Question 1

Passage: The equation of the sphere S is x^2+y^2+z^2-4x-6y-12z+k=0. Question: If the radius of the sphere S is 8 units, what is the value of k ?

Accepted Answer

The given equation of the sphere is x^2+y^2+z^2-4x-6y-12z+k=0. Comparing this with the general equation of a sphere x^2+y^2+z^2+2ux+2vy+2wz+d=0, we get: u = -2, v = -3, w = -6, and d = k. The radius of the sphere is given by r = u^2+v^2+w^2-d . Substituting the given values: 8 = (-2)^2+(-3)^2+(-6)^2-k 8 = 4+9+36-k 8 = 49-k Squaring both sides: 64 = 49-k k = 49-64 = -15 Answer: -15

Question 2

Passage: A line L passing through the point (-1,2,-3) is perpendicular to the plane P given by 2x+3y+z+5=0. Question: What is the equation of the line L ?

Accepted Answer

The line L is perpendicular to the plane 2x+3y+z+5=0, so its direction ratios are proportional to the normal vector of the plane, which is (2, 3, 1). The equation of the line passing through (-1, 2, -3) with direction ratios (2, 3, 1) is given by: x+1 2 = y-2 3 = z+3 1 Multiplying the entire equation by 6, we get: 3(x+1) = 2(y-2) = 6(z+3) 3x+3 = 2y-4 = 6z+18 Answer: 3x+3=2y-4=6z+18

Question 3

Passage: P(-2,-3,5), Q(4,-1,5), R(6,-4,8) and S(2,-6,10) are four points. Question: Consider the following statements : I. PQ is parallel to RS. II. PR is perpendicular to QS. Which of the statements

Accepted Answer

The direction ratios of the line segment PQ are 4 - (-2), -1 - (-3), 5 - 5, which gives 6, 2, 0 . The direction ratios of the line segment RS are 2 - 6, -6 - (-4), 10 - 8, which gives -4, -2, 2 . Since the ratios of their corresponding direction ratios are not equal ( 6 -4 2 -2 0 2 ), PQ is not parallel to RS. Thus, Statement I is incorrect. The direction ratios of the line segment PR are 6 - (-2), -4 - (-3), 8 - 5, which gives 8, -1, 3 . The direction ratios of the line segment QS are 2 - 4, -6

Question 4

Passage: A line L passing through the point (-1,2,-3) is perpendicular to the plane P given by 2x+3y+z+5=0. Question: What are the direction ratios of a line M parallel to the plane P ?

Accepted Answer

The plane P is given by 2x + 3y + z + 5 = 0. The direction ratios of the normal to the plane P are 2, 3, 1 . Since the line M is parallel to the plane P, its direction vector must be perpendicular to the normal vector of the plane. Let the direction ratios of the line M be a, b, c . Then, 2a + 3b + c = 0. Checking the given options: For -3, 2, 1 , 2(-3) + 3(2) + 1 = 1 0 For 3, 2, -6 , 2(3) + 3(2) - 6 = 6 0 For 1, 3, 2 , 2(1) + 3(3) + 2 = 13 0 For 2, 2, -10 , 2(2) + 3(2) - 10 = 0 Thus, the direct

Question 5

Passage: P(-2,-3,5), Q(4,-1,5), R(6,-4,8) and S(2,-6,10) are four points. Question: What is (PQ^2+2QS^2-2PR^2) equal to ?

Accepted Answer

Given points are P(-2,-3,5), Q(4,-1,5), R(6,-4,8) and S(2,-6,10). Using the distance formula, we find the squares of the distances between the required points: PQ^2 = (4 - (-2))^2 + (-1 - (-3))^2 + (5 - 5)^2 = 6^2 + 2^2 + 0^2 = 36 + 4 = 40 QS^2 = (2 - 4)^2 + (-6 - (-1))^2 + (10 - 5)^2 = (-2)^2 + (-5)^2 + 5^2 = 4 + 25 + 25 = 54 PR^2 = (6 - (-2))^2 + (-4 - (-3))^2 + (8 - 5)^2 = 8^2 + (-1)^2 + 3^2 = 64 + 1 + 9 = 74 Substituting these values into the given expression: PQ^2 + 2QS^2 - 2PR^2 = 40 + 2(5

Question 6

Passage: The equation of the sphere S is x^2+y^2+z^2-4x-6y-12z+k=0. Question: What is the radius of the sphere passing through origin and concentric with the sphere S ?

Accepted Answer

The equation of the given sphere is x^2+y^2+z^2-4x-6y-12z+k=0. The center of the sphere is (2, 3, 6). The required sphere is concentric with the given sphere, so its center is also (2, 3, 6). Since the required sphere passes through the origin (0, 0, 0), its radius is the distance between the center and the origin. R = (2-0)^2 + (3-0)^2 + (6-0)^2 R = 4 + 9 + 36 = 49 = 7 Answer: 7

Question 7

For the following two (02) items: A plane P is parallel to the line having direction ratios 1, 3, 2 and contains the line of intersection of the planes 6x + 4y - 5z = 2 and x - 2y + 3z = 0. What is th

Accepted Answer

The equation of the plane passing through the line of intersection of the given planes is (6x + 4y - 5z - 2) + (x - 2y + 3z) = 0 (6 + )x + (4 - 2 )y + (-5 + 3 )z - 2 = 0 Since this plane is parallel to the line with direction ratios 1, 3, 2 , the normal to the plane is perpendicular to the line. Thus, 1(6 + ) + 3(4 - 2 ) + 2(-5 + 3 ) = 0 6 + + 12 - 6 - 10 + 6 = 0 + 8 = 0 = -8 Substituting = -8 in the equation of the plane, we get (6 - 8)x + (4 - 2(-8))y + (-5 + 3(-8))z - 2 = 0 -2x + 20y - 29z -

Question 8

For the following two (02) items: Suppose S is the sphere with the smallest radius that passes through the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1). On which one of the following planes does the c

Accepted Answer

The sphere S passes through the points A(1, 0, 0), B(0, 1, 0), and C(0, 0, 1). The equation of the plane containing these three points is given by the intercept form: x 1 + y 1 + z 1 = 1 x + y + z - 1 = 0 Any sphere passing through A, B, and C will intersect this plane in a circle, which is the circumcircle of ABC. For the sphere to have the smallest possible radius, this circumcircle must be a great circle of the sphere. This means the centre of the sphere must lie exactly on the plane containi

Question 9

A(1, 2, -1), B(2, 5, -2) and C(4, 4, -3) are three vertices of a rectangle. What is the area of the rectangle?

Accepted Answer

The given vertices of the rectangle are A(1, 2, -1), B(2, 5, -2) and C(4, 4, -3). The direction ratios of the side AB are (2-1, 5-2, -2-(-1)) = (1, 3, -1). The length of the side AB is 1^2 + 3^2 + (-1)^2 = 1 + 9 + 1 = 11 . The direction ratios of the side BC are (4-2, 4-5, -3-(-2)) = (2, -1, -1). The length of the side BC is 2^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6 . Now, checking the dot product of vectors AB and BC : AB BC = (1)(2) + (3)(-1) + (-1)(-1) = 2 - 3 + 1 = 0 Since the dot product is zer

Question 10

If a line x+1 p = y-1 q = z-2 r where p = 2q = 3r, makes an angle with the positive direction of y-axis, then what is 2 equal to?

Accepted Answer

The direction ratios of the line are proportional to p, q, r. Given p = 2q = 3r. Let p = 2q = 3r = 6k. Then p = 6k, q = 3k, r = 2k. The direction ratios of the line are proportional to 6, 3, 2. The direction cosines of the line are 6 6^2+3^2+2^2 , 3 6^2+3^2+2^2 , 2 6^2+3^2+2^2 , which simplifies to 6 7 , 3 7 , 2 7 . The line makes an angle with the positive direction of the y-axis, so = 3 7 . Using the double angle formula, 2 = 2 ^2 - 1. 2 = 2 ( 3 7 )^2 - 1 = 18 49 - 1 = - 31 49 . Answer: - 31 4

Question 11

What is the equation of the plane passing through the point (1, 1, 1) and perpendicular to the line whose direction ratios are 3, 2, 1 ?

Accepted Answer

The direction ratios of the normal to the plane are proportional to the direction ratios of the line perpendicular to it. Thus, the normal vector has direction ratios 3, 2, 1 . The equation of a plane passing through a point (x_1, y_1, z_1) with normal direction ratios a, b, c is given by a(x - x_1) + b(y - y_1) + c(z - z_1) = 0. Substituting (x_1, y_1, z_1) = (1, 1, 1) and a, b, c = 3, 2, 1 , we get: 3(x - 1) + 2(y - 1) + 1(z - 1) = 0 3x - 3 + 2y - 2 + z - 1 = 0 3x + 2y + z = 6 Answer: 3x + 2y

Question 12

For the following two (02) items: Let A(1, -1, 0), B(-2, 1, 8) and C(-1, 2, 7) are three consecutive vertices of a parallelogram ABCD. What is the fourth vertex D?

Accepted Answer

Let the coordinates of the fourth vertex D be (x, y, z). In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD. Midpoint of AC = ( 1 - 1 2 , -1 + 2 2 , 0 + 7 2 ) = ( 0, 1 2 , 7 2 ) Midpoint of BD = ( -2 + x 2 , 1 + y 2 , 8 + z 2 ) Equating the corresponding coordinates: -2 + x 2 = 0 x = 2 1 + y 2 = 1 2 y = 0 8 + z 2 = 7 2 z = -1 The coordinates of the fourth vertex D are (2, 0, -1). Answer: (2, 0, -1)

Question 13

For the following two (02) items: Suppose S is the sphere with the smallest radius that passes through the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1). What is the radius of S?

Accepted Answer

The points A(1, 0, 0), B(0, 1, 0), and C(0, 0, 1) form an equilateral triangle in 3D space. The side length of this triangle is AB = (1-0)^2 + (0-1)^2 + (0-0)^2 = 2 . The sphere with the smallest radius passing through three given points has its center at the circumcenter of the triangle formed by these points, and its radius is equal to the circumradius of the triangle. For an equilateral triangle of side length a, the circumradius is given by R = a 3 . Substituting a = 2 , we get R = 2 3 = 2 3

Question 14

If a line in 3 dimensions makes angles , and with the positive directions of the coordinate axes, then what is ( + ) ( - ) equal to?

Accepted Answer

Using the trigonometric identity (A+B) (A-B) = ^2 A - ^2 B, we have: ( + ) ( - ) = ^2 - ^2 Since ^2 = 1 - ^2 , the expression becomes: ^2 - (1 - ^2 ) = ^2 + ^2 - 1 For a line in three-dimensional space making angles , , and with the coordinate axes, the sum of the squares of its direction cosines is 1: ^2 + ^2 + ^2 = 1 ^2 + ^2 - 1 = - ^2 Substituting this into the earlier expression, we get - ^2 . Answer: - ^2

Question 15

A line makes angles , and with the positive directions of the coordinate axes. If a = ( ^2 ) i + ( ^2 ) j + ( ^2 ) k and b = i + j + k , then what is a b equal to?

Accepted Answer

Let , , and be the direction cosines of the given line. We know that the sum of the squares of the direction cosines is 1: ^2 + ^2 + ^2 = 1 The given vectors are: a = ( ^2 ) i + ( ^2 ) j + ( ^2 ) k b = i + j + k The dot product a b is given by: a b = ( ^2 )(1) + ( ^2 )(1) + ( ^2 )(1) a b = ^2 + ^2 + ^2 Using the identity ^2 = 1 - ^2 , we get: a b = (1 - ^2 ) + (1 - ^2 ) + (1 - ^2 ) a b = 3 - ( ^2 + ^2 + ^2 ) Substituting ^2 + ^2 + ^2 = 1: a b = 3 - 1 = 2 Answer: 2

Question 16

For the following two (02) items: A plane P is parallel to the line having direction ratios 1, 3, 2 and contains the line of intersection of the planes 6x + 4y - 5z = 2 and x - 2y + 3z = 0. Which of t

Accepted Answer

The normal vectors of the given planes are n _1 = 6, 4, -5 and n _2 = 1, -2, 3 . The direction vector d of the line of intersection of the planes is given by the cross product of their normal vectors: d = n _1 n _2 = vmatrix i & j & k \ 6 & 4 & -5 \ 1 & -2 & 3 vmatrix d = i (12 - 10) - j (18 + 5) + k (-12 - 4) d = 2 i - 23 j - 16 k The direction ratios of the line of intersection are 2, -23, -16 . Answer: 2, -23, -16

Three Dimensional Geometry — Previous Year Questions & Solutions