Q: A line ℓ passing through the origin is perpendicular to the lines ℓ₁: (3 + t) i + (-1 + 2t) j + (4 + 2t) k, -∞ < t < ∞ ℓ₂: (3 + 2s) i + (3 + 2s) j + (2 + s) k, -∞ < s < ∞ Then the coordinates(s) of the point(s) on ℓ₂ at a distance of √17 from the point of intersection of ℓ and ℓ₁ is(are)

(-1, -1, 0). The line ℓ is perpendicular to ℓ₁, and its intersection point with ℓ₁ is determined. Using the distance condition of √17 from this point, the coordinates on ℓ₂ are calculated as (-1, -1, 0).

Q: In three-dimensional space, consider the planes P1: y = 0 and P2: x + z = 1. Let P3 represent a plane distinct from both P1 and P2, passing through the line formed by the intersection of P1 and P2. If the point (0, 1, 0) lies at a distance of 1 unit from P3, and the point (α, β, γ) lies at

2α - β + 2γ + 4 = 0. The plane P₃ passes through the intersection of P₁ and P₂ and satisfies the given distance conditions. Substituting the point (α, β, γ) into the plane equations confirms that only 2α - β + 2γ + 4 = 0 is valid.

Q: If the point (3, 1, 7) is reflected across the plane x − y + z = 3, and the resulting image point is P, what is the equation of the plane that passes through P and includes the line given by x/1 = y/2 = z/1?

x − 4y + 7z = 0. The reflection of the point (3, 1, 7) across the plane x − y + z = 3 gives the image point P. The plane passing through P and containing the given line is determined to have the equation x − 4y + 7z = 0.

Q: What is the equation of a plane that passes through the point (1, 1, 1) and is orthogonal to the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7?

4x + 2y + 15z = 31. To find the equation of the plane, we first determine the direction ratios of the normal to the required plane, which are found to be proportional to 4, 2, and 15. Using the point (1, 1, 1) that the plane passes through, we can then derive the equation of the plane as 4x + 2y + 15z = 31.

Q: Consider a cube Q defined by the vertices {(x1, x2, x3) ∈ R³: x1, x2, x3 ∈ {0, 1}}. Let F represent the collection of all twelve lines formed by the diagonals of the six faces of the cube. Additionally, let S denote the group of four lines that pass through the main diagonals of the cube,

1/√6. The maximum distance between a face diagonal and a main diagonal of the cube is determined geometrically. Using the properties of the cube and the shortest distance formula for skew lines, the maximum value is found to be 1/√6.

Q: In three-dimensional space R³, consider the points P = (1, 2, 3) and Q = (4, 2, 7). The distance between two points X and Y is denoted by dist(X, Y). Define the set S as {X ∈ R³: (dist(X, P))² − (dist(X, Q))² = 50} and the set T as {Y ∈ R³: (dist(Y, Q))² − (dist(Y, P))² = 50}. Which of the

A triangle with an area of 1 can be formed using only points from S as its vertices.. A triangle with an area of 1 can be formed using only points from S as its vertices because the set S is defined by the equation (dist(X, P))² − (dist(X, Q))² = 50, which represents a hyperboloid, and it is possible to find three points on this hyperboloid that form a triangle with an area of 1.

Q: Two lines in 3D space are given: M1 passes through the origin with direction ratios (1,1,1), i.e. x = y = z, and M2 passes through the origin with direction ratios (1,2,3), i.e. x/1 = y/2 = z/3. A third line M3 passes through the point A = (1,1,1) and forms a triangle together with M1 and

(2, 4, 6). The cross product OA x OB = (1,1,1) x (s,2s,3s) = (s,-2s,s) has magnitude s*sqrt(6). Setting (1/2)*s*sqrt(6) = sqrt(6) gives s = 2, so B = (2,4,6).

Q: The line x = y = z (parametric: x=t, y=t, z=t) intersects the pair of planes x*sinA + y*sinB + z*sinC = 18 and x*sin2A + y*sin2B + z*sin2C = 9, where A, B, C are angles of a triangle. Find the value of 80*(sinA/2 * sinB/2 * sinC/2).

5. Substituting x=y=z=t: t*(sinA+sinB+sinC)=18...(1) and t*(sin2A+sin2B+sin2C)=9...(2). For triangle angles: sinA+sinB+sinC = 4*cos(A/2)*cos(B/2)*cos(C/2) and sin2A+sin2B+sin2C = 4*sinA*sinB*sinC. Divide (2) by (1): (sin2A+sin2B+sin2C)/(sinA+sinB+sinC) = 9/18 = 1/2. So 4*sinA*sinB*sinC / (4*cos(A/2)*cos(B/2)*cos(C/2)) = 1/2. Now sinA*sinB*sinC = 8*sin(A/2)*cos(A/2)*sin(B/2)*cos(B/2)*sin(C/2)*cos(C/2). So [8*sin(A/2)*cos(A/2)*sin(B/2)*cos(B/2)*sin(C/2)*cos(C/2)] / [cos(A/2)*cos(B/2)*cos(C/2)] = 1/2. This gives 8*sin(A/2)*sin(B/2)*sin(C/2) = 1/2. So sin(A/2)*sin(B/2)*sin(C/2) = 1/16. Therefore 8

Question 1

Perpendiculars are drawn from points on the line (x+2)/(2) = (y+1)/(-1) = (z)/(3) to the plane x + y + z = 3. The feet of perpendiculars lie on the line -

Accepted Answer

(x)/(2) = (y-1)/(-7) = (z-2)/(5). The feet of perpendiculars from the given line to the plane lie on a line whose direction ratios are derived from the normal vector of the plane and the line's direction. This matches the equation (x)/(2) = (y-1)/(-7) = (z-2)/(5).

Question 2

A line ℓ passing through the origin is perpendicular to the lines ℓ₁: (3 + t) i + (-1 + 2t) j + (4 + 2t) k, -∞ < t < ∞ ℓ₂: (3 + 2s) i + (3 + 2s) j + (2 + s) k, -∞ < s < ∞ Then the coordinates(s) of the point(s) on ℓ₂ at a distance of √17 from the point of intersection of ℓ and ℓ₁ is(are)

Accepted Answer

(-1, -1, 0). The line ℓ is perpendicular to ℓ₁, and its intersection point with ℓ₁ is determined. Using the distance condition of √17 from this point, the coordinates on ℓ₂ are calculated as (-1, -1, 0).

Question 3

In three-dimensional space, consider the planes P1: y = 0 and P2: x + z = 1. Let P3 represent a plane distinct from both P1 and P2, passing through the line formed by the intersection of P1 and P2. If the point (0, 1, 0) lies at a distance of 1 unit from P3, and the point (α, β, γ) lies at

Accepted Answer

2α - β + 2γ + 4 = 0. The plane P₃ passes through the intersection of P₁ and P₂ and satisfies the given distance conditions. Substituting the point (α, β, γ) into the plane equations confirms that only 2α - β + 2γ + 4 = 0 is valid.

Question 4

If the point (3, 1, 7) is reflected across the plane x − y + z = 3, and the resulting image point is P, what is the equation of the plane that passes through P and includes the line given by x/1 = y/2 = z/1?

Accepted Answer

x − 4y + 7z = 0. The reflection of the point (3, 1, 7) across the plane x − y + z = 3 gives the image point P. The plane passing through P and containing the given line is determined to have the equation x − 4y + 7z = 0.

Question 5

What is the equation of a plane that passes through the point (1, 1, 1) and is orthogonal to the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7?

Accepted Answer

4x + 2y + 15z = 31. To find the equation of the plane, we first determine the direction ratios of the normal to the required plane, which are found to be proportional to 4, 2, and 15. Using the point (1, 1, 1) that the plane passes through, we can then derive the equation of the plane as 4x + 2y + 15z = 31.

Question 6

Consider a cube Q defined by the vertices {(x1, x2, x3) ∈ R³: x1, x2, x3 ∈ {0, 1}}. Let F represent the collection of all twelve lines formed by the diagonals of the six faces of the cube. Additionally, let S denote the group of four lines that pass through the main diagonals of the cube,

Accepted Answer

1/√6. The maximum distance between a face diagonal and a main diagonal of the cube is determined geometrically. Using the properties of the cube and the shortest distance formula for skew lines, the maximum value is found to be 1/√6.

Question 7

In three-dimensional space R³, consider the points P = (1, 2, 3) and Q = (4, 2, 7). The distance between two points X and Y is denoted by dist(X, Y). Define the set S as {X ∈ R³: (dist(X, P))² − (dist(X, Q))² = 50} and the set T as {Y ∈ R³: (dist(Y, Q))² − (dist(Y, P))² = 50}. Which of the

Accepted Answer

A triangle with an area of 1 can be formed using only points from S as its vertices.. A triangle with an area of 1 can be formed using only points from S as its vertices because the set S is defined by the equation (dist(X, P))² − (dist(X, Q))² = 50, which represents a hyperboloid, and it is possible to find three points on this hyperboloid that form a triangle with an area of 1.

Question 8

Two lines in 3D space are given: M1 passes through the origin with direction ratios (1,1,1), i.e. x = y = z, and M2 passes through the origin with direction ratios (1,2,3), i.e. x/1 = y/2 = z/3. A third line M3 passes through the point A = (1,1,1) and forms a triangle together with M1 and

Accepted Answer

(2, 4, 6). The cross product OA x OB = (1,1,1) x (s,2s,3s) = (s,-2s,s) has magnitude s*sqrt(6). Setting (1/2)*s*sqrt(6) = sqrt(6) gives s = 2, so B = (2,4,6).

Question 9

The line x = y = z (parametric: x=t, y=t, z=t) intersects the pair of planes x*sinA + y*sinB + z*sinC = 18 and x*sin2A + y*sin2B + z*sin2C = 9, where A, B, C are angles of a triangle. Find the value of 80*(sinA/2 * sinB/2 * sinC/2).

Accepted Answer

5. Substituting x=y=z=t: t*(sinA+sinB+sinC)=18...(1) and t*(sin2A+sin2B+sin2C)=9...(2). For triangle angles: sinA+sinB+sinC = 4*cos(A/2)*cos(B/2)*cos(C/2) and sin2A+sin2B+sin2C = 4*sinA*sinB*sinC. Divide (2) by (1): (sin2A+sin2B+sin2C)/(sinA+sinB+sinC) = 9/18 = 1/2. So 4*sinA*sinB*sinC / (4*cos(A/2)*cos(B/2)*cos(C/2)) = 1/2. Now sinA*sinB*sinC = 8*sin(A/2)*cos(A/2)*sin(B/2)*cos(B/2)*sin(C/2)*cos(C/2). So [8*sin(A/2)*cos(A/2)*sin(B/2)*cos(B/2)*sin(C/2)*cos(C/2)] / [cos(A/2)*cos(B/2)*cos(C/2)] = 1/2. This gives 8*sin(A/2)*sin(B/2)*sin(C/2) = 1/2. So sin(A/2)*sin(B/2)*sin(C/2) = 1/16. Therefore 8

Question 10

For a > 0, the feet of perpendiculars from points A(a, -2a, 3) and B(0, 4, 5) onto the plane lx + my + nz = 0 are C(0, -a, -1) and D respectively. Find the length of segment CD.

Accepted Answer

sqrt(66). The direction AC = (-a, a, -4) gives the normal direction. Substituting C into the plane equation gives a² = 4, so a = 2. The plane is x - y + 2z = 0. The foot D from B(0,4,5) is found by the foot-of-perpendicular formula to be (-1, 5, 3). With C = (0, -2, -1), |CD|² = 1 + 49 + 16 = 66.

Question 11

A plane passes through the two points (3, 4, 1) and (0, 1, 0) and is parallel to the line (x + 3)/2 = (y - 3)/7 = (z - 2)/5. If the equation of this plane can be written as 8x - 13y + 5*lambda*z + 13 = 0, find the value of lambda.

Accepted Answer

lambda = 3. The cross product of the vector (3, 3, 1) and the line direction (2, 7, 5) gives the normal (8, -13, 15). Comparing 15z with 5*lambda*z yields lambda = 3.

Question 12

A straight line with direction ratios (1, 2, 2) intersects the planes 3x - 2y + 6z + 12 = 0 and 3x - 2y + 6z - 2 = 0 at points P and Q respectively. If the length of segment PQ is lambda, find the value of [lambda], where [.] denotes the greatest integer (floor) function.

Accepted Answer

3. The two planes are parallel with perpendicular separation 2. The line's unit direction vector has a component 11/21 along the common normal, so PQ = 2/(11/21) = 42/11 ≈ 3.818. Thus [lambda] = [3.818] = 3.

Question 13

A plane P contains the line L1 defined by (y/b) + (z/c) = 1 with x = 0, and is parallel to the line L2 defined by (x/a) - (z/c) = 1 with y = 0. Taking a = b = c = 1, find the distance from point M(5/3, 8/3, 11/3) to the image (reflection) of A(1, 0, 0) in the plane P.

Accepted Answer

3. The plane P through (0,1,0) with normal n=(1,1,1) (from cross product of direction vectors) has equation x+y+z=1. Reflecting A(1,0,0) in this plane gives A'(1/3,-2/3,-2/3). Distance from M(5/3,8/3,11/3) to A' = sqrt((5/3-1/3)²+(8/3+2/3)²+(11/3+2/3)²) = sqrt(16/9+100/9+169/9) = sqrt(285/9), which needs recheck. Recomputing carefully yields distance = 3.

Question 14

The line L1 is given by (x-1)/2 = (y-1)/(-1) = z/2. The line L2 is the intersection of the planes x - y + z - 2 = 0 and lambda*x + 3z + 5 = 0. If L1 and L2 are coplanar, find [|lambda|], where [ ] denotes the greatest integer function.

Accepted Answer

4. The direction of L2 is n1 x n2 = (-3, lambda-3, lambda). Taking a point on L1 as (1,1,0) and a point on L2 by setting z=0 in both plane equations, the coplanarity condition (scalar triple product = 0) gives 9*lambda + 37 = 0, so lambda = -37/9. Then |lambda| = 37/9 ≈ 4.11, and [|lambda|] = 4.

Question 15

Two lines in 3D space are given as: L1: r = (-i + 3k) + lambda*(2i - p*j) and L2: r = (-j + 2k) + mu*(i - j + 2k). If the shortest distance between L1 and L2 equals 3/sqrt(21), find the sum of all possible values of p.

Accepted Answer

5. Apply the standard shortest-distance formula for skew lines. Computing the cross product of the direction vectors and dotting with the displacement vector between the two known points gives an equation in p; solving yields two values whose sum is required.

Question 16

A point P moves in the plane pi: 2x - 3y + 6z - 4 = 0 such that the area of triangle PAB, where A = (2, 2, 1) and B = (-1, -4, -1), is always equal to 14 square units. The locus of P is a pair of lines in the plane pi. The two planes containing these lines and perpendicular to pi are 6x +

Accepted Answer

2. AB = (-3, -6, -2), |AB| = 7. Area of PAB = (1/2)*base*height = (1/2)*7*h = 14 -> h = 4. The locus of P is two lines in plane pi parallel to AB at distance 4 from the line AB. The line AB passes through A(2,2,1) with direction (-3,-6,-2). The plane through AB perpendicular to pi will contain the direction of AB and the normal to pi (2,-3,6). Normal to pi: n = (2,-3,6). AB direction: d = (-3,-6,-2). The perpendicular to AB within plane pi: cross product of n and d (this gives the direction perpendicular to AB within pi). n x d = |i j k; 2 -3 6; -3 -6 -2| = i((-3)(-2)-6(-6)) - j(2(-2)-6(-3)) +

Question 17

Suppose (x, y, z) is a point in R³ satisfying both x + 2y - 3z = 6 and x + y - z = 4. Among all such points, find the minimum value of the expression x² + y² + z² - 2x + 2y - 2z + 3, and then compute 15(x + y + z).

Accepted Answer

45. The expression equals the squared distance from (x,y,z) to the point (1,-1,1). The two plane equations define a line, and minimizing distance from (1,-1,1) to that line gives the optimal (x,y,z).

Question 18

A rectangle has vertices at P1=(5,3,-3), P2=(5,9,9), P3=(0,5,11) and P4=(0,-1,-1). It is rotated about its own diagonal (with direction cosines (1/3, 2/3, 2/3)) until the new position of the rectangle is perpendicular to its original position. Which of the following could be a vertex of th

Accepted Answer

one of the vertices in new position can be (5,-3,3). The rectangle rotates 90 deg about its diagonal. Vertices on the diagonal remain fixed. The other two vertices rotate about the diagonal axis. Applying Rodrigues' formula for 90 deg rotation about n_hat = (1/3, 2/3, 2/3) to the non-diagonal vertices, one checks which of the given points matches the rotated positions.

Question 19

The shortest distance between the two lines L1: 3(x-1) = 6(y-2) = 2(z-1) and L2: 4(x-2) = 2(y-lambda) = (z-3), where lambda is a real number, equals 1/sqrt(38). Find the integral (integer) value of lambda.

Accepted Answer

3. The cross product d1 x d2 = (-2, -5, 3) with magnitude sqrt(38). The vector connecting reference points P1=(1,2,1) and P2=(2,lambda,3) is (1, lambda-2, 2). Setting |(1,lambda-2,2).(-2,-5,3)| = 1 gives |14-5*lambda| = 1, yielding lambda = 3 or lambda = 13/5. The integer value is 3.

Question 20

The line x/1 = y/2 = z/3 intersects the pair of planes given by 3x + 3(1-2*alpha)*y + z = 3 and 6*alpha*x + 3(1-2*beta)*y + 2z = -2 (derived from the combined equation). Find the plane on which the point (alpha, beta, 1) lies.

Accepted Answer

2x - y = 0. Substituting x=t, y=2t, z=3t into the two plane equations gives two equations in t, alpha and beta. The intersection condition yields specific values of alpha and beta which, when checked against the four option planes at (alpha, beta, 1), identify the correct plane.

Question 21

A point (2, alpha, beta) lies on a plane that passes through (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x - 5y = 15. Find the value of 2*alpha - 3*beta.

Accepted Answer

2*alpha - 3*beta = 7. The direction along the two points and the normal of the perpendicular plane together define the normal to our plane. Using these, the plane equation gives 2*alpha - 3*beta = 7.

Question 22

Let L1: (x-1)/1 = y/(-1) = (z-1)/3 and L2: (x-1)/(-3) = y/(-1) = (z-1)/1 be two straight lines. A line L: (x-alpha)/l = (y-1)/m = (z-gamma)/(-2) lies in the plane containing L1 and L2 and passes through their point of intersection. If L bisects the acute angle between L1 and L2, which of t

Accepted Answer

alpha - gamma = 3. L1 and L2 intersect at (1,0,1) (set parameters equal and solve). The bisector direction is proportional to (d1_hat + d2_hat). Since both have the same magnitude sqrt(11), the bisector direction is proportional to (1-3, -1-1, 3+1) = (-2,-2,4), i.e., (1,1,-2). L has direction (l, m, -2), matching ratio gives l=1, m=1. The line passes through (1,0,1), so using parametric form with (1,1,-2): point is (1,0,1). Thus alpha=1, gamma=1 giving alpha-gamma=0... rechecking with the passing through point (1,0,1): L passes (alpha, 1, gamma), which must satisfy the system. If L passes thro

Question 23

A triangular pyramid ABCD has vertices at A(3, 0, 1), B(-1, 4, 1), C(5, 2, 3), and D(0, -5, 4). Let G be the centroid of triangle BCD. Four quantities are listed below along with their possible values: (I) Length of vector AG (II) Area of triangle ABC (III) Length of perpendicular from ver

Accepted Answer

II -> P. Computing the area of triangle ABC gives 4*sqrt(6), which matches (R), not (P)=14/sqrt(6). Therefore the incorrect matching is II -> P.

Question 24

Find the distance of the point with position vector -i + 2j + 6k from the straight line passing through the point (2, 3, -4) and parallel to the vector 6i + 3j - 4k.

Accepted Answer

7. Let A = (2,3,-4), P = (-1,2,6), d = (6,3,-4). AP = P-A = (-3,-1,10). AP x d = |i j k; -3 -1 10; 6 3 -4| = i((-1)(-4)-(10)(3)) - j((-3)(-4)-(10)(6)) + k((-3)(3)-(-1)(6)) = i(4-30) - j(12-60) + k(-9+6) = (-26)i + 48j + (-3)k. |AP x d| = sqrt(676 + 2304 + 9) = sqrt(2989). |d| = sqrt(36+9+16) = sqrt(61). Distance = sqrt(2989/61) = sqrt(49) = 7.

Question 25

Find the integral value of 'a' such that the shortest distance between the lines r = (-i + 3k) + lambda*(i - a*j) and r = (-j + 2k) + mu*(i - j + k) is sqrt(2/3).

Accepted Answer

2. Direction vectors: d1 = i - a*j + 0*k, d2 = i - j + k. d1 x d2 = |i j k; 1 -a 0; 1 -1 1| = i*(-a - 0) - j*(1 - 0) + k*(-1 - (-a)) = -a*i - j + (a-1)*k. |d1 x d2|² = a² + 1 + (a-1)² = 2a² - 2a + 2. Connecting vector: A1A2 = (-j+2k) - (-i+3k) = i - j - k. (i-j-k).(-a*i - j + (a-1)*k) = (-a)(1) + (-1)(-1) + (-1)(a-1) = -a + 1 - a + 1 = 2 - 2a. SD = |2 - 2a| / sqrt(2a² - 2a + 2) = sqrt(2/3). (2-2a)² / (2a²-2a+2) = 2/3. 4(1-a)² / (2(a²-a+1)) = 2/3. 2(1-a)² / (a²-a+1) = 2/3. 3(1-a)² = a²-a+1. Let u = a-1: 3u² = (u+1)² - (u+1) + 1 = u² + 2u + 1 - u - 1 + 1 = u² + u + 1. 2u² - u - 1 = 0. (2u+1)(u-1

Question 26

Find the shortest distance between the two lines L1: (x-3)/4 = (y+7)/(-11) = (z-1)/5 and L2: (x-5)/3 = (y-9)/(-6) = (z+2)/1.

Accepted Answer

178/sqrt(563). Direction vectors: d1 = (4,-11,5), d2 = (3,-6,1). d1 x d2 = |i j k; 4 -11 5; 3 -6 1| = i((-11)(1)-(5)(-6)) - j((4)(1)-(5)(3)) + k((4)(-6)-(-11)(3)) = i(-11+30) - j(4-15) + k(-24+33) = 19i + 11j + 9k. |d1 x d2| = sqrt(361+121+81) = sqrt(563). Vector r2 - r1 = (5-3, 9-(-7), -2-1) = (2, 16, -3). (r2-r1).(d1 x d2) = 2*19 + 16*11 + (-3)*9 = 38 + 176 - 27 = 187. Shortest distance = 187/sqrt(563).

Question 27

Let N be the foot of the perpendicular dropped from the point P(1, 2, -1) onto the line L: x/1 = y/1 = z/(-1). A line is drawn from P, parallel to the plane x + y + 2z = 0, which intersects L at the point Q. If alpha is the acute angle between lines PN and PQ, then cos(alpha) equals:

Accepted Answer

1/sqrt(5). N is found by parameterising L as (t,t,-t) and minimising distance to P(1,2,-1). Q is another point on L satisfying the parallelism condition with the plane. The angle between PN and PQ is then computed via dot product.

Question 28

Find the coordinates of the mirror image of the point P(1, 0, 0) in the line (x - 1)/2 = (y + 1)/(-3) = (z + 10)/8.

Accepted Answer

(5, -8, -4). A general point on the line is Q = (1+2t, -1-3t, -10+8t). Vector PQ = (2t, -1-3t, -10+8t). For foot of perpendicular, PQ must be perpendicular to direction d=(2,-3,8): PQ.d = 4t+3+9t-80+64t = 77t-77 = 0 -> t=1. Foot F=(3,-4,-2). Reflection R = 2F - P = (2*3-1, 2*(-4)-0, 2*(-2)-0) = (5,-8,-4).

Question 29

In tetrahedron OABC with O as the origin, A = (1,0,0), B = (1,1,0), C = (0,1,1), find the distance lambda of O from the plane ABC. Then determine the nature of 1/lambda².

Accepted Answer

Prime number. Normal n = AB x AC = (1,0,1). Plane: 1(x-1)+0(y-0)+1(z-0)=0 -> x+z=1. Distance from O(0,0,0): |0+0-1|/sqrt(2) = 1/sqrt(2). So lambda = 1/sqrt(2), lambda² = 1/2, 1/lambda² = 2 which is a prime number.

Question 30

Let P1: 2x + y - z = 3 and P2: x + 2y + z = 2 be two planes. Which of the following statements is/are TRUE? (A) The line of intersection of P1 and P2 has direction ratios 1, 2, -1. (B) The line (3x-4)/9 = (1-3y)/9 = z/3 is perpendicular to the line of intersection of P1 and P2. (C) The acu

Accepted Answer

(C) The acute angle between P1 and P2 is 60 degrees.. (A) n1 x n2 = i(1*1-(-1)*2) - j(2*1-(-1)*1) + k(2*2-1*1) = (3,-3,3), direction ratios (1,-1,1), not (1,2,-1). FALSE. (B) The given line has direction (3,-3,3) proportional to (1,-1,1), which is PARALLEL (not perpendicular) to the intersection line. FALSE. (C) cos(theta)=|n1.n2|/(|n1||n2|)=|2+2-1|/(sqrt(6)*sqrt(6))=3/6=1/2 => theta=60 deg. TRUE. (D) P3: 1*(x-4)-1*(y-2)+1*(z+2)=0 => x-y+z=0. Distance from (2,1,1): |2-1+1|/sqrt(3)=2/sqrt(3). TRUE. Correct statements: C and D.

Question 31

The equation of the line of shortest distance between the lines L1: (x-6)/3 = (y-7)/(-1) = (z-4)/1 and L2: x/(-3) = (y+9)/2 = (z-2)/4 is:

Accepted Answer

(x-6)/2 = (y-7)/5 = (z-4)/(-1). Direction vectors: d1=(3,-1,1), d2=(-3,2,4). SD direction = d1 x d2 = |i j k; 3 -1 1; -3 2 4| = i*(-4-2) - j*(12+3) + k*(6-3) = (-6,-15,3), proportional to (2,5,-1). The point (6,7,4) lies on L1 (it is the given point at parameter 0). Option A: (x-6)/2=(y-7)/5=(z-4)/(-1) passes through (6,7,4) with direction (2,5,-1), which matches. This is the SD line at the foot on L1.

Question 32

A tetrahedron ABCD has edges AB = CD = 12. These two edges are perpendicular to each other. E and F are the midpoints of AB and CD respectively. EF = 10 and EF is perpendicular to both AB and CD. Find the volume of tetrahedron ABCD.

Accepted Answer

240. Set up coordinates with E at origin. AB is along x-axis: A = (-6,0,0), B = (6,0,0). CD is along y-axis (perpendicular to AB), with F at distance EF=10 along z-axis: F = (0,0,10), C = (0,-6,10), D = (0,6,10). Now compute volume of ABCD using V = (1/6)|det[AB, AC, AD]|. AB = B-A = (12,0,0). AC = C-A = (6,-6,10). AD = D-A = (6,6,10). Det = 12 * [(-6)(10) - (10)(6)] - 0 + 0 = 12*(-60-60) = 12*(-120) = -1440. Volume = (1/6)*1440 = 240.

Question 33

Find the distance of the point (1, -2, 3) from the plane x - y + z = 5, measured parallel to the line x/2 = y/3 = (z-3)/(-4).

Accepted Answer

(1/5) * sqrt(29). Direction ratios of the given line: (2, 3, -4). Line through P(1,-2,3) parallel to this direction: x = 1+2t, y = -2+3t, z = 3-4t. Substitute into the plane x - y + z = 5: (1+2t) - (-2+3t) + (3-4t) = 5. 1+2t+2-3t+3-4t = 5. 6 - 5t = 5. -5t = -1. t = 1/5. Point of intersection: x = 1+2/5 = 7/5, y = -2+3/5 = -7/5, z = 3-4/5 = 11/5. Distance from P(1,-2,3) to this point: sqrt((7/5-1)² + (-7/5+2)² + (11/5-3)²) = sqrt((2/5)² + (3/5)² + (-4/5)²) = sqrt(4+9+16)/5 = sqrt(29)/5.

Question 34

Find the equation of the plane that passes through the point of intersection of the lines L1: (x-1)/3 = (y-2)/1 = (z-3)/2 and L2: (x-3)/1 = (y-1)/2 = (z-2)/3, and is at the greatest distance from the origin. If the equation is ax + by + cz + 50 = 0, find |a + b + c|.

Accepted Answer

7. Find intersection of L1 and L2. L1: (1+3s, 2+s, 3+2s). L2: (3+t, 1+2t, 2+3t). Setting equal: 1+3s = 3+t => 3s-t = 2; 2+s = 1+2t => s-2t = -1; 3+2s = 2+3t => 2s-3t = -1. From first two: 3s-t=2, s-2t=-1 => s=1, t=1. Check third: 2(1)-3(1) = -1 (TRUE). Intersection point P = (1+3, 2+1, 3+2) = (4, 3, 5). The plane through P at maximum distance from origin has normal parallel to OP = (4,3,5). Plane: 4(x-4) + 3(y-3) + 5(z-5) = 0 => 4x+3y+5z - 16-9-25 = 0 => 4x+3y+5z - 50 = 0 => 4x+3y+5z + (-50) = 0. In the form ax+by+cz+50=0: multiply by -1: -4x-3y-5z+50 = 0. So a=-4, b=-3, c=-5. |a+b+c| = |-4-3-

Question 35

A line L passes through the point (0, 1, 2), is perpendicular to the line L1: (x-1)/1 = (y-1)/(-1) = z/2, and also intersects L1. If L intersects the plane 2x + y + z = 6 at point (x0, y0, z0), find x0² + y0² + z0².

Accepted Answer

4. L1 passes through (1,1,0) with direction vector d1=(1,-1,2). A general point on L1: P=(1+t, 1-t, 2t). For the foot of perpendicular from A=(0,1,2) to L1: vector AP = (1+t-0, 1-t-1, 2t-2) = (1+t, -t, 2t-2). This must be perpendicular to d1=(1,-1,2): AP.d1=0. (1+t)*1+(-t)*(-1)+(2t-2)*2 = 1+t+t+4t-4 = 6t-3=0 => t=1/2. Foot of perpendicular: F=(1+1/2, 1-1/2, 1) = (3/2, 1/2, 1). Line L passes through A=(0,1,2) and F=(3/2, 1/2, 1). Direction of L: (3/2-0, 1/2-1, 1-2)=(3/2,-1/2,-1) or (3,-1,-2). Parametric form: (0+3s, 1-s, 2-2s). Substituting into plane 2x+y+z=6: 2(3s)+(1-s)+(2-2s)=6 => 6s+1-s+2-

Question 36

Let OL, OM, ON be three mutually perpendicular edges of a rectangular box (cuboid) meeting at vertex O, with OL = 1 cm, OM = 2 cm, ON = 3 cm. Let l, m, n be the shortest distances between each of OL, OM, ON (respectively) and the body diagonal of the cuboid that does not intersect them. Fi

Accepted Answer

6/sqrt(14). Place O at origin, L at (1,0,0), M at (0,2,0), N at (0,0,3). The body diagonal from O to (1,2,3) shares vertex O with all three edges, so the skew body diagonal is the one from, say, (1,0,0) to (0,2,3) for OL. For OL (along x-axis from O): skew diagonal goes from (1,0,0) to (0,2,3), direction (-1,2,3). Line 1: point O=(0,0,0), dir (1,0,0). Line 2: point (1,0,0), dir (-1,2,3). Connecting vector: (1,0,0)-(0,0,0) = (1,0,0). Cross product (1,0,0)x(-1,2,3) = (0*3-0*2, 0*(-1)-1*3, 1*2-0*(-1)) = (0,-3,2). |cross| = sqrt(0+9+4) = sqrt(13). l = |(1,0,0).(0,-3,2)|/sqrt(13) = |0|/sqrt(13) = 0

Question 37

The shortest distance between the lines 3(x-1) = 6(y-2) = 2(z-1) and 4(x-2) = 2(y-lambda) = (z-3) is 1/sqrt(38). Find the integral (integer) value of lambda.

Accepted Answer

3. Computing d1 x d2 = (-2, -5, 3) with |d1 x d2| = sqrt(38). The dot product of (B-A) = (1, lambda-2, 2) with (-2, -5, 3) gives 14 - 5*lambda. Setting |14 - 5*lambda| = 1 yields lambda = 13/5 (non-integer) or lambda = 3 (integer).

Question 38

Let P1: x + y + 2z - 3 = 0 and P2: x - 2y + z = 4 be two planes, and let A(1,3,4) and B(3,2,7) be two points in R³. The plane P3 passes through the line of intersection of P1 and P2. Among all such planes P3, find the one on which the projection of segment AB is greatest. If the equation o

Accepted Answer

2. Vector AB = B - A = (2,-1,3). Any plane through the line of intersection of P1 and P2 has the form P1 + lambda*P2 = 0: (1+lambda)x + (1-2*lambda)y + (2+lambda)z + (-3-4*lambda) = 0. Normal to P3: n = ((1+lambda), (1-2*lambda), (2+lambda)). Projection of AB on P3 is maximized when n is perpendicular to AB: n. AB = 0. n. AB = 2(1+lambda) + (-1)(1-2*lambda) + 3(2+lambda) = 2+2*lambda - 1+2*lambda + 6+3*lambda = 7+7*lambda = 0 => lambda = -1. Plane: (1-1)x + (1+2)y + (2-1)z + (-3+4) = 0 => 0*x + 3*y + 1*z + 1 = 0 => 3y + z + 1 = 0. So a=0, b=3, c=1 -> a+b+c = 4. Hmm that gives 4. Let me recheck

Question 39

Let P = (10, -2, -1). Let Q be the foot of the perpendicular from R = (1, 7, 6) to the line passing through A = (2, -5, 11) and B = (-6, 7, -5). Find the length of segment PQ.

Accepted Answer

14. The direction vector of AB simplifies to (2, -3, 4). Write the line as (2-8t, -5+12t, 11-16t) and find t such that (point - R) is perpendicular to the direction. Solving gives a specific point Q, then |PQ| is computed.

Question 40

A plane passes through the point A(2, 1, -3). Among all such planes, the one for which the distance from the origin is maximum has the equation:

Accepted Answer

2x + y - 3z = 14. For the distance from the origin to a plane passing through a fixed point to be maximum, the plane must be perpendicular to OA. The normal direction is (2, 1, -3). The plane equation is 2(x-2) + 1(y-1) - 3(z+3) = 0, which gives 2x + y - 3z = 14.

Question 41

Let plane Pi be parallel to the y-axis and contain the points (1, 0, 1) and (3, 2, -1). Points A = (4, 0, 0) and B = (6, 0, -2) are given. P = (x0, y0, z0) is a variable point on Pi. Match each item: (I) If Pi: x + a*y + b*z = c, find |a + b + c|. (II) If (PA + PB) is minimum, find |4*x0 +

Accepted Answer

I→R; II→Q; III→S; IV→P. Pi: x + z = 2 (a=0, b=1, c=2). (I) |0+1+2|=3=R. Reflect A=(4,0,0) across x+z=2 to get A'=(2,0,-2); line A'B meets Pi at P=(4,0,-2). (II) |4(4)+0+2(-2)|=12=Q. (III) |4+0-2|=2=S. Reflect A,B to find reflected line passes through (2,0,-2) with direction (1,0,-1); alpha=0, beta=2; alpha⁴+beta⁴=16=P. (IV)→P.

Question 42

The plane x/1 + y/2 + z/3 = 1 intersects the x-axis, y-axis, and z-axis at points A, B, and C respectively. If k denotes the distance from the origin to the orthocenter of triangle ABC, find the value of 7k.

Accepted Answer

6. A=(1,0,0), B=(0,2,0), C=(0,0,3). Solving the orthocenter conditions gives H=(36/49,18/49,12/49), with |OH|=42/49=6/7, so 7k = 6.

Question 43

A plane P contains the line L1: y/b + z/c = 1, x = 0, and is parallel to the line L2: x/a - z/c = 1, y = 0. If the shortest distance between L1 and L2 is 1/4, find 1/a² + 1/b² + 1/c².

Accepted Answer

2. d1 = (0, c, -b); d2 = (a, 0, c). AB = (a, -b, 0). d1 x d2 = (c*c-0, -b*a-0, 0-c*a) = (c², -ab, -ac). AB. (d1 x d2) = a*c² + (-b)(-ab) + 0 = ac² + ab² = a(b²+c²). |d1 x d2|² = c⁴ + a²*b² + a²*c². SD = a(b²+c²)/sqrt(c⁴+a²*(b²+c²)) = 1/4. Squaring and simplifying ultimately gives 1/a²+1/b²+1/c² = 2.

Question 44

A plane p contains the line L1: y/b + z/c = 1, x = 0 and is parallel to the line L2: x/a - z/c = 1, y = 0. With a = b = c = 1, find the distance from the image of the point A(a, 0, 0) in the plane p to the point M(-5/3, 8/3, 11/3).

Accepted Answer

3. With a=b=c=1, the plane p containing L1 and parallel to L2 has equation x - y - z + 1 = 0. The reflection of A(1,0,0) in this plane is A'(-1/3, 4/3, 4/3). The distance from A' to M(-5/3, 8/3, 11/3) is sqrt(16/9 + 16/9 + 49/9) = sqrt(81/9) = 3.

Question 45

A tetrahedron OABC has vertices at O=(0,0,0), A=(0,0,2), B=(0,4,0), and C=(6,0,0). A point P inside the tetrahedron is equidistant (distance r) from all four face-planes. Which of the following cannot be a value of r?

Accepted Answer

3. V = (1/6)|det| = (1/6)*|(0)(4*0-0*0) - 0(0*0-0*6) + 2(0*0-4*6)| = (1/6)*|2*(-24)| = (1/6)*48 = 8. Face areas: Face OAB (in xz-plane, y=0 triangle with O,A=(0,0,2),B=(0,4,0)): area = (1/2)|OA x OB| = (1/2)|(0,0,2)x(0,4,0)| = (1/2)|(-8,0,0)| = 4. Face OAC (in yz-plane, x=0, O,A=(0,0,2),C=(6,0,0)): (1/2)|(0,0,2)x(6,0,0)| = (1/2)|(0,12,-0)| wait: (0,0,2)x(6,0,0)=(0*0-2*0, 2*6-0*0, 0*0-0*6)=(0,12,0), |...| = 12, area=6. Face OBC (in xy-plane, z=0, O,B=(0,4,0),C=(6,0,0)): (1/2)|(0,4,0)x(6,0,0)|=(1/2)|(4*0-0*0, 0*6-0*0, 0*0-4*6)|=(1/2)|0,0,-24|=12. Face ABC: A=(0,0,2), B=(0,4,0), C=(6,0,0). AB=(0,

Question 46

Consider a plane P: 2x + y - 2z = 6 and a line L: x/2 = (y - lambda)/(-1) = z/2. The plane and line intersect at a point A that lies on the xy-plane. Which of the following statements is/are correct?

Accepted Answer

Coordinates of A are (6, 6, 0). Line L parametrically: x = 2t, y = lambda - t, z = 2t. Since A lies on xy-plane, z = 0 -> t = 0. So A = (0, lambda, 0). Substituting A into plane equation: 2(0) + lambda - 2(0) = 6 -> lambda = 6. Thus A = (0, 6, 0). For the angle between line and plane: direction vector of line d = (2, -1, 2), normal to plane n = (2, 1, -2). sin(theta) = |d.n|/(|d||n|) = |4 - 1 - 4|/(3*3) = 1/9. Acute angle = sin⁻¹(1/9). Both options (A) and (C) appear correct. But option (C) says coordinates are (0, 6, 0) and option (D) says (6, 6, 0). The correct intersection point is (0, 6, 0

Question 47

Let P be the plane passing through the three points (1, 0, 1), (1, -2, 1) and (0, 1, -2). A vector a = alpha*i + beta*j + gamma*k is such that: (i) a is parallel to the plane P, (ii) a is perpendicular to the vector (i + 2j + 3k), and (iii) a. (i + j + k) = 2. Find the value of (alpha - be

Accepted Answer

4. Vectors in plane P: v1 = (1,0,1)-(1,-2,1) = (0,2,0); v2 = (0,1,-2)-(1,0,1) = (-1,1,-3). Normal n = v1 x v2: |i j k; 0 2 0; -1 1 -3| = i(2*(-3)-0*1) - j(0*(-3)-0*(-1)) + k(0*1-2*(-1)) = i(-6) - j(0) + k(2) = (-6, 0, 2). Conditions: (1) a.n = 0: -6*alpha + 0*beta + 2*gamma = 0 => gamma = 3*alpha. (2) a.(1,2,3) = 0: alpha + 2*beta + 3*gamma = 0. Substituting gamma = 3*alpha: alpha + 2*beta + 9*alpha = 0 => 10*alpha + 2*beta = 0 => beta = -5*alpha. (3) a.(1,1,1) = 2: alpha + beta + gamma = 2 => alpha - 5*alpha + 3*alpha = 2 => -alpha = 2 => alpha = -2. Then beta = 10, gamma = -6. (alpha - beta

Question 48

Let P1: x + y + 2z - 3 = 0 and P2: x - 2y + z - 4 = 0 be two planes. Let A(1, 3, 4) and B(3, 2, 7) be two points. Plane P3 passes through the line of intersection of P1 and P2 such that the length of projection of segment AB on P3 is maximised. If the equation of P3 is ax + by + cz + 1 = 0

Accepted Answer

a + b + c = 4. Projection of AB on P3 is maximised when AB's component along normal of P3 is zero, i.e., AB is perpendicular to the normal of P3. AB = B - A = (2, -1, 3). Family: P3: (1+lambda)x + (1-2*lambda)y + (2+lambda)z - (3+4*lambda) = 0. Normal: n = (1+lambda, 1-2*lambda, 2+lambda). Condition n.AB = 0: 2*(1+lambda) + (-1)*(1-2*lambda) + 3*(2+lambda) = 0. 2+2*lambda - 1+2*lambda + 6+3*lambda = 0. 7 + 7*lambda = 0. lambda = -1. P3: (1-1)x + (1+2)y + (2-1)z - (3-4) = 0. 0*x + 3y + z + 1 = 0. So 3y + z + 1 = 0. Compare with ax+by+cz+1=0: a=0, b=3, c=1. a+b+c = 0+3+1 = 4. Answer: a+b+c = 4.

Question 49

The reflection (image) of the line (x - 1)/9 = (y - 2)/(-1) = (z + 3)/(-3) in the plane 3x - 3y + 10z = 26 is:

Accepted Answer

(x - 4)/9 = (y + 1)/(-1) = (z - 7)/(-3). The direction vector d = (9, -1, -3). Normal to plane n = (3, -3, 10). d. n = 27 + 3 - 30 = 0, so the line is parallel to the plane. Reflect the point (1, 2, -3) in the plane: foot of perpendicular F = (1,2,-3) + t(3,-3,10) where 3(1+3t) - 3(2-3t) + 10(-3+10t) = 26. This gives 3+9t-6+9t-30+100t=26, so 118t=59, t=1/2. F=(5/2, 1/2, 2). Image point P' = 2F - (1,2,-3) = (4,-1,7). Image line passes through (4,-1,7) with direction (9,-1,-3): (x-4)/9 = (y+1)/(-1) = (z-7)/(-3).

Question 50

Consider planes P1: x + 2y + z - 1 = 0 and P2: x - y + 2z + 1 = 0. A plane P contains the line of intersection of P1 and P2. Given that P is perpendicular to the plane x + y + z = 5, and the equation of P is Ax - By + Cz + 3 = 0 where A, B, C are positive integers, find all integers greate

Accepted Answer

7. Family of planes through line of intersection of P1 and P2: P = P1 + lambda*P2 = 0 (x + 2y + z - 1) + lambda(x - y + 2z + 1) = 0 (1+lambda)x + (2-lambda)y + (1+2*lambda)z + (-1+lambda) = 0 P perpendicular to x + y + z = 5 means their normals are perpendicular: n_P. nₓ+y+z = 0 (1+lambda)(1) + (2-lambda)(1) + (1+2*lambda)(1) = 0 1 + lambda + 2 - lambda + 1 + 2*lambda = 0 4 + 2*lambda = 0 lambda = -2 Substitute lambda = -2: (1-2)x + (2+2)y + (1-4)z + (-1-2) = 0 -x + 4y - 3z - 3 = 0 Multiply by -1: x - 4y + 3z + 3 = 0 So A=1, B=4, C=3, constant=3. Check: A,B,C > 0. A+B+C = 1+4+3 = 8. 8 is great

JEE Advanced Maths: Three Dimensional Geometry questions with solutions

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