Question 1

Consider the following two statements:

Statement 1: If A, B and C are points with position vectors a = 2î + ĵ + k̂, b = 3î - ĵ + 3k̂ and c = î + 7ĵ - 5k̂, then the figure OABC forms a tetrahedron.

Statement 2: If the position vectors a, b and c of points A, B and C are non-coplanar

Accepted Answer

Statement 1 is true, Statement 2 is true, and Statement 2 correctly explains Statement 1. The scalar triple product [a b c] equals the determinant of rows (2,1,1),(3,-1,3),(1,7,-5) = 2(5-21)-1(-15-3)+1(21+1) = 8, which is nonzero. So a,b,c are non-coplanar, OABC is a tetrahedron (Statement 1 true), and Statement 2 is the correct general reason.

Question 2

A moving plane always contains the fixed point (1, 2, 3). The set of points that are the perpendicular projections of the origin onto this plane is described by

Accepted Answer

x² + y² + z² − x − 2y − 3z = 0. The correct option represents a sphere centered at the point (1, 2, 3) with a radius that ensures the origin's perpendicular projection lies on the plane defined by the fixed point. This relationship is established by the equation of a sphere, which includes the coordinates of the fixed point and accounts for the distance from the origin.

Question 3

The direction cosines l, m, n of one of the two lines satisfying the relations l - 5m + 3n = 0 and 7l² + 5m² - 3n² = 0 are

Accepted Answer

1/√14, 2/√14, 3/√14. The correct option satisfies both given equations for direction cosines, ensuring they maintain the necessary relationships between l, m, and n while also adhering to the constraint that the sum of the squares of the direction cosines equals 1.

Question 4

A straight line is equally inclined to the x-axis and the y-axis, making an angle α with each. If its angle θ with the z-axis satisfies sin²θ = 2 sin²α, determine α.

Accepted Answer

π/4. Direction cosines (cos a, cos a, cos th) give 2cos^2 a + cos^2 th = 1. Using sin^2 th = 2 sin^2 a gives cos^2 th = 2cos^2 a - 1. Equating: 2cos^2 a - 1 = 1 - 2cos^2 a => cos^2 a = 1/2 => alpha = pi/4.

Question 5

Let Q be the reflection of the point P(2, 3, 4) across the plane x − 2y + 5z = 6. The equation of the line joining P and Q is

Accepted Answer

(x − 2)/1 = (y − 3)/−2 = (z − 4)/5. The correct option represents the direction vector of the line connecting points P and Q, where the reflection across the plane results in a change in the y-coordinate while keeping the x and z coordinates consistent with the original point. This indicates that the line's slope in the y-direction is negative, confirming the correct ratios in the equation.

Question 6

Find the point where the perpendicular drawn from the point (2, 4, −1) meets the line x + 5 = (1/4)(y + 3) = (−1/9)(z − 6).

Accepted Answer

(−4, 1, −3). Points on the line are (-5+t, -3+4t, 6-9t) with direction (1,4,-9). Requiring the vector from P(2,4,-1) to be perpendicular to (1,4,-9) gives t=1, so the foot is (-4,1,-3).

Question 7

Find the equation of the plane that cuts the coordinate axes so as to form a triangle whose centroid is at (α, β, γ).

Accepted Answer

x/α + y/β + z/γ = 3. The correct option represents a plane that intersects the coordinate axes at points proportional to α, β, and γ, forming a triangle with a centroid at (α, β, γ). The equation x/α + y/β + z/γ = 3 indicates that the intercepts are scaled such that the centroid aligns with the specified coordinates.

Question 8

Find the equations of the two straight lines passing through the origin that make an angle of π/3 with the line (x-3)/2 = (y-3)/1 = z/1.

Accepted Answer

None of the above. The correct option is 'None of the above' because the equations of the lines that make a specific angle with the given line must be derived using the direction ratios of the line and the angle condition, which are not represented in any of the provided options.

Question 9

The two planes 3x - y + z + 1 = 0 and 5x + y + 3z = 0 meet along the line PQ. Find the equation of the plane that passes through the point (2, 1, 4) and is perpendicular to PQ.

Accepted Answer

x + y - 2z = -5. The line PQ has direction n1 x n2 = (-4,-4,8), proportional to (1,1,-2). The required plane has this as its normal and passes through (2,1,4): 1(x-2)+1(y-1)-2(z-4)=0 -> x+y-2z = -5.

Question 10

What is the nature of the intersection between the line (x-1)/1 = (y-2)/-2 = (z-1)/3 and the plane x + 2y + z = 6?

Accepted Answer

They intersect in infinitely many points. The line's direction (1,-2,3) is perpendicular to the plane's normal (1,2,1) (dot product 0), and the point (1,2,1) satisfies x+2y+z=6. So the line lies entirely in the plane, giving infinitely many intersection points.

Question 11

For the three planes given by bx - ay - n = 0, cy - bz = l, and az - cx = m, what condition must hold for them to have a common line of intersection?

Accepted Answer

a + bm + cn = 0. Taking c*(bx-ay-n) + a*(cy-bz-l) + b*(az-cx-m), all the x, y, z coefficients cancel, leaving the consistency condition -(cn+al+bm)=0, i.e. al+bm+cn=0. This corresponds to the option containing bm+cn (al+bm+cn=0).

Question 12

Find the distance of the point (1, -2, 3) from the plane x - y + z = 5 when the measurement is taken along a line parallel to x/2 = y/3 = (z-1)/6.

Accepted Answer

1. Take the line through (1,-2,3) parallel to (2,3,-6): (1+2t, -2+3t, 3-6t). Substituting into x-y+z=5 gives 6-7t... solving 7-? gives t=1/7, and the distance is |t|*|(2,3,-6)| = (1/7)*7 = 1.

Question 13

Find the radius of the sphere represented by x² + y² + z² = 49 and the plane 2x + 3y - z - 5√14 = 0.

Accepted Answer

2√6. The equation of the sphere indicates that its radius is the square root of 49, which is 7. The distance from the center of the sphere to the plane is calculated using the formula for the distance from a point to a plane, and this distance is found to be 2√6. Therefore, the radius of the sphere that intersects with the plane is 2√6.

Question 14

Find the equation of the line lying in the plane π: 2x - y + z - 4 = 0, perpendicular to the line (x-2)/(1) = (y-2)/(-1) = (z-3)/(-2), and passing through the point where this line meets π.

Accepted Answer

(x-1)/(3) = (y-3)/(5) = (z-5)/(-1). The given line meets the plane at (1,3,5). The required line lies in the plane and is perpendicular to the given line, so its direction is n_plane x d_line = (2,-1,1)x(1,-1,-2)=(3,5,-1). The line is (x-1)/3=(y-3)/5=(z-5)/(-1).

Question 15

A plane given by 2ax - 3ay + 4az + 6 = 0 passes through the midpoint of the segment joining the centres of the spheres x² + y² + z² + 6x - 8y - 2z = 13 and x² + y² + z² - 10x + 4y - 2z = 8. The value of a is

Accepted Answer

-2. Centres are (-3,4,1) and (5,-2,1); midpoint (1,1,1). Substituting in 2ax-3ay+4az+6=0: (2-3+4)a + 6 = 0 => 3a = -6 => a = -2.

Question 16

Find the equation of a plane that contains the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3, and is at a distance 2/√3 from the point (3, 1, -1).

Accepted Answer

5x - 11y + z = 17. Take (x+2y+3z-2)+lambda(x-y+z-3)=0. Setting its distance from (3,1,-1) equal to 2/sqrt3 yields lambda=-7/2, giving 5x-11y+z=17. This plane contains the line of intersection and is at the required distance.

Question 17

Statement 1: If θ denotes the angle between the line (x-2)/2=(y-1)/(-3)=(z+2)/(-2) and the plane x+y-z=5, then θ=sin⁻¹(1/√(51)).
Statement 2: The angle made by a line with a plane is the complementary angle of the angle between that line and the normal to the plane.

Accepted Answer

Statement 1 is true, Statement 2 is true, and Statement 2 correctly explains Statement 1. Line direction (2,-3,-2), plane normal (1,1,-1): sin(theta)=|2-3+2|/(sqrt(17)*sqrt(3))=1/sqrt(51), so Statement 1 is true. Statement 2 (line-plane angle is complement of line-normal angle) is true and correctly explains Statement 1.

Question 18

The vertices of a parallelogram in consecutive order are A(2, −1, 4), B(1, 0, −1), C(1, 2, 3), and D. What is the perpendicular distance of the point P(8, 2, −12) from the plane containing this parallelogram?

Accepted Answer

32√6/9. The correct option is right because the distance from a point to a plane can be calculated using the formula involving the normal vector of the plane and the coordinates of the point. In this case, the calculations yield the distance as 32√6/9, confirming that this option accurately represents the perpendicular distance from point P to the plane of the parallelogram.

Question 19

A tetrahedron is formed by the points O(0, 0, 0), A(1, 2, 1), B(2, 1, 3), and C(−1, 1, 2). The angle between the planes OAB and ABC is

Accepted Answer

cos−1(19/35). The angle between two planes can be determined using the normal vectors of the planes. By calculating the normal vectors for planes OAB and ABC and then using the dot product formula, we find that the angle between these planes is cos−1(19/35), confirming option B as the correct answer.

Question 20

Find the equation of the plane that contains the point (3, 2, 0) and also includes the line given by (x-4)/1 = (y-7)/5 = (z-4)/4.

Accepted Answer

x - y + z = 1. The equation of the plane must satisfy the point (3, 2, 0) and be parallel to the direction vector of the given line. The correct option, x - y + z = 1, can be verified by substituting the point into the equation, confirming it lies on the plane.

Question 21

Find the direction ratios of a line normal to the plane passing through the points (1, 0, 0) and (0, 1, 0), if this plane makes an angle of π/4 with the plane x + y = 3.

Accepted Answer

1, 1, √2. The correct option represents the direction ratios of the normal vector to the plane formed by the given points, which is orthogonal to the plane defined by the equation x + y = 3. The angle of π/4 indicates a specific relationship between the normals of the two planes, leading to the conclusion that the direction ratios must satisfy this geometric condition.

Question 22

What is the least distance between the plane 12x + 4y + 3z = 327 and the sphere x² + y² + z² + 4x - 2y - 6z = 155?

Accepted Answer

13. The least distance between the plane and the sphere is determined by finding the distance from the center of the sphere to the plane and then subtracting the radius of the sphere. In this case, the calculations yield a minimum distance of 13 units.

Question 23

The two straight lines given by x = ay + b, z = cy + d and x = a'y + b', z = c'y + d' are mutually perpendicular if and only if

Accepted Answer

aa' + cc' + 1 = 0. The condition for two lines to be mutually perpendicular in a three-dimensional space involves their directional coefficients. The equation aa' + cc' + 1 = 0 captures the necessary relationship between the slopes of the lines in the x-z plane, ensuring that their angles of intersection are 90 degrees.

Question 24

For the two lines [0m(x-2)/1 = (y-3)/1 = (z-4)/(-k)[0m and [0m(x-1)/k = (y-4)/2 = (z-5)/1[0m to lie in the same plane, the value of [0m k [0m must be

Accepted Answer

k = 0 or -3. The two lines must be coplanar, which occurs when the scalar triple product of their direction vectors and the vector connecting a point from each line is zero. By calculating the necessary conditions for coplanarity, we find that the values of k that satisfy this condition are 0 or -3.

Question 25

A sphere is given by x² + y² + z² + 2x - 2y - 4z - 19 = 0. If it is intersected by the plane x + 2y + 2z + 7 = 0, what is the radius of the resulting circular section?

Accepted Answer

3. To find the radius of the circular section formed by the intersection of the sphere and the plane, we first rewrite the sphere's equation in standard form to identify its center and radius. The center is at (-1, 1, 2) and the radius is 5. The distance from the center to the plane is calculated, and using the Pythagorean theorem, we find the radius of the circular section to be 3.

Question 26

Find the angle formed by the two lines given by 2x = 3y = -z and 6x = -y = -4z.

Accepted Answer

90°. The two lines are represented by their direction vectors, which can be derived from the given equations. Since the direction vectors are orthogonal, the angle between them is 90°, indicating that the lines intersect at a right angle.

Question 27

Find the shortest distance between the line r = 2î - 2ĵ + 3k̂ + λ(î - ĵ + 4k̂) and the plane r · (î + 5ĵ + k̂) = 5.

Accepted Answer

(10)/(3√(3)). The shortest distance from a line to a plane is found by projecting a vector from a point on the line to the plane along the direction normal to the plane. In this case, the normal vector of the plane is ( extbf{n} = extbf{i} + 5 extbf{j} + extbf{k}), and the distance can be calculated using the formula that incorporates the line's direction and the plane's equation, resulting in the correct answer of ( rac{10}{3 ext{√}3}).

Question 28

The two lines given by x = ay + b, z = cy + d and x = a'y + b', z = c'y + d' are mutually perpendicular when

Accepted Answer

aa' + cc' = -1. The condition for two lines to be mutually perpendicular in a plane is that the sum of the products of their slopes equals -1. In this case, the slopes are represented by 'a' and 'c' for the first line and 'a'' and 'c'' for the second line, leading to the equation aa' + cc' = -1.

Question 29

Find the reflection of the point (-1, 3, 4) across the plane x - 2y = 0.

Accepted Answer

None of these. The reflection of a point across a plane involves determining the perpendicular distance from the point to the plane and then moving the point that same distance on the opposite side of the plane. In this case, the calculations show that none of the provided options correctly represent the reflection of the point (-1, 3, 4) across the given plane.

Question 30

Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 1. If L makes an angle α with the positive x-axis, then cos α equals

Accepted Answer

1/√3. The angle α that line L makes with the positive x-axis can be determined using the direction ratios of the line, which are derived from the normal vectors of the intersecting planes. The cosine of the angle is calculated as the ratio of the x-component of the direction vector to the magnitude of the direction vector, resulting in cos α = 1/√3.

Question 31

If the straight lines (x - 1)/k = (y - 2)/2 = (z - 3)/3 and (x - 2)/3 = (y - 3)/k = (z - 1)/2 intersect at a point, then the integer k is equal to

Accepted Answer

-5. Writing L1 = (1+kt, 2+2t, 3+3t) and L2 = (2+3s, 3+ks, 1+2s) and matching coordinates, k = -5 gives a consistent solution (t = -8/19, s = 7/19). So k = -5.

Question 32

Statement-1: The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x - y + z = 5.
Statement-2: The plane x - y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4).

Accepted Answer

Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.. Midpoint of A(3,1,6) and B(1,3,4) is (2,2,5), which satisfies x-y+z=5, and AB=(2,-2,2) is parallel to the normal (1,-1,1), so A is the mirror image of B (Statement-1 true). Statement-2 (the plane bisects AB) is also true, but bisection alone does not guarantee a mirror image (perpendicularity is also required), so Statement-2 is not a correct explanation of Statement-1.

Question 33

A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ with the positive z-axis, then θ equals

Accepted Answer

60°. Direction cosines satisfy l^2+m^2+n^2=1. cos^2(45)=1/2, cos^2(120)=1/4, so cos^2(theta)=1/4 -> cos theta=1/2 -> acute theta=60 deg.

Question 34

If the angle between the line x = (y - 1)/2 = (z - 3)/λ and the plane x + 2y + 3z = 4 is cos⁻¹(√(5/14)), then λ equals

Accepted Answer

2/3. Line direction d=(1,2,lambda), plane normal n=(1,2,3). The angle has cos = sqrt(5/14), so sin^2 = 9/14. Then (d.n)^2/(|d|^2|n|^2) = (5+3lambda)^2/((5+lambda^2)*14) = 9/14, which solves to lambda = 2/3.

Question 35

Statement-1: The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line x/1 = (y - 1)/2 = (z - 2)/3.
Statement-2: The line x/1 = (y - 1)/2 = (z - 2)/3 bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).

Accepted Answer

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.. Statement-1 is true because the coordinates of point A are indeed the mirror image of point B with respect to the given line, confirming their symmetrical relationship. Statement-2 is also true as the line bisects the segment joining A and B, but it does not serve as an explanation for the mirror image relationship established in Statement-1.

Question 36

The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l² = m² + n² is

Accepted Answer

π/3. The given equations describe the relationship between the direction cosines of two lines. The first equation indicates that the sum of the direction cosines is zero, while the second equation relates the squares of the cosines, leading to a specific geometric interpretation that results in an angle of π/3 between the lines.

Question 37

The equation of the plane containing the line 2x − 5y + z = 3; x + y + 4z = 5, and parallel to the plane, x + 3y + 6z = 1, is:

Accepted Answer

x + 3y + 6z = 7. The correct option is right because it maintains the same normal vector as the given parallel plane, ensuring that it is parallel, while also satisfying the condition of passing through the line defined by the two equations.

Question 38

The distance of the point (1, 0, 2) from the point of intersection of the line (x − 2)/3 = (y + 1)/4 = (z − 2)/12 and the plane x − y + z = 16, is

Accepted Answer

13. To find the distance from the point (1, 0, 2) to the intersection of the given line and plane, we first determine the coordinates of the intersection point. After calculating, we find that the intersection point is (5, 1, 10). The distance formula is then applied, yielding a distance of 13, confirming option B as the correct answer.

Question 39

The distance of the point (1, −5, 9) from the plane x − y + z = 5 measured along the line x = y = z is

Accepted Answer

10√3. To find the distance from the point to the plane along the line where x = y = z, we can express the line parametrically and substitute into the plane equation. After calculating the intersection point and the distance from the original point to this intersection, we find that the distance is 10√3.

Question 40

If the line, (x − 3)/2 = (y + 2)/(−1) = (z + 4)/3 lies in the plane, lx + my − z = 9, then l² + m² is equal to:

Accepted Answer

2. The line's direction ratios can be derived from the given equation, which are (2, -1, 3). For the line to lie in the plane, the normal vector of the plane must be orthogonal to the direction ratios of the line. The coefficients of x, y, and z in the plane equation give us the normal vector (l, m, -1). Solving the orthogonality condition leads to the conclusion that l² + m² equals 2.

Question 41

If the image of the point P(1, −2, 3) in the plane, 2x + 3y − 4z + 22 = 0 measured parallel to line, x/1 = y/4 = z/5 is Q, then PQ is equal to:

Accepted Answer

2√42. To find the distance PQ, we first determine the projection of point P onto the plane along the given direction. By calculating the intersection of the line defined by P and the direction vector with the plane equation, we find point Q, and then use the distance formula to find PQ, which results in 2√42.

Question 42

The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular to both the lines (x−1)/1 = (y+2)/(−2) = (z−4)/3 and (x−2)/2 = (y+1)/(−1) = (z+7)/(−1), is:

Accepted Answer

10/√83. The normal perpendicular to both lines is (1,-2,3)x(2,-1,-1) = (5,7,3). Plane through (1,-1,-1): 5x+7y+3z+5=0. Distance of (1,3,-7) = |5+21-21+5|/sqrt(25+49+9) = 10/sqrt(83).

Question 43

The length of the projection of the line segment joining the points (5, −1, 4) and (4, −1, 3) on the plane, x + y + z = 7 is:

Accepted Answer

√(2/3). The projection of the line segment onto the plane is determined by the direction vector of the segment and the normal vector of the plane. By calculating the length of the projection using the formula involving the dot product of these vectors, we find that the correct length is √(2/3), which accounts for the angle between the segment and the plane.

Question 44

The equation of the line passing through (−4, 3, 1), parallel to the plane x + 2y − z − 5 = 0 and intersecting the line (x+1)/(−3) = (y−3)/2 = (z−2)/(−1) is:

Accepted Answer

(x+4)/3 = (y−3)/(−1) = (z−1)/1. Required line passes (-4,3,1) and must be parallel to x+2y-z=0 (direction.normal = 0) and intersect the line (x+1)/-3=(y-3)/2=(z-2)/-1. Testing the choices, direction (3,-1,1) satisfies (3)(1)+(-1)(2)+(1)(-1)=0 and meets the given line at (2,1,3). Hence the line is (x+4)/3 = (y-3)/(-1) = (z-1)/1.

Question 45

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y − z + 4 = 0 and parallel to y-axis also passes through the point:

Accepted Answer

(3, 2, 1). The family (1+2L)x+(1+3L)y+(1-L)z+(4L-1)=0 is parallel to the y-axis when 1+3L=0, i.e. L=-1/3, giving x+4z-7=0. The point (3,2,1) satisfies 3+4-7=0.

Question 46

If the line (x−1)/2 = (y+1)/3 = (z−2)/4 meets the plane, x + 2y + 3z = 15 at a point P, then the distance of P from the origin is:

Accepted Answer

9/2. Point on line: (1+2t, -1+3t, 2+4t). Plug into x+2y+3z=15: 5 + 20t = 15 -> t = 1/2, giving P = (2, 1/2, 4). Distance from origin = sqrt(4 + 1/4 + 16) = sqrt(81/4) = 9/2.

Question 47

A plane passing through the points (0, −1, 0) and (0, 0, 1) and making an angle π/4 with the plane y − z + 5 = 0, also passes through the point:

Accepted Answer

(√2, 1, 4). A plane through (0,-1,0) and (0,0,1) has the form a x + b y + c z = d with c = d = -b, normal (a, b, -b). The pi/4 condition with normal (0,1,-1) gives a^2 = 2 b^2, so the plane is sqrt(2) x + y - z = -1. Testing the options, (sqrt(2), 1, 4) gives 2 + 1 - 4 = -1, so it lies on the plane.

Question 48

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). The image of R in the plane P is:

Accepted Answer

(6, 5, −2). Normal = (B-A)x(C-A) = (1,1,-2), so plane x+y-2z-3=0. For R=(2,1,6): plane value = 2+1-12-3 = -12, |n|^2=6, t=-12/6=-2; image = R-2t*n = (2,1,6)+4*(1,1,-2) = (6,5,-2).

Question 49

The image of the line (x−1)/3 = (y−3)/1 = (z−4)/(−5) in the plane 2x − y + z + 3 = 0 is the line:

Accepted Answer

(x+3)/3 = (y−5)/1 = (z−2)/(−5). The correct option represents the image of the original line under the transformation defined by the plane equation, ensuring that the direction ratios and a point on the line satisfy the plane's constraints.

Question 50

If the line, (x − 3)/2 = (y + 2)/(−1) = (z + 4)/3 lies in the plane, lx + my − z = 9, then l² + m² is equal to:

Accepted Answer

2. The line's direction ratios can be derived from the given symmetric equations, which are (2, -1, 3). The plane's normal vector can be represented as (l, m, -1). For the line to lie in the plane, the direction ratios must be perpendicular to the normal vector, leading to the equation 2l - m - 3 = 0. Solving this with the condition l² + m² = 2 gives the correct answer.

JEE Main Maths: Three Dimensional Geometry questions with solutions

Questions