Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301 Website:www.askiitians.com Email. [email protected]Tel:0120-4616500 Ext - 204 SOLVED PROBLEMS OBJECTIVE 1. The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l 2 + m 2 + n 2 = 0 is (A) π/3 (B) 2π/3 (C) π/4 (D) None of these hb : Eliminating n between the two relations, we have l 2 + m 2 – (l + m) 2 = 0 or 2lm = 0 => either l = 0 or m = 0 if l = 0, then m + n = 0 i,e. m = – n => l/0 = m/1 = n/–1, giving the direction ratios of one line. If m = 0, then l + n = 0 i.e. l = – n => l/0 = m/1 = n/–1, giving direction ratios of the other lines. The angles between these lines is 2. The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is: (A) x – 2y + 11 = 0 (B) x + 2y + 11 = 0 (C) x + 2y – 11 = 0 (D) x – 2y – 11 = 0 hb : Equation of the required plane is (x + y + z – 6) + λ(2x + 3y + z + 5) = 0 i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0 This plane is perpendicular to xy plane whose equation is z = 0
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Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
1. The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l2 + m2 + n2 = 0 is
(A) π/3 (B) 2π/3
(C) π/4 (D) None of these
hb : Eliminating n between the two relations, we have
l2 + m2 – (l + m)2 = 0 or 2lm = 0 => either l = 0 or m = 0
if l = 0, then m + n = 0 i,e. m = – n
=> l/0 = m/1 = n/–1, giving the direction ratios of one line.
If m = 0, then l + n = 0 i.e. l = – n
=> l/0 = m/1 = n/–1, giving direction ratios of the other lines.
The angles between these lines is
2. The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:
(A) x – 2y + 11 = 0 (B) x + 2y + 11 = 0
(C) x + 2y – 11 = 0 (D) x – 2y – 11 = 0
hb : Equation of the required plane is (x + y + z – 6) + λ(2x + 3y + z + 5) = 0
3. The coordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0 are :
(A) (–3/61, 4/61, 6/61) (B) (3/61, –4/61, 6/61)
(C) (–3/61, –4/61, 6/61) (D) (3/61, 4/61, 6/61)
hb : The equation of the plane is 3x + 4y – 6z + 1 = 0 ……(1)
The direction ratios of the normal to the plane (1) are 3, 4, –6. So equation of the line through (0, 0, 0) and perpendicular to the plane (1) are
x/3 = y/4 = z/–6 = r (say) ……(2)
The coordinates of any point P on (2) are (3r, 4r, –6r). If this point lie on the plane (1), then
3(3r) + 4(4r) – 6(–6r) + 1 = 0 i.e. r = –1/61
Putting the value of r coordinates of the foot of the perpendicular P are (–3/61, –4/61, 6/61).
4. 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/–6 is:
(A) 7 unit (B) 4 unit
(C) 1 unit (D) 2 unit
hb : Here we are not to find perpendicular distance of the point from the plane but distance measured along with the given line. The method is as follow:
The equation of the line through the point (1, –2, 3) and parallel to given line is
x–1/2 = y+2/ 3 = z–3/–6 = r (say)
Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
10. The points (0, -1, -1), (-4, 4, 4), (4, 5, 1) and (3, 9, 4) are
(A) collinear (B) coplanar
(C) forming a square (D) none of these hb : Equation of the plane passing through the points (0, -1, -1), (-4, 4, 4) and (4, 5, 1)
is = 0 …. (1)
The point (3, 9, 4) satisfies the equation (1).
Hence (B) is the correct answer. 11. A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. The locus of the point common to plane through A, B, C parallel to coordinate planes is (A) ayz + bzx + cxy = xyz (B) axy + byz + czx = xyz (C) axy + byz + czx = abc (D) bcx + acy + abz = abc hb : Let the equation to the plane be x/α + y/β + z/γ = 1 => x/α + y/β + z/γ = 1 (the plane passes through a, b, c) Now the points of intersection of the plane with the coordinate axes are A(α, 0, 0), B(0, β, 0) & C(0, 0, γ)
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=> Equation to planes parallel to the coordinate planes and passing through A, B & C are x = α, y = β and z = γ. ∴ The locus of the common point is a/x + b/y + c/z = 1 (by eliminating α, β, γ from above equation) Hence (A) is the correct answer. 12. Consider the following statements: Assertion (A): the plane y + z + 1 = 0 is parallel to x-axis. Reason (R): normal to the plane is parallel to x-axis. Of these statements: (A) both A and R are true and R is the correct explanation of A (B) both A and R are true and R is not a correct explanation of A (C) A is true but R is false (D) A is false but R is true hb : Given plane y + z + 1 = 0 is parallel to x-axis as 0.1 + 1.0 + 1.0 = 0 but normal to the plane will be perpendicular to x-axis. Hence (C) is the correct answer. 13. The equation of the plane containing the line x = α/l, y = β/m and z = γ/n is a(x – α) + b(y – β) + c(z – γ) = 0, where al + bm + cn is equal to (A) 1 (B) –1 (C) 2 (D) 0 hb : Since straight line lies in the plane so it will be perpendicular to the normal at the given plane. Since direction cosines of straight line are l, m, n and direction ratios of normal to the plane are a, b, c. So, al + bm + cn = 0. Hence (D) is the correct answer. 14. The shortest distance between the two straight lines x–4/3 / 2 = y+6/5 / 3 = z–3/2 / 4 and 5y+6/8 = 2z–3/9 = 3x–4/5 is (A) √29 (B) 3 (C) 0 (D) 6√10 hb : Since these two lines are intersecting so shortest distance between the lines will be 0. Hence (C) is the correct answer. 15. A straight line passes through the point (2, –1, –1). It is parallel to the plane 4x + y + z + 2 = 0 and is perpendicular to the line x/1 = y/–2 = z–5/1. The equations of the straight line are (A) x–2/4 = y+1/1 = z+1/1 (B) x+2/4 = y–1/1 = z–1/1 (C) x–2/–1 = y+1/1 = z+1/3 (D) x+2/–1 = y–1/1 = z–1/3 hb : Let direction cosines of straight line be l, m, n ∴ 4l + m + n = 0 l – 2m + n = 0 => l/3 = m/–3 = n/–9 => l/–1 = m/+1 = n/3 ∴ Equation of straight line is x–2/–1 = y+1/1 = z+1/3.
Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
Hence (C) is the correct choice. 16. If centre of a sphere is (1, 4, –3) and radius is 3 units, then the equation of the sphere is (A) x2 + y2 + z2 – 2x – 8y + 6z + 17 = 0 (B) 2(x2 + y2 + z2) – 2x – 8y + 6z + 17 = 0 (C) x2 + y2 + z2 – 4x + 16y + 12z + 17 = 0 (D) x2 + y2 + z2 + 2x + 8y – 6z – 17 = 0 hb : Equation of sphere will be (x – 1)2 + (y – 4)2 + (z + 3)2 = 9 Hence (A) is the correct answer. 17. If equation of a sphere is 2(x2 + y2 + z2) – 4x – 8y + 12z – 7 = 0 and one extremity of its diameter is (2, –1, 1), then the other extremity of diameter of the sphere will be (A) (2, 9, –13) (B) (0, 9, 7) (C) (0, 5, 7) (D) (2, 5, –13) hb : The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then x1+2/2 = 1, y1–1/2 = 2, z1+1/2 = –3 ∴ Required point is (0, 5, 7). Hence (C) is the correct answer. 18. The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to (1, – 2, – 2), (0, 2, 1) is (A) (2/3, 1/3, 2/3) (B) (2/3, –1/3, –2/3) (C) (2/3, 1/3, –2/3) (D) (–2/3, –1/3, –2/3) hb : Let direction ratios of the required line be Therefore a - 2 b - 2 c = 0 And 2 b + c = 0 => c = - 2 b a - 2 b + 4b = 0 => a = - 2 b Therefore direction ratios of the required line are <- 2b, b, - 2b> = <2, - 1, 2> direction cosines of the required line
= 19. The points (4, 7, 8), (2, 3, 4), (–1, –2, 1) and (1, 2, 5) are : (A) the vertices of a parallelogram (B) collinear (C) the vertices of a trapezium (D) concyclic hb : Let A Ξ (4, 7, 8), B Ξ (2, 3, 4), C Ξ (- 1, - 2, 1), D Ξ (1, 2, 5) Direction cosines of AB Ξ (2/6, 4/6, 4/6) = (1/3, 2/3, 2/3) Direction cosines of CD Ξ (–2/6, –4/6, –4/6) = (1/3, 2/3, 2/3) So, AB parallel to CD Direction cosines of AD Ξ (3/√43, 5/√43, 3/√43) Direction cosines of BC Ξ (–3/√43, –5/√43, –3/√43) = (3/√43, 5/√43, 3/√43) so, AD is parallel to BC.
Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
hb . Foot of perpendicular drawn from P to x-axis will have its coordinates as (x, 0, 0).
Required distance = √y12 + z1
2
36. The point of intersection of the xy plane and the line passing through the points and is:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.4)
(A) (–13/5, 23/5, 0) (B) (13/5, 23/5, 0)
(C) (13/5, –23/5, 0) (D) (–13/5, –23/5, 0)
Ans. (B)
hb . Direction ratios of AB are 2, -3, 5.
Thus equation of AB is, x–3/2 = y–4/–3 = z–1/5
For the point of intersection of this line with xy–plane, we have
Z = 0
=> x–3/2 = y–4/–3 = –1/5
=> x=3 –2/5 = 13/5, y = 4 + 3/5 = 23/5
Hence, the required point is (15/5, 23/5, 0)
37. The projections of the line segment AB on the coordinate axes are –9, 12, -8 respectively. The direction cosines of the line segment AB are:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.5)
38. The direction cosines of two mutually perpendicular lines are l1,m1,n1 and l2,m2,n2. The direction cosines of the line perpendicular to both the given lines will be:(Ref.P.K.Sharma_Three Dimen._P.C6.1Q.6)
(A) meet in a unique point (B) meet in a unique line
(C) are mutually perpendicular (D) none of these
Ans. (B) Clearly, given planes have a common line of intersection namely the z-axis.
43. The equation of a plane passing through (1, 2, -3), (0, 0, 0) and perpendicular to the plane 3x – 5y + 2z = 11, is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.13)
(A) 3x + y + 5/3 z = 0 (B) 4x + y + 2z = 0
(C) 3x – y + z/3 (D) x + y + z = 0
Ans. (D)
Let the required plane be
Ax + by + cz = 0.
We have 3a – 5b + 2c = 0, a + 2b – 3c = 0
=> a/15–4 = b/2+9 = c/6+5
=> a : b : c = 11 : 11 : 11
Thus plan is x + y + z = 0
44. The direction ration of a normal to the plane passing through (1, 0, 0), (0, 1, 0) and making an angle π/4 with the plane x + y = 3 are:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.15)
(A) (1, √2, 1) (B) (1, 1,√2)
(C) (1,1,2) (D) (√2, 1, 1)
Ans. (B)
Let the plane be x/a + y/b + z/c = 1
=> 1/a = 1, 1/b = 1
.=> a = b = 1
Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
45. The equation of a plane passing through the line of intersection of the planes x + y + z = 5, 2x – y + 3z = 1 and parallel to the line y = z = 0 is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.16)
(A) 3x – z = 9 (B) 3y – z = 9
(C) x – 3z = 9 (D) y – 3z = 9
Ans. (B)
Plane will be in the form
(x + y + z – 5) + a(2x – y + 3z) –1) = 0 i.e., x(1 + 2a) + y(1 – a) + z(1 + 3a) = 5 + a
It is parallel to the line y = z = 0.
Since, (1 + 2a) = 0
∴ a = – 1/2
Thus required plane is
3/2y – 1/2 z = 9/2
i.e., 3y – z = 9
46. The angle between lines whose direction cosines are given by l + m + n = 0, l2 + m2 – n2= 0, is:(Ref.P.K.Sharma_Three Dimen._P.C6.2Q.17)
(A) π/2 (B) π/3
(C) π/6 (D) None of these
Ans. (D)
Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
48. The equation of the plane passing through the points (2, -1, 0), (3, -4, 5) and parallel to the line 2x = 3y = 4z is:(Ref.P.K.Sharma_Three Dimen._P.C6.3Q.20)
49. A plane passes through the point . If the distance of this plane from the origin is maximum, then it’s equation is:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.1)
(A) x + 2y – 3z + 4 = 0 (B) x + 2y + 3z = 0
(C) 2y - x + 3z = 0 (D) x - 2y + 3z = 0
Ans. (B) Clearly in this case OA will be a normal to the plane.
Direction cosine of segment OA are
Trans Web Educational Services Pvt. Ltd B – 147,1st Floor, Sec-6, NOIDA, UP-201301
52. The straight lines x–2/1 = y–3/1 = z–4/–k and x–1/k = y–4/2 = z–5/1, will intersect provided:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.6)
(A) k = {3, -3} (B) k = {0, -1}
(C) k = {-1, 1} (D) k = {0, -3}
Ans. (D)
Any point on the first line can be takes as
P1 = (r1 + 2, r1 + 3, – kr1+4)
These lines will intersect if for some r1 and r2 we have
r1 + 2 = kr2 + 1,
r1 + 3 = 2r2 + 1,
–kr1 + 4 = r2 + 5,
∴ r1 + kr2 + 1 = 0, r1 = 2r2 + 1,
=> r2 = 2/k–2, r1 = k+2 / k–2k
putting these values in the last condition, we get
k2 + 3k = 0
.=> k = {–3,0}
53. The plane ax + by + cz = d, meets the coordinate axes at the points, A, B and C respectively. Area of triangle ABC is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.4Q.7)
Equation of any plane containing the line x/1 = y–2/3 = z+4/–1 is ax + b (y–2) + c (z+4) = 0
where a + 3b – c = 0
This plane will also contain the second line if
2a – 3b + c = 0
and 4a b(0 – 2) + c(0 + 4) = 0
Solving these equation, we get
a = 0, b = 0, c = 0.
That means the given lines are non–coplanar.
56. Equation of the plane such that foot of altitude drawn from (-1, 1, 1) to the plane has the coordinate (3, -2, -1), is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.11)
(A) x + y + z = 0 (B) 4x - 3y – 2z = 20
(C) 3x + y – z = 8 (D) 4x + 3y – z = 7
Ans. (B)
Clearly, the direction ratio of the plane are
4, –3, –2
Thus equation of plane will be
4x – 3y – 2z = d
It will necessarily pass through (3, –2, –1)
i.e,, d = 12 + 6 + 2 = 20
Thus the equation of plane is
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59. The shortest distance between the line x + y + 2z – 3 = 2x + 3y + 4z – 4 = 0 and the z-axis is:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.15)
(A) 1 unit (B) 2 units
(C) 3 units (D) 4 units
Ans. (B)
We have,
x + y + 2z – 3 = 0, x + 2z – 2 + 3/2 y = 0
Solving these equations, we get
Y = –2.
Thus required shortest distance is 2 units.
60. The length of projection, of the line segment joining the points (1, -1, 0) and (-1, 0, 1), to the plane 2x + y + 6z = 1, is equal to:(Ref.P.K.Sharma_Three Dimen._P.C6.5Q.17)
(A) √255/61 (B) √237/61
(C) √137/61 (D) √155/61
Ans. (B)
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