and normal to the plane. Thus, `cos(90^(@)-theta)=((1)(2)+(2)(-1)+(2)(sqrtlamda))/(sqrt(1+4+4)sqrt(4+1+lamda))` or `sintheta=(2-2+2sqrtlamda)/(3sqrt5+lamda)or(1)/(3)=(2sqrtlamda)/(3sqrt5+lamda)` or `sqrt(5+lamda)=2sqrtlamda` or `5+lamda=4lamda` or `3lamda=5` or `lamda=(5)/(3)` |
|
| 153. |
The length of theperpendicular drawn from `(1,2,3)`to the line `(x-6)/3=(y-7)/2=(z-7)/(-2)`isa. `4`b. `5`c. `6`d. `7`A. 4B. 5C. 6D. 7 |
|
Answer» Correct Answer - d Let P be the point (1,2,3) and PN be the length of the perpendicular from P on the given line. Coordinates of point N are `(3lamda+6,2lamda+7,-2lamda+7)`. Now PN is perpendicular to the given line of vector `3veci+2vecj-2veck`. Thus, `3(3lamda+6-1)+2(2lamda+7-2)-2(-2lamda+7-3)=0orlamda=-1` Then, point N is (3,5,9) `impliesPN=7` |
|
| 154. |
LetL be the line of intersection of the planes `2x""+""3y""+""z""=""1`and `x""+""3y""+""2z""=""2`. If L makes an angles ` alpha `withthe positive x-axis, then cos` alpha `equals`1/(sqrt(3))``1/2`1`1/(sqrt(2))`A. `(1)/(2)`B. 1C. `(1)/(sqrt2)`D. `(1)/(sqrt3)` |
|
Answer» Correct Answer - d Since line of intersection is perpendicular to both the planes, direction rations of the line of intersection is ltbgt `|[hati,hatj,hatk],[2,3,1],[1,3,2]|=3hati-3hatj+3hatk` Hence, `cosalpha=(3)/(sqrt(9+9+9))=(1)/(sqrt3)` |
|
| 155. |
If the angle between theplane `x-3y+2z=1`and the line `(x-1)/2=(y-1)/1=(z-1)/(-3)i stheta,`then the find the value of `cos e cthetadot` |
|
Answer» Correct Answer - `2` Vector normal to the plane is `vecn=hati-3hatj+2hatk` and vector along the line is `vecv= 2hati+hatj- 3hatk` Now `" "sintheta = (vecx*vecv)/(|vecx||vecv|)= |(2-3-6)/(sqr(14)sqrt(14))|=|(7)/(14)|` Hence, `cosec theta = 2` |
|
| 156. |
Let the equation of the plane containing the line `x-y - z -4=0=x+y+ 2z-4` and is parallel to the line of intersection of the planes `2x + 3y +z=1` and `x + 3y + 2z = 2` be `x + Ay + Bz + C=0` Compute the value of `|A+B+C|`. |
|
Answer» Correct Answer - `6` A plane containing the line of intersection of the given planes is `" "x-y-z-4+lamda(x+y+2z-4)=0` i.e., `(lamda+1)x+ (lamda-1)y+ (2lamda-1)z-4(lamda+1)=0` vector normal to it `" "V=(lamda+1)hati+ (lamda-1)hatj+ (2lamda-1)hatkS" "` (i) Now the vector along the line of intersection of the planes `2x+3y+z-1=0 and x+3y+2z-2=0` is given by `" "vecn= |{:(hati,,hatj,,hatk), (2,,3,,1), (1,,3,,2):}|= 3(hati-hatj+hatk)` As `vecn` is parallel to the plane (i), we have `" "vecn*vecV=0` `" "(lamda+1)- (lamda-1)+ (2lamda-1)=0` `" "2+2lamda-1=0 or lamda = (-1)/(2)` Hence, the required plane is `" "(x)/(2)-(3y)/(2)-2z-2=0` `" "x-3y-4z-4=0` Hence, `|A+B+C|=6` |
|
| 157. |
Let `A_(1), A_(2), A_(3), A_(4)` be the areas of the triangular faces of a tetrahedron, and `h_(1), h_(2), h_(3), h_(4)` be the corresponding altitudes of the tetrahedron. If the volume of tetrahedron is `1//6` cubic units, then find the minimum value of ` (A_(1) +A_(2) + A_(3) + A_(4))(h_(1)+ h_(2)+h_(3)+h_(4))` (in cubic units). |
|
Answer» Correct Answer - `8` Volume `(V)= (1)/(3) A_(1)h_(1) or h_(1) = (3V)/(A_(1))` Similarly `h_(2)= (3V)/(A_(2)), h_(3)= (3V)/(A_(3)) and h_(4)= (3V)/(A_(4))` So `" "(A_(1)+ A_(2)+ A_(3)+A_(4))(h_(1)+h_(2)+ h_(3)+h_(4))` `" "= (A_(1)+A_(2)+A_(3)+A_(4))((3V)/(A_(1))+ (3V)/(A_2)+ (3V)/(A_3)+ (3V)/(A_4))` `" "= 3V(A_(1)+ A_(2)+A_(3)+A_(4))((1)/(A_1)+ (1)/(A_2)+ (1)/(A_3)+ (1)/(A_4))` Now using A.M.-H.M inequality in `A_(1), A_(2), A_(3), A_(4)`, we get `" "(A_1+A_2+A_3+A_4)/(4) ge (4)/(((1)/(A_1)+ (1)/(A_2)+ (1)/(A_3)+ (1)/(A_4)))` or `" "(A_(1) + A_(2) + A_(3) + A_(4))((1)/(A_1)+ (1)/(A_2)+ (1)/(A_3)+ (1)/(A_4)) ge 16` Hence, the minimum value of `(A_(1)+ A_(2)+ A_(3)+A_(4))(h_(1)+h_(2)+h_(3)+h_(4)) = 3V(16)= 48V= 48(1//6)=8` |
|
| 158. |
If `P(x ,y ,z)`is a point on the linesegment joining `Q(2,2,4)a n d R(3,5,6)`such that the projectionsof ` vec O P`on te axes are 13/5, 19/5and 26/5, respectively, then find the ratio in which `P`divides `Q Rdot` |
|
Answer» Since `vec(OP)` has projections `(13)/(5), (19)/(5) and (26)/(5)` on the coordinates axes, `vec(OP)=(13)/(5)hati+(19)/(5)hati+(26)/(5)hatk`. Suppose P divides the join of `Q(2, 2, 4) and R(3, 5, 6)` in the ratio `lamda:1`. Then the position vector of P is `" "((3lamda+2)/(lamda+1))hati+((5lamda+2)/(lamda+1))hatj+((6lamda+4)/(lamda+1))hatk` `therefore" "(13)/(5)hati+(19)/(5) hatj+(26)/(5) hatk` `" "((3lamda+2)/(lamda+1))hati+((5lamda+2)/(lamda+1))hatj+((6lamda+4)/(lamda+1))hatk` Thus, we have `" "(3lamda+2)/(lamda+1)=(13)/(5), (5lamda+2)/(lamda+1)=(19)/(5) and (6lamda+4)/(lamda+1)=(26)/(5)` `rArr" "2lamda=3 or lamda = 3//2` Hence, P divides QR in the ratio 3 : 2. |
|
| 159. |
Show that the line through the points `(4, 7, 8), (2, 3, 4)`is parallel to the line through the points `( 1, 2, 1)`,`(1, 2, 5)`. |
|
Answer» Direction ratios of the line passing through the points `(4,7,8),(2,3,4)` are `a_(1),b_(1),c_(1) = 2-4,3-7,4-8` `= - 2,-4,-4` Direction of the line passing through the points `(-1,-2,1),(1,2,5)` `a_(2),b_(2),c_(2)=1+1,2+2,5-1=2,4,4` Now, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))` Therefore, two lines are parallel . |
|
| 160. |
Find the vector and the cartesian equations of the line that passes through the points`(3, 2, 5), (3, 2, 6)`. |
|
Answer» `vecb=vec(pq)=0hati+ohatj+11hatk` `vecr=3hati-2hatj-5hatk+lambda(11hatk)` `(x-3)/0=(y+2)/0=(z+5)/11` |
|
| 161. |
If the lines `(x-1)/(-3)=(y-2)/(2k)=(z-3)/(-2)a n d(x-1)/(3k)=(y-5)/1=(z-6)/(-5)`are at right angel, thenfind the value of `kdot` |
|
Answer» The lines are perpendicular if `a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`. Hence, `-3(3k)+2k(1)+2(-5)=0 or k=-(10)/(7)`. |
|
| 162. |
Find the angel between thelines `2x=3y=-za n d6x=-y=-4zdot` |
|
Answer» The lines are `(x)/(3)=(y)/(2)=(z)/(-6) and (x)/(2)=(y)/(-12) = (z)/(-3)`. Since `a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=6-24+18=0`, we have `" "theta=90^(@)` |
|
| 163. |
The straight line `(x-3)/3=(y-2)/1=(z-1)/0`isParallel to x-axisParallel to the y-axisParallel to the z-axisPerpendicular to the z-axis |
| Answer» The line is along the vector `3hati+hatj` which is perpendicular to the z-axis as `(3hati+hatj)*hatk=0`. | |
| 164. |
The cartesian equation of a line is `(x-5)/3=(y+4)/7=(z-6)/2`. Write its vector form. |
|
Answer» `(x-5)/3=(y+4)/7=(z-6)/2` `vecr=veca+lambdavecb` `(5hati-4hatj+6hatk)+lambda(3hati+7hatj+2hatk)` |
|
| 165. |
Find the equation of theline passing through the intersection `(-1,2,3)`and perpendicular to thelines `(x-1)/2=(y-2)/(-3)=(z-3)/4a n d(x-4)/5=(y-1)/2=zdot`and also through the point `(2,1,-2)dot` |
|
Answer» Intersection point of the lines `" "(x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-4)/(5)=(y-1)/(2)=z` is (-1, -1, -1) (on solving). Therefore, the equation of the line passing through the points(-1, -1, -1) and (2, 1, -2) is `(x+1)/(3)=(y+1)/(2)=(x+1)/(-1)`. |
|
| 166. |
Find the point where line which passes through point (1, 2, 3) and is parallel to line `vecr=hati-hatj+2hatk+lamda(hati-hatj+3hatk)` meets the `xy`-plane. |
|
Answer» Line is passing through the point (1, 2, 3) and parallel to the line `vecr=hati-hatj+2hatk+lamda(hati-2hatj+3hatk)` or parallel to the vector `hati-2hatj+3hatk`. Hence, equation of line is `" "(x-1)/(1)=(y-2)/(-2)=(z-3)/(3)` It meets `xy`-plane, where `z`=0 Then from the equation of line, we have `" "(x-1)/(1)=(y-2)/(-2)=(0-3)/(3)` `rArr" "x=0, y=4`. Hence, line meets`xy`-plane at (0, 4, 0). |
|
| 167. |
If the coordinates of the points `A,B,C,D be 91,2,3),(4,5,7),(-4,3,-6) and (2,9,2)` respectively then find the angle between AB and CD. |
|
Answer» `:.` Direction ratios of AB `(4-1),(5-2),(7-3)= (3,3,4)` Direction ratios of CD `(2+4),(9-3),(7-3)=(3,3,4)` Direction of CD `(2+4),(9-3),(2+6)=(6,6,8)` Let the directi on ratios of AB and CD are `(a_(1),b_(1),c_(1))` and `(a_(2),b_(2),c_(2))` respectively. Then, `(a_(1))/(a_(2)) = 3/6 = 1/2 , (b_(1))/(b_(2)) = 3/6 = 1/2 , (c_(1))/(c_(2)) = 4/8 = 1/2` `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))` `:. (a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2))` Therefore, line AB is parallel to line CD, so the angle between AB and CD is `0^(@)`. |
|
| 168. |
Find the equation of theline passing through the point `(-1,2,3)`and perpendicular to thelines `x/2=(y-1)/(-3)=(z+2)/(-2)a n d(x+3)/(-1)=(y+3)/2=(z-1)/3dot` |
|
Answer» The line through point (-1, 2, 3) is perpendicular to the lines `(x)/(2)=(y-1)/(-3)=(2+2)/(-2) and (x+3)/(-1)= (y+3)/(-1)=(y+3)/(2)=(z-1)/(3)`. Therefore, the line is along the vector `|{:(hati,,hatj,,hatk),(2,,-3,,-2),(-1,,2,,3):}|` or `-5hati-4hatj+hatk`. Hence, equation of the line is `(x+1)/(5)=(y-2)/(4)=(z-3)/(-1)`. |
|
| 169. |
Find the equation of the line passing through the points (1, 2, 3) and (-1, 0, 4). |
|
Answer» Since line is passing through the points `A(1, 2, 3) and B(-1, 0, 4)`, it is along the vecor `vec(AB)= -2hati-2hatj+hatk`. Hence, equation of line is `" "vecr=hati+2hatj+3hatk+lamda(-2hati-2hatj+hatk)` or `" "vecr=-hati+4hatk+lamda(-2hati-2hatj+hatk)` `rArr" "(x-1)/(-2)=(y-2)/(-2)=(z-3)/(1)` or `" "(x+1)/(-2)=(y-0)/(-2)=(z-4)/(1)` |
|
| 170. |
If the angle between the lines with direction ratios 2, -1, 1 and 1, k, 2 is `60^(@)`, then k =A. 1, 17B. `-1, 17`C. `1, -17`D. `-1, -17` |
| Answer» Correct Answer - C | |
| 171. |
If the line with direction ratios 2, -1, 2 is perpendicular to the line with direction ratios 1, k, -3, then k =A. -4B. 4C. -8D. 8 |
| Answer» Correct Answer - A | |
| 172. |
The cartesian equation of a line is `6x+1=3y-2 = 3-2x`. Find its direction ratios. |
| Answer» Correct Answer - `1,2,-3` | |
| 173. |
The Cartesian equation of aline is `(x-3)/2=(y+1)/(-2)=(z-3)/5`. Find the vector equationof the line. |
|
Answer» The given line is `(x-3)/(2)=(y+1)/(-2)=(z-3)/(5)` Note that it passes through (3, -13) and is parallel to the line whose direction ratios are 2, -2 and 5. Therefore, its vector equation is `vecr=3hati-hatj+3hatk+lamda(2hati-2hatj+5hatk)`, where `lamda` is a parameter. |
|
| 174. |
Find the vector equation of the line passingthrough the point `(2,-1,-1)`which is parallel to the line `6x-2=3y+1=2z-2.` |
|
Answer» The given line is `-6x-2=3y+1=2z-2` or `" "(x+(1//3))/(-1//6)=(y+(1//3))/(1//3)=(z-1)/(1//2)` The direction ratios are `-(1)/(6), (1)/(3) and (1)/(2) or -1, 2 and 3`. The required equation is `(x-2)/(-1)=(y+1)/(2)=(z+1)/(3)` |
|
| 175. |
Find the vector and cartesian equations of the plane which passes through the point `(5, 2, 4)`and perpendicular to the line with direction ratios `(2, 3, 1)`. |
|
Answer» Let `A(5,2,4)` is the point passing through given plane. In vector form, `vecA = 5hati+2hatj+4hatk` Let `vecr` is another vector on that plane, then, line joining `vecr` and `veca` will be, `vecr - vecA` We are given another line with direction cosines (2,3,1) that is perpendicular to the first line. So, dot product of these two lines will be `0`. `:. (vecr - (5hati+2hatj+4hatk)).(2hati+3hatj+hatk) = 0` `=>vecr.(2hati+3hatj+hatk) - 10-6-4 = 0` `=>vecr.(2hati+3hatj+hatk) = 20` We can take, `vecr = ahati+bhatj+chatk ` Then, our equation becomes, `2a+3b+c = 20`, which is required vector equation. Cartesian equation will be, `2x+3y+z = 20` |
|
| 176. |
The two values of k for which the lines with direction ratios `k, -6, -2` and `k -1, k, 4` are perpendicular to each other areA. 1, 8B. 1, -8C. `-1, 8`D. `-1, -8` |
| Answer» Correct Answer - C | |
| 177. |
Which of the triplet can not represent direction cosines of a line ?A. `((2)/(5), (3)/(5), (4)/(5))`B. `((1)/(sqrt(3)),(1)/(sqrt(3)),(1)/(sqrt(3)))`C. `((3)/(sqrt(50)),(3)/(sqrt(50)),(3)/(sqrt(50)))`D. `((4)/(sqrt(77)),(4)/(sqrt(77)),(4)/(sqrt(77)))` |
| Answer» Correct Answer - A | |
| 178. |
Given `P = (3,-6,10)and PQ = sqrt(17)` . If direction cosines of line PQ are ` (-2)/(sqrt(17)),3/(sqrt(17)),(-2)/(sqrt(17))` , then point Q can beA. `(5,12,-19)`B. `(1,-3,8)`C. `(-9,5,12)`D. `(-2,3,-2)` |
| Answer» Correct Answer - b | |
| 179. |
If OP = 21 and direction cosines of `bar(OP)` are `(2)/(7),(6)/(7), (-3)/(7)`, then the co-ordinates of P areA. (6, 18, -9)B. (-6, -12, 9)C. (-6, 18, -9)D. (6, 18, 9) |
| Answer» Correct Answer - A | |
| 180. |
If the direction cosines of `bar(AB)` are `(-2)/(sqrt(17)), (3)/(sqrt(17)), (-2)/(sqrt(17))` such that `A-=(3, -6, 10)` and `l(AB)=sqrt(17)`, then the co-ordinates of point B areA. (1, 3, -8) or (5, -9, 12)B. (1, -3, 8) or (5, -9, 12)C. (1, -3, 8) or (5, 9, -12)D. (1, 3, -8) or (5, 9, -12) |
| Answer» Correct Answer - B | |
| 181. |
If `P-=(3, 4, -12)`, then the direction cosines of `bar(OP)` areA. 6, 6, -3B. `(1)/(5), (6)/(5), (7)/(5)`C. `(-2)/(7), (3)/(7), (6)/(7)`D. `(3)/(13), (4)/(13), (-12)/(13)` |
| Answer» Correct Answer - D | |
| 182. |
If the direction cosines of a line are `1/c,1/c,1/c` rthen (A) `c.0` (B) `0ltclt1` (C) `c=+-sqrt(3)` (D) `cgt2`A. `0 lt c lt 1`B. `c gt 2`C. ` c gt2`D. ` c = pm sqrt(3)` |
| Answer» Correct Answer - d | |
| 183. |
find the equation of a plane passing through the points `(2,-1,0)` and `(3,-4,5)` and parallel to the line `2x = 3y=4z`A. (A) 29(x-2)+27(y+1)-22zB. (B) 29(x-2)-27(y+1)-22zC. (C) 29(x-2)+27(y+1)+22zD. (D) 29(x+2)+27(y-1)-2x=3 |
|
Answer» Correct Answer - (B)=> 29(x-2)-27(y+1)-22z=0 |
|
| 184. |
Find the ratio in which the join of the points `P(2,-1,3) and Q( 4,3,1)` is divided by the point `((20)/(7),(5)/(7),(15)/(7))` |
|
Answer» Let he required ratio be `lambda:1` . Then , the coordinates of R are `((4lambda+2)/(lambda+1),(3lamda-1)/(lambda+1),(lambda+3)/(lambda+1))` But, the coordinates or R are `((20)/(7),(5)/(7),(15)/(7))` `:. (4lambda+2)/(lambda+1)=(20)/(7) or lambda=(3)/(4)` So, the required ratio is `(3)/(4) :1`, i.e., 3:4 . |
|
| 185. |
Find the ratio in which the join the `A(2,1,5)a n dB(3,4,3)`is divided by the plane `2x+2y-2z=1.`Also, find the coordinates of the point ofdivision. |
|
Answer» Suppose the given plnae intesects AB at a point C and let the required ratior be `lambda:1` Then, the coordinates of C are `((3lambda+2)/(lambda+1),(4lambda+1)/(lambda+1),(3lambda+5)/(lambda+1))` Since C lies on the plane `2x+2y-2z=1`, this point must satisfy tyhe equation of the plane. `:. 2((3lambda+2)/(lambda+1))+2((4lambda+1)/(lambda+1))-2((3lambda+5)/(lambda+1))=1 or lambda=(5)/(7)` So, the required ratio is `(5)/(7) :1 , i.e., 5:7` Putting `lambda=(5)/(j7)` in (i), then required point of division is `C((29)/(12),(9)/(4),(25)/(6))` |
|
| 186. |
Find the angles between the line whose direction ratios are `4, -3, 5 and 3, 4, 5`.A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - C | |
| 187. |
Find the ratio in which the line segment, joining the points, `P(2,3,4) and Q(-3,5,-4)` is divided by the yz-plane. Also, find the points of intersection. |
|
Answer» Let PQ be the divided by the zy-plane at a point R in the ratio `lambda:1`. Then, the coordinate of R are `((-3lambda+2)/(lambda+1),(5lambda+3)/(lambda+1),(-4lambda+4)/(lambda+1))` Since R lies on the yz-plane, the x-coordinate or R is thereofore, 0. `:. (-3lambda+2)/(lambda+1)=0, or lambda=(2)/(3)` So, the required ratio is `(2)/(3) : 1,i.e., 2:3` Putting `lambda=(2)/(3)` in (i), the point of intersection of the line segment PQ any the yz- plane is `(0,(19)/(5),(4)/(5))` |
|
| 188. |
If a straight line in space is equally inclined to the co-rodinate axes, then the cosine of its angle of inclination to any one of the axes isA. `(1)/(2)`B. `(1)/(3)`C. `(1)/(sqrt(2))`D. `(1)/(sqrt(3))` |
| Answer» Correct Answer - D | |
| 189. |
If a point lies on yz-plane then what is its x-coordinate? |
| Answer» Correct Answer - `x=0` | |
| 190. |
A line lies in YZ- plane and makes angle of `30^(@)` with Y-axis, then its inclination to Z-axis isA. `30^(@) or 60^(@)`B. `30^(@) or 150^(@)`C. `60^(@) or 90^(@)`D. `60^(@) or 120^(@)` |
| Answer» Correct Answer - D | |
| 191. |
The co-ordinate of the point in which the line joining the points (3, 5, -7) and (-2, 1, 8) is inscribed by YZ-plane areA. `(0, (13)/(5), -2)`B. `(0, (13)/(5), 2)`C. `(0, (-13)/(5), 2)`D. `(0, (-13)/(5), -2)` |
| Answer» Correct Answer - B | |
| 192. |
The ratio in which the line joining points (2, 4, 5) and (3, 5, -4) divide YZ-plane isA. `-2:3`B. `2:3`C. `-3:2`D. `3:2` |
| Answer» Correct Answer - A | |
| 193. |
If the angle between the lines with direction ratios a, 3, 5 and 2, -1, 2 is `45^(@)`, then a =A. 2, 26B. 4, 52C. 2, 104D. 8, 26 |
| Answer» Correct Answer - B | |
| 194. |
If the cosine of the angle between the lines with direction ratios 1, -1, 2 and 0, 1, k is `(sqrt(3))/(2)`, then k =A. `-1, -7`B. `1, -7`C. `-1, 7`D. 1, 7 |
| Answer» Correct Answer - A | |
| 195. |
The acute angle between the lines joining points (2, 1, 3) and (1, -1, 2) and the line having direction ratios 2, 1, -1 isA. `30^(@) `B. `45^(@)`C. `60^(@)`D. `120^(@)` |
| Answer» Correct Answer - C | |
| 196. |
If `A-=(3,4-2),B-=(1,-1,2),C-=(0,3,2)` and `D-=(3,5,6)`, then the angle between `bar(AB)` and `bar(CD)` isA. `45^(@)`B. `90^(@)`C. `30^(@) `D. `60^(@)` |
| Answer» Correct Answer - B | |
| 197. |
An angle between the lines whose direction number are 1, -2, 1 and -6, -1, 4 isA. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
| Answer» Correct Answer - D | |
| 198. |
Show that the lines `(x+3)/(-3)=(y-1)/1=(z-5)/5`and `(x+1)/(-1)=(y-2)/2=(z-5)/5`are coplanar. |
| Answer» Correct Answer - `x-2y+z=0` | |
| 199. |
then image of the point `(-1,3,4)` in the plane `x-2y=0`A. `(-(17)/(3),(19)/(3),4)`B. (15,11,4)C. `(-(17)/(3),(19)/(3),1)`D. `((9)/(5),-(13)/(5),4)` |
|
Answer» Correct Answer - d Let `P(alpha,beta,gamma)` be the image of the point Q(-1,3,4) Midpoint of PQ lies on x-2y=0. Then, `(alpha-1)/(2)-2((beta-3)/(2))=0` or `alpha-1-2beta-6=0oralpha-2beta=7" "(i)` Also PQ is perpendicular to the plane. Then, `(alpha+1)/(1)=(beta-3)/(-2)=(gamma-4)/(0)" "(ii)` Solving (i) and (ii) we get `alpha=(9)/(5)beta,-(13)/(5),gamma=4` Therefore, image is `((9)/(5),(13)/(5),4)` Alternate method: For image, `(alpha-(-1))/(1)=(beta-3)/(-2)=(gamma-4)/(0)=(-2(-1-2(3)))/((1)^(2)+(-2)^(2))` or `alpha=(9)/(5),beta=-(13)/(5),gamma=4` |
|
| 200. |
Show that the lines `(x-3)/2=(y+1)/(-3)=(z+2)/1 dn (x-7)/(-3)=y/1=(z+7)/2` are coplanar. Also find the equation of the plane containing them. |
| Answer» Correct Answer - `x+y+z=0` | |