Q: Let L_1 be the line of intersection of the planes given by the equation- 2 x+3 y+z=4 and x+2 y+z=5 . Let L_2 be the line passing through the point P(2,-1,3) and parallel to L_1. Let M denote the plane

Let L_1: r = a +t b b = | array ccc i & j & k \\ 2 & 3 & 1 \\ 1 & 2 & 1 array |= i (1)- j (1)+ k (1) Dr's of L_1: aligned & L_1: x+1 1 = y-0 -1 = z-6 1 \\ & \& L_2: x-2 1 = y+1 -1 = z-3 1 = \\ & M: 2 x+y-2 z-6=0 aligned Let point on L _2( +2,- -1, +3) aligned & 2( +2)- -1-2 -6-6=0 \\ & 2 +4-3 -13=0 \\ & =-9 \\ & Q(-7,8,-6) aligned Line PR: x-2 2 = y+1 1 = z-3 -2 = R (2 +2, -1,-2 +3) Put in plane aligned & 2(2 +2)+ -1-2(-2 +3)-6=0 \\ & 4 +4+ -1+4 -6-6=0 \\ & 9 -9=0 =1 \\ & R (4,0,1) \\ & P (2,-1,

Q: A straight line drawn from the point P(1,3,2), parallel to the line x-2 1 = y-4 2 = z-6 1 , intersects the plane L_1: x-y+3 z=6 at the point Q. Another straight line which passes through Q and is perp

aligned & line : x-1 1 = y-3 2 = z-2 1 \\ & (x, y, z)=( +1,2 +3, +2) aligned Put in L_1: x-y+3 z=6 aligned & ( +1)-(2 +3)+3( +2)=6 \\ & 2 =2 =1 \\ & Q=(2,5,3) aligned line : x-2 1 = y-5 -1 = z-3 3 (x, y, z)=(t+2,5-t, 3 t+3) Put in L_2: 2 x-y+z=-4 aligned & 2(t+4)-(5-t)+(3 t+3)=-4 \\ & 6 t=-6 t=-1 \\ & R=(1,6,0) aligned aligned & Perimeter = 6 + 13 + 11 \\ & Centroid = ( 4 3 , 14 3 , 5 3 ) aligned

Q: Let R ^3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7). Let dist (X, Y) denote the distance between two points X and Y in R ^3. Let gathered S= \ X R ^3:( dist (X, P))^2-

aligned & S = \ X :( XP )^2-( XQ )^2=50 \ \\& T = \ Y :( YQ )^2-( YP )^2=50 \ aligned for finding S X(x, y, z) and for T Y(x, y, z) aligned & ((x-1)^2+(y-1)^2+( z -1)^2 )- (( x -4)^2+( y -2)^2+( z -7)^2 )=50 \\& S =\ ( x , y , z ): 6 x +8 z =105\ \\& T =\ ( x , y , z ): 6 x +8 z =5\ aligned Since S and T both are plane ; (1) There exist a triangle in plane S whose area =1 (always) (2) L ~\&~ M lies on plane T , hence line segment joining L \& M will lie on plane T . (3) Distance between S \& ~T

Q: Let R be such that the lines L _1: x+11 1 = y +21 2 = z +29 3 and L_2: x+16 3 = y+11 2 = z+4 intersect. Let R _1 be the point of intersection of L _1 and L _2. Let O =(0,0,0), and n denote a unit norm

aligned & L _1: x +11 1 = y +21 2 = z +29 3 = a \\ & L _2: x +16 3 = y +11 2 = z +4 = b aligned aligned & x=a-11=3 b-16 a-3 b=-5...(1) \\ & y=2 a-21=2 b-11 2 a-2 b=10...(2) \\ & z=3 a-29=b r-4 3 a-b =25...(3) aligned from (1) & (2) a =10, ~b =5 Now from (3) 3(10)-5 =25 =1 aligned & R _1 (-1,-1,1) \\ & OR _1=- i - j + k \\ & n = | array lrr i & j & k \\ 1 & 2 & 3 \\ 3 & 2 & 1 array |=-4 i -(-8) j -4 k \\ & n =-4 i +8 j +4 k =-4( i -2 j + k ) \\ & n = 4( i -2 j + k ) 4 6 = ( i -2 j + k ) 6 \\ & OR

Q: Let ℓ 1 and ℓ 2 be the lines r → 1 = λ i ^ + j ^ + k ^ and r → 2 = j ^ - k ^ + μ i ^ + k ^ , respectively, Let X be the set of all the planes H that conta

Given, H 0 will be the plane containing the line ℓ 1 and parallel to ℓ 2 . So, the normal vector of plane parallel to ℓ 1 and ℓ 2 is given by, i ^ j ^ k ^ 1 1 1 1 0 1 = j ^ 1 - j ^ 1 - 1 + k ^ - 1 = i ^ - k ^ Hence, the equation of plane H 0 will be, H 0 : x - z = C which passes through origin, So, C = 0 ∴ H 0 : x - z = 0 Now solving, P d H 0 = 1 distance of point 0 , 1 , - 1

Q: Let S be the reflection of a point Q with respect to the plane given by r → = - t + p i ^ + t j ^ + 1 + p k ^ where t , p are real parameters and i ^ , j ^ , k ^ ar

Given, Equation of plane is r → = - t + p i ^ + t j ^ + 1 + p k ^ On rearranging we get, r → = k ^ + t - i ^ + j ^ + p - i ^ + k ^ So, equation of plane in standard form is given by r → - k ^ - i ^ + j ^ - i ^ + k ^ = 0 ⇒ x y z - 1 - 1 1 0 - 1 0 1 = 0 ⇒ x + y + z = 1 ⋯ 1 Now given coordinate of Q = 10 , 15 , 20 And coordinates of S = α , β , γ Now using the image formula of point and plane we

Question 1

Let P be the plane such that it contains the straight line x-1 2 = y-3 3 = z+2 1 and is perpendicular to the plane x + 2y + 3z = 4. Let P_1 be the plane which passes through the point (4, 2, 2) and is

Accepted Answer

Let the normal vector to the plane P be n . Since the plane P contains the line x-1 2 = y-3 3 = z+2 1 , its normal vector n is perpendicular to the line's direction vector b = 2 i + 3 j + k . Since P is perpendicular to the plane x + 2y + 3z = 4, n is perpendicular to its normal vector n _1 = i + 2 j + 3 k . n = (2 i + 3 j + k ) ( i + 2 j + 3 k ) = i (9-2) - j (6-1) + k (4-3) = 7 i - 5 j + k Plane P passes through the point (1, 3, -2) which lies on the given line. Equation of plane P is 7(x-1) -

Question 2

Let L be the straight line joining the points P(1, 2, -1) and Q(2, 3, 1). Let S be the foot of the perpendicular drawn from the point R(4, -1, 5) to the line L. Another line passing through R intersec

Accepted Answer

The direction ratios of line L passing through P(1, 2, -1) and Q(2, 3, 1) are (2-1, 3-2, 1-(-1)) = (1, 1, 2). The equation of line L is x-1 1 = y-2 1 = z+1 2 = . Any point on L can be written as S( +1, +2, 2 -1). Since S is the foot of the perpendicular from R(4, -1, 5) to L, the direction ratios of RS are ( -3, +3, 2 -6). As RS L, the dot product of their direction ratios is zero: 1( -3) + 1( +3) + 2(2 -6) = 0 6 - 12 = 0 = 2 Thus, the coordinates of S are (3, 4, 3). The length of PS is (3-1)^2

Question 3

Let L_1 be the line of intersection of the planes given by the equation- 2 x+3 y+z=4 and x+2 y+z=5 . Let L_2 be the line passing through the point P(2,-1,3) and parallel to L_1. Let M denote the plane

Accepted Answer

Let L_1: r = a +t b b = | array ccc i & j & k \ 2 & 3 & 1 \ 1 & 2 & 1 array |= i (1)- j (1)+ k (1) Dr's of L_1: aligned & L_1: x+1 1 = y-0 -1 = z-6 1 \ & \& L_2: x-2 1 = y+1 -1 = z-3 1 = \ & M: 2 x+y-2 z-6=0 aligned Let point on L _2( +2,- -1, +3) aligned & 2( +2)- -1-2 -6-6=0 \ & 2 +4-3 -13=0 \ & =-9 \ & Q(-7,8,-6) aligned Line PR: x-2 2 = y+1 1 = z-3 -2 = R (2 +2, -1,-2 +3) Put in plane aligned & 2(2 +2)+ -1-2(-2 +3)-6=0 \ & 4 +4+ -1+4 -6-6=0 \ & 9 -9=0 =1 \ & R (4,0,1) \ & P (2,-1,

Question 4

A straight line drawn from the point P(1,3,2), parallel to the line x-2 1 = y-4 2 = z-6 1 , intersects the plane L_1: x-y+3 z=6 at the point Q. Another straight line which passes through Q and is perp

Accepted Answer

aligned & line : x-1 1 = y-3 2 = z-2 1 \ & (x, y, z)=( +1,2 +3, +2) aligned Put in L_1: x-y+3 z=6 aligned & ( +1)-(2 +3)+3( +2)=6 \ & 2 =2 =1 \ & Q=(2,5,3) aligned line : x-2 1 = y-5 -1 = z-3 3 (x, y, z)=(t+2,5-t, 3 t+3) Put in L_2: 2 x-y+z=-4 aligned & 2(t+4)-(5-t)+(3 t+3)=-4 \ & 6 t=-6 t=-1 \ & R=(1,6,0) aligned aligned & Perimeter = 6 + 13 + 11 \ & Centroid = ( 4 3 , 14 3 , 5 3 ) aligned

Question 5

Let R ^3 denote the three-dimensional space. Take two points P=(1,2,3) and Q=(4,2,7). Let dist (X, Y) denote the distance between two points X and Y in R ^3. Let gathered S= \ X R ^3:( dist (X, P))^2-

Accepted Answer

aligned & S = \ X :( XP )^2-( XQ )^2=50 \ \& T = \ Y :( YQ )^2-( YP )^2=50 \ aligned for finding S X(x, y, z) and for T Y(x, y, z) aligned & ((x-1)^2+(y-1)^2+( z -1)^2 )- (( x -4)^2+( y -2)^2+( z -7)^2 )=50 \& S =\ ( x , y , z ): 6 x +8 z =105\ \& T =\ ( x , y , z ): 6 x +8 z =5\ aligned Since S and T both are plane ; (1) There exist a triangle in plane S whose area =1 (always) (2) L ~\&~ M lies on plane T , hence line segment joining L \& M will lie on plane T . (3) Distance between S \& ~T

Question 6

Let R be such that the lines L _1: x+11 1 = y +21 2 = z +29 3 and L_2: x+16 3 = y+11 2 = z+4 intersect. Let R _1 be the point of intersection of L _1 and L _2. Let O =(0,0,0), and n denote a unit norm

Accepted Answer

aligned & L _1: x +11 1 = y +21 2 = z +29 3 = a \ & L _2: x +16 3 = y +11 2 = z +4 = b aligned aligned & x=a-11=3 b-16 a-3 b=-5...(1) \ & y=2 a-21=2 b-11 2 a-2 b=10...(2) \ & z=3 a-29=b r-4 3 a-b =25...(3) aligned from (1) & (2) a =10, ~b =5 Now from (3) 3(10)-5 =25 =1 aligned & R _1 (-1,-1,1) \ & OR _1=- i - j + k \ & n = | array lrr i & j & k \ 1 & 2 & 3 \ 3 & 2 & 1 array |=-4 i -(-8) j -4 k \ & n =-4 i +8 j +4 k =-4( i -2 j + k ) \ & n = 4( i -2 j + k ) 4 6 = ( i -2 j + k ) 6 \ & OR

Question 7

Let Q  be the cube with the set of vertices x 1 , &#160; x 2 , &#160; x 3 &#8712; &#8477; 3 &#160; : &#160; x 1 , &#160; x 2 , &#160; x 3 &#8712; 0 , &#160; 1 . Let F be the set of all twelve lin

Accepted Answer

Plotting the diagram of cube we get, Now equation of O D line will be, r &#8594; = 0 &#8594; + &#955; i ^ + j ^ And equation of diagonal B E  will be, r &#8594; 1 = j ^ + &#956; i ^ - j ^ + k ^ Now finding the shortest distance between line  O D &#160; &#38; &#160; B E  we get, S . D = j ^ - 0 &#183; i ^ + j ^ &#215; i ^ - j ^ + k ^ 1 2 + 1 2 + 0 2 &#183; 1 2 + 1 2 + 1 2 &#8658; S . D = j ^ &#183; i ^ - j ^ - 2 k ^ 6 = 1 6 Now in other case shortest distance will be zero

Question 8

Let &#8467; 1  and &#8467; 2  be the lines r &#8594; 1 = &#955; i ^ + j ^ + k ^ and r &#8594; 2 = j ^ - k ^ + &#956; i ^ + k ^ , respectively, Let X be the set of all the planes H that conta

Accepted Answer

Given, H 0  will be the plane containing the line &#8467; 1 and parallel to &#8467; 2 . So, the normal vector of plane parallel to  &#8467; 1 and &#8467; 2 is given by, i ^ j ^ k ^ 1 1 1 1 0 1 = j ^ 1 - j ^ 1 - 1 + k ^ - 1 = i ^ - k ^ Hence, the equation of plane  H 0 &#160;  will be, H 0 &#160; : &#160; x - z = C  which passes through origin, So, C = 0 &#8756; &#160; H 0 &#160; : &#160; x - z = 0 Now solving, P d H 0 = 1  distance of point 0 , &#160; 1 , &#160; - 1

Question 9

Let  P 1  and  P 2  be two planes given by P 1 : 10 x + 15 y + 12 z - 60 = 0 , P 2 : - 2 x + 5 y + 4 z - 20 = 0 . Which of the following straight lines can be an edge of some tetra

Accepted Answer

Given, P 1  and  P 2  be two planes given by P 1 : 10 x + 15 y + 12 z - 60 = 0 , P 2 : - 2 x + 5 y + 4 z - 20 = 0 . Now finding line of intersection of both the planes, Let  z = &#955;  then we have equations, 10 x + 15 y = 60 - 12 &#955; &#160; . . . 1 - 2 x + 5 y = 20 - 4 &#955; &#160; . . . . . 2 Now on solving equation  1 &#160; &#38; &#160; 2  we get, x 0 = y - 4 - 4 = z 5   Replacing&#160; &#955; &#160; with&#160; z Now any skew line with the line of

Question 10

Let S  be the reflection of a point  Q with respect to the plane given by r &#8594; = - t + p i ^ + t j ^ + 1 + p k ^ where  t , p  are real parameters and i ^ , j ^ , k ^  ar

Accepted Answer

Given, Equation of plane is r &#8594; = - t + p i ^ + t j ^ + 1 + p k ^ On rearranging we get, r &#8594; = k ^ + t - i ^ + j ^ + p - i ^ + k ^ So, equation of plane in standard form is given by r &#8594; - k ^ - i ^ + j ^ - i ^ + k ^ = 0 &#8658; x y z - 1 - 1 1 0 - 1 0 1 = 0 &#8658; x + y + z = 1 &#160; &#160; &#160; &#160; &#160; &#160; &#8943; 1 Now given coordinate of  Q = 10 , 15 , 20 And coordinates of  S = &#945; , &#946; , &#947; Now using the image formula of point and plane we

Question 11

Let &#945; , &#160; &#946; , &#160; &#947; , &#160; &#948; be real numbers such that &#945; 2 + &#946; 2 + &#947; 2 &#8800; 0 and &#945; + &#947; = 1 . Suppose the point 3 , 2 , - 1 is the mirror imag

Accepted Answer

From the diagram, Mid point of A & A ' lies on the plane, i.e., B 2 , 1 , - 1 2 α + β - γ = δ … 1 α + γ = 1 … . 2 Direction Ratio of AA ' 2 , 2 , 0 will be proportional to normal to the plane, α 2 = β 2 = γ 0 = λ α = 2 λ , β = 2 λ , γ = 0 ∴ 2 λ + 0 = 1 ⇒ λ = 1 2 ∴ α = 1 , β = 1 , γ = 0 , δ = 3 Now, (A) α + β + γ = 1 + 1 + 0 = 2 ≠ δ (B) α + β = 1 + 1 = 2 (C) δ + β = 3 + 1 = 4 (D) δ - γ = 3

Question 12

Let L 1 and L 2 be the following straight lines. L 1 : x &#8722; 1 1 = y &#8722; 1 = z &#8722; 1 3 and L 2 : x &#8722; 1 &#8722; 3 = y &#8722; 1 = z &#8722; 1 1 Suppose the straight line L : x &#8722;

Accepted Answer

Given the lines  L 1 : x &#8722; 1 1 = y &#8722; 1 = z &#8722; 1 3  and L 2 : x &#8722; 1 - 3 = y &#8722; 1 = z &#8722; 1 1 . From the given lines, we can say that both the lines pass through the point  1 , &#160; 0 , &#160; 1 . Now, let  a &#8594;  be the direction vector of line  L 1  i.e.,  a &#8594; = i - j + 3 k  and let  b &#8594;  be the direction vector of line  L 2  i.e.,  b &#8594; = - 3 i - j + k . We also know

Question 13

Three lines L 1 : r &#8594; = &#955; i ^ , &#160; &#955; &#8712; R , L 2 : r &#8594; = k ^ + &#956; j ^ , &#160; &#956; &#8712; R and L 3 : &#160; r &#8594; = i ^ + j ^ + &#957; k ^ , &#160; &#957; &#

Accepted Answer

As given L 1 &#8658; r &#8594; = &#955; i , &#160; &#955; &#8712; R L 2 &#8658; r &#8594; = k + &#956; j , &#160; &#956; &#8712; R and L 3 &#8658; r &#8594; = i + j + &#957; k , &#160; &#957; &#8712; R Let point P &#955; , &#160; 0 , &#160; 0 , &#160; Q 0 , &#160; &#956; , &#160; 1 and R 1 , &#160; 1 , &#160; &#957; and P Q &#8594; = - &#955; i ^ + &#956; j ^ + k ^ , &#160; P R &#8594; = 1 - &#955; i ^ + j ^ + &#957; k ^ as P , &#160; Q and R are co-linear &#8658; P Q &#8594; is parallel to P R

Question 14

Let L 1 and L 2 denotes the lines r &#8594; = i ^ + &#955; - i ^ + 2 j ^ + 2 k ^ , &#8201; &#955; &#8712; R and r &#8594; = &#956; 2 i ^ - j ^ + 2 k ^ , &#956; &#8712; R respectively. If L 3 is a line

Accepted Answer

L 1 : r &#8594; = i ^ + &#955; - i ^ + 2 j ^ + 2 k ^ , &#160; &#160; &#955; &#8712; R Let a general point on line  L 1  be  A 1 - &#955; , &#160; 2 &#955; , &#160; 2 &#955; Also  L 2 : r &#8594; = &#956; 2 i ^ - j ^ + 2 k ^   let a general point on line L 2  be  B 2 &#956; , &#160; - &#956; , &#160; 2 &#956; Then   A B &#8594; = 2 &#956; + &#955; - 1 i ^ + - &#956; - 2 &#955; &#160; j ^ + 2 &#956; - 2 &#955; k ^ Now, A B &#8594; is perpendicular to bo

Question 15

Three lines are given by r &#8594; = &#955; i ^ , &#160; &#955; &#8712; R r &#8594; = &#956; i ^ + j ^ , &#956; &#8712; R and r &#8594; = &#957; i ^ + j ^ + k ^ , &#957; &#8201; &#8712; R . Let the li

Accepted Answer

Finding point &#8216; A &#8217; Line r &#8594; = &#955; i ^ &#160; and plane x + y + z = 1 General point of line &#955; , 0 , &#160; 0 , satisfy the plane &#955; + 0 + 0 = 1 &#955; = 1 Hence, A ( 1 , &#160; 0 , &#160; 0 ) ...(i) For point &#8216; B &#8217; General point of line r &#8594; = &#956; i ^ + j ^ , &#160; &#160; B &#956; , &#956; , 0 Satisfy the plane and &#956; + &#956; + 0 = 1 &#956; = 1 2 Point B 1 2 , 1 2 , 0 ...(ii) Similarly for point &#8216; C &#8217; General point of line r &#8

Three Dimensional Geometry — Previous Year Questions & Solutions