FAQ THREE DIMENSIONAL GEOMETRY

1 Find the distance of the point (2, 3, 4) from the plane

ANS: Distance of the point (2, 3, 4) from the plane

2 Write the intercept cut off by the plane 2x + y

ANS: For intercept on the x-axis, put y = 0 and

3 Find the angle between the planes

ANS:

THREE DIMENSIONAL GEOMETRY

Find the distance of the point (2, 3, 4) from the plane

Distance of the point (2, 3, 4) from the plane is

y – z = 5 on the x-axis.

= 0 and z = 0

1

1

1

4 If a line has direction ratios 2, –1, –2, then what are its direction cosines?

ANS: Dr’s are 2, – 1, – 2 Dividing by

Dc’s are

5 Find the vector normal to the plane

ANS:

6 Write the vector equation of the plane passing through the point (

ANS: Plane parallel to the plane

If this plane passes through the point (a, b, c

[from (i)]

a + b + c + l = 0 = – (a + b + c)

From (i), Plane is

2, then what are its direction cosines?

Write the vector equation of the plane passing through the point (a, b, c) and parallel to the plane

is ...(i)

c), i.e. point with position vector , then

1

1

1

7 Find the sum of the intercepts cut off by the plane 2

ANS: Plane is 2x + y – z = 5 Intercept form is

8 Find distance of a point (2, 5, –3) from the plan

ANS: Distance of point (2, 5, –3) from plane

= units

9

Write the direction ratios of the following line:

ANS: Given line is x = –3,

the sum of the intercepts cut off by the plane 2x + y – z = 5, on the coordinate axes.

= 5 Intercept form is Sum of intercepts =

3) from the plan

3) from plane = 4 is

Write the direction ratios of the following line: x = –3,

Direction ratios are 0, 3, –1.

1

1

=

1

10 Find the Cartesian equation of the line which passes through the point (

ANS: As given line is parallel to the line

direction ratios of the line are 3, –5, 6. Line passes through the point (

11

Write the vector equation of the following line:

ANS: The line passes through point (5, – 4, 6) and dr’s of the line are 3, 7,

vector equation is

12 What are the direction cosines of a line, which makes equal angles with the coordinate axes?

ANS:

Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is parallel to the line

5, 6. Line passes through the point (–2, 4, –5) Cartesian equation of the line is

Write the vector equation of the following line:

4, 6) and dr’s of the line are 3, 7, – 2.

cosines of a line, which makes equal angles with the coordinate axes?

2

Cartesian equation of the line is

2

2

13

Find the angle between the line

ANS:

14 Using dire ction ratios, show that the points (2, 3, 4), (

ANS: Let points be A(2, 3, 4), B(– 1, – 2, 1) and

Direction ratios of AB are 2 + 1, 3 + 2, 4 – 1, i.e. 3, 5, 3;

Direction ratios of BC are 5 + 1, 8 + 2, 7 – 1, i.e. 3, 5, 3 As

AB is parallel to BC, B is common. Hence, the given points are collinear.

and the plane 10x + 2y – 11z = 3.

ratios, show that the points (2, 3, 4), (–1, –2, 1) and (5, 8, 7) arecollinear.

2, 1) and C(5, 8, 7)

1, i.e. 3, 5, 3;

1, i.e. 3, 5, 3 As

is common. Hence, the given points are collinear.

2

2

15 Find the coordinates of a point, where the line

ANS: General point on the line is (λ – 2, 3λ +

If it lies on the YZ-plane then λ – 2 = 0 λ = 2.

point of intersection is (0, 6 + 5, 10 – 1), i.e. (0, 11, 9).

16 If a line makes ang le 90°, 60° and θ with x, y

ANS:

:

17 If α, β, γ are the angles that a line makes with

ANS: α, β, γ are the angles which a line makes with

cos2α + cos2β + cos2γ = 1

1 – sin2α + 1 – sin2β + 1 – sin2γ = 1

sin2α + sin2β + sin2γ = 2

Find the coordinates of a point, where the line cuts YZ-plane.

2, 3λ + 5, 5λ – 1).

λ = 2.

1), i.e. (0, 11, 9).

y and z-axes respectively, where θ is acute, then find θ.

If α, β, γ are the angles that a line makes with x, y and z-axes respectively then find the value of sin2α + sin2

α, β, γ are the angles which a line makes with x, y and z – axes

2

2

2β + sin2γ. 2

18

Find the angle between the straight lines

ANS:

.

2

19 Show that the line through the points (1, – 1, 2) and (3, 4,

points (0, 3, 2) and (3, 5, 6).

ANS: Direction ratios of line through the points (1 ,

i.e. 2, 5, – 4 ...(i)

Direction ratios of line through the points (0, 3, 2) and (3, 5, 6) are 3 From (i) and (ii), as 3 × 2 + 5 × 2 – 4 × 4 = 0.

Hence, lines are perpendicular.

20 The x-coordinate of a point P on the line joining the points

ANS: Line is

General point on the line is (3λ + 2, – λ + 2, –

λ = ; z-coordinate is – 2 + 1 = – 1.

1, 2) and (3, 4, – 2) is perpendicular to the line through the

Direction ratios of line through the points (1 , – 1, 2) and (3, 4, – 2) are 3 – 1, 4 + 1,– 2 – 2,

3, 2) and (3, 5, 6) are 3 – 0, 5 – 3, 6 – 2, i.e. 3, 2, 4 4 × 4 = 0.

on the line joining the points Q(2, 2, 1) and R(5, 1, –2) is 4. Find its z-coordinate.

– 3λ + 1) 3λ + 2 = 4

2

...(ii)

coordinate. 2

21 The equations of a line are 5x – 3 = 15y + 7 = 3

ANS:

22 Find the distance between the parallel planes 2

ANS: Parallel planes are 2x – y + 3z – 4 = 0

And 6x – 3y + 9z + 13 = 0 or 2x – y + 3z +

Distance between plane (i) and (ii)

+ 7 = 3 – 10z. Write the direction cosines of the line.

Find the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0.

4 = 0 ...(i)

= 0 ...(ii)

2

2

23. Show that the line

ANS:

24. Find the point on the line

ANS: General point on the line

Distance of point A from point (1, 2, 3) is

Squaring and solving for λ, we get λ = 0 or λ =

Substituting in (i), we get point as A(

lies in the plane

at a distance from the point (1, 2, 3).

= λ is A(3λ – 2, 2λ – 1, 2λ + 3) ...(

Squaring and solving for λ, we get λ = 0 or λ =

(–2, –1, 3) or A .

2

4

...(i)

25. Find the length and the foot of the perpendicular drawn from the point (2,

ANS:

General point on the line

(say) is

Q(10λ + 11, –4λ – 2, –11λ – 8)

Direction ratios of PQ are

10λ + 11 – 2, –4λ – 2 + 1, –11λ – 8 – 5

i.e. 10λ + 9, –4λ – 1, –11λ – 13

If PQ is perpendicular to the given line, then

10(10λ + 9) – 4 (–4λ – 1) – 11 (–11λ – 13) = 0

237λ = – 237 λ = –1

Substituting in (i), we get the foot of perpendicular as

Length of perpendicular PQ =

Find the length and the foot of the perpendicular drawn from the point (2, –1, 5) on the line

= λ

...(i)

), we get the foot of perpendicular as Q(1, 2, 3).

4

26.

Find the value of λ, so that the lines each other.

ANS: Consider, line

Direction ratios of line are –105, 10λ, 77.

Now consider, line

Direction ratios of line are 3λ, 35 If lines are perpendicular, then (

– 315λ – 70λ + 2695 = 0 385λ = 2695

and are perpendicular to

105, 10λ, ...(i)

Direction ratios of line are 3λ, –7, ...(ii)

If lines are perpendicular, then (–105) (3λ) + (10λ) (–7) + 77 × 35 = 0

385λ = 2695 λ = 7.

are perpendicular to 4

27. Find the equation of a line passing through the point

and

ANS:

28. Find the equation of the plane which is perpendicular to the intersection of the planes x + 2y + 3z – 4 = 0 and 2

ANS: General equation of plane is x + 2y

or (1 + 2λ)x + (2 + λ)y + (3 – λ)z + (– 4 + 5λ) = 0

Plane (i) is perpendicular to the plane 5x + 3

Find the equation of a line passing through the point P(2, – 1, 3) and perpendicular to the lines:

.

Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of 4 = 0 and 2x + y – z + 5 = 0.

y + 3z – 4 + λ(2x + y – z + 5) = 0

4 + 5λ) = 0 ...(i)

+ 3y + 6z + 8 = 0 5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0

4

+ 8 = 0 and which contains the line of 4

7λ = – 29 λ = – Substituting in (i), we get 51x

29. Find the image of the point having position vector

ANS:

x + 15y – 50z + 173 = 0 as the required equation of the plane.

Find the image of the point having position vector in the plane = 0.

4

30. Find the equation of the line passing through the point

and the plane x + y – z = 8.

ANS:

Find the equation of the line passing through the point P(4, 6, 2) and the point of intersection of the line

= 8.

4

31. Find the Cartesian equation of the plane passing through the points

.

ANS: Plane passing through the point A(0, 0, 0) is a(

Plane (i) passes through the point (3, –1, 2)

3a – b + 2c = 0

Plane (i) is parallel to the line

a – 4b + 7c = 0

Eliminating a, b, c from (i), (ii) and (iii), we get

x(–7 + 8) – y(21 – 2) + z(–12 + 1) = 0 x – 19y – 11z = 0 is the required equation.

Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, –1, 2) and parallel to the line

(0, 0, 0) is a(x – 0) + b(y – 0) + c(z – 0) = 0 ...(i)

1, 2)

...(ii)

...(iii)

), we get = 0

12 + 1) = 0 = 0 is the required equation.

1, 2) and parallel to the line

4

32. Find the shortest distance between the lines whose vector equations are

ANS:

Find the shortest distance between the lines whose vector equations are

4

33 If a line makes angles α, β and γ with x, y and

(i) sin2 α + sin2 β + sin2 γ = 2

(ii) cos 2α + cos 2β + cos 2γ = –1.

and z-axes respectively, prove that

4

ANS: α, β and γ are the angles with x, y and

Therefore, cos α, cos β, cos γ are direction cosines.

cos2 α + cos2 β + cos2 γ = 1.

Also sin2 α + cos2 α = 1 and cos 2α = 2 cos

(i) sin2 α + sin2 β + sin2 γ = (1 – cos2 α) + (1

= 3 – (cos2 α + cos2 β + cos2 γ) = 3 – 1 = 2.

(ii) cos 2α + cos 2β + cos 2γ = (2 cos2 α – 1) + (2 cos

= 2(cos2 α + cos2 β + cos2 γ) – 3 = 2(1) –

34. Find the value of λ for which the points with

the plane .

ANS: We have

|5 – 2 – 21 + 9| = |15 + 2λ – 21 + 9| 2λ + 3 = ±9 λ = 3, –6

35. Find the distance between the point (7, 2, 4) and the plane determined by the points

A(2, 5, –3), B(–2, –3, 5) and C(5, 3, –3).

and z-axes.

γ are direction cosines.

α = 1 and cos 2α = 2 cos2 α – 1

α) + (1 – cos2 β) + (1 – cos2 γ)

1 = 2.

1) + (2 cos2 β – 1) + (2 cos2 γ – 1).

3 = –1

Find the value of λ for which the points with position vectors and are equidistant from

=

21 + 9| |–9| = |2λ + 3| 6

Find the distance between the point (7, 2, 4) and the plane determined by the points

are equidistant from 4

4

ANS: Plane determined by points A(2, 5,

= 0

(x – 2)(16) – (y – 5)(–24) + (z + 3)(32) = 0 16x – 32 + 24y – 120 + 32z + 96 = 0

16x + 24y + 32z – 56 = 0

Distance of the point (7, 2, 4) from the plane 2

36

Find the coordinates of the point, where the line the angle between the line and the plane.

(2, 5, –3), B(–2, –3, 5) and C(5, 3, –3) is

+ 3)(32) = 0 + 96 = 0

2x + 3y + 4z – 7 = 0

Distance of the point (7, 2, 4) from the plane 2x + 3y + 4z – 7 = 0 is

Find the coordinates of the point, where the line intersects the plane x – y + z – 5 = 0. Also find

= 0

5 = 0. Also find 4

ANS: General point on the line

A(3λ + 2, 4λ – 1, 2λ + 2) on the plane x – y + z – 5 = 0,

then 3λ + 2 – 4λ + 1 + 2λ + 2 – 5 = 0

λ = 0 coordinates of point of intersection are

37. Find the distance of the point (2, 12, 5) from the point of intersection of the line

the plane .

= λ (say) is

...(i) If point

coordinates of point of intersection are A(2, –1, 2). Angle between the line and the plane is

Find the distance of the point (2, 12, 5) from the point of intersection of the line

If point A lies

1, 2). Angle between the line and the plane is

and 4

ANS:

38 Find the distance of the point (– 1, – 5, – 10) from the plane

.

10) from the plane x – y + z = 5 measured parallel to the line

4

ANS: Let AP be parallel to the given line, equation of

General point P on AP is (3λ – 1, 4λ – 5, 12λ

If this lies on the plane then 3λ – 1 – 4λ + 5 + 12λ

λ = 1 Substituting in (i) we get point of intersection as (2,

Distance AP =

39. Find the image of the point (1, 6, 3) in the line

ANS: Let B(α, β, γ) be the image of point

be parallel to the given line, equation of AP is

5, 12λ – 10) ...(i)

4λ + 5 + 12λ – 10 = 5

) we get point of intersection as (2, – 1, 2)

= 13 units

Find the image of the point (1, 6, 3) in the line .

(α, β, γ) be the image of point A(1, 6, 3) in the line .

4

Let AB meets the line at R Then R is mid-point of

Coordinates of R are

General point on line is R(λ, 2λ + 1, 3λ + 2)

Direction ratios of AR are λ – 1, 2λ + 1 – 6, 3λ + 2

As AR is perpendicular to the line then 1(λ –

λ = 1 Substituting in (ii), we get foot of perpendicular as

From (i),

α = 1, β = 0, γ = 7. Image is B(1, 0, 7)

40. Find the equation of a plane which passes through the point (3, 2, 0) and contains the line

ANS: General equation of the plane passes through the point (3, 2, 0) is

point of AB and R is foot of the perpendicular from A to the line.

...(i)

...(ii)

6, 3λ + 2 – 3, i.e., λ – 1, 2λ – 5, 3λ – 1.

– 1) + 2(2λ – 5) + 3(3λ – 1) = 0 14λ = 14

), we get foot of perpendicular as R(1, 3, 5)

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line

General equation of the plane passes through the point (3, 2, 0) is

. 4

a(x – 3) + b(y – 2) + c(z – 0) = 0

As it contains the line

point (3, 6, 4) on line lies on the plane a(0) + b(4) + c(4) = 0 [from (i)] or 0 Line is perpendicular to normal to the plane

a + 5b + 4c = 0

Eliminating a, b and c from (i), (ii) and (iii

(x – 3) (–1) – (y – 2) (–1) + z(–1)= 0 the plane.

41. Find the equation of the plane passing through the point (3, 2, 1) and parallel to the plane

Find the distance of the point (2, 3, 2) from the plane.

ANS: Equation of plane parallel to x + y –

...(i)

point (3, 6, 4) on line lies on the plane )] or 0a + 1b + 1c = 0 ...(ii

Line is perpendicular to normal to the plane

...(iii)

iii), we get

1)= 0 –x + 3 + y – 2 – z = 0 x – y + z – 1 = 0 is the required equation of

Find the equation of the plane passing through the point (3, 2, 1) and parallel to the plane x + y – 2z = 2.

Find the distance of the point (2, 3, 2) from the plane.

– 2z = 2 is x + y – 2z = λ ...(i)

ii)

1 = 0 is the required equation of

4

If this passes through (3, 2, 1), we have 3 + 2

λ = 3

Equation of plane is x + y – 2z = 3

Distance of point (2, 3, 2) from plane =

42. Find the equations of the line through the point (3, 0, 1) and parallel to the planes

ANS: Equation of line through point (3, 0, 1) is

Line is parallel to planes x + 2y = 0 and 3y –

a + 2b + 0c = 0

0a + 3b – c = 0

a = –2, b = 1, c = 3 Substituting in (

43. Show that the lines

Hence, find their point of intersection.

ANS:

If this passes through (3, 2, 1), we have 3 + 2 – 2 = λ

Distance of point (2, 3, 2) from plane = = = units

Find the equations of the line through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.

Equation of line through point (3, 0, 1) is ...(i)

– z = 0

= 3 Substituting in (i), we get the equation of line as

are intersecting.

= 0. 4

6

44. Show that the lines

ANS:

are coplanar.

6

45. Find the vector equation of the plane passing through the intersection of the planes

ANS:

46. Find the equation of the plane passing through the line of intersection of the planes

Find the vector equation of the plane passing through the intersection of the planes

and the point (1, 1, 1).

Find the equation of the plane passing through the line of intersection of the planes

6

6

ANS:

47. Find the distance of the point (1, –2, 3) from the plane

ANS:

and parallel to the x-axis.

2, 3) from the plane x – y + z = 5 measured parallel to the line

. 6

48: Show that the line whose vector equation is

Also, find the distance between them.

ANS:

Show that the line whose vector equation is is parallel to the plane

.

6

49. Show that the lines containing the lines.

ANS:

50. Find the equation of the plane passing through the line of intersection of the planes

are coplanar. Also find the equation of the plane

Find the equation of the plane passing through the line of intersection of the planes

are coplanar. Also find the equation of the plane 6

6

ANS: General equation of a plane passing through the line of intersection of planes

whose perpendicular distance from origin is unit.

General equation of a plane passing through the line of intersection of planes

51. If the lines equation of plane containing these lines.

ANS:

are perpendicular, find the value of k and hence find the

are perpendicular, find the value of k and hence find the 6

52. Find the distance between the point (–1, –5, x – y + z = 5.

53 The distance of point (2, 5, 7) from the x-axis is

(a) 2 (b) (c)

ANS: (b), as distance of point (2, 5, 7) from the

54 P is a point on the line segment joining the pointswill be

(a) 2 (b) (c) (

5, –10) and the point of intersection of the line

axis is

(d)

), as distance of point (2, 5, 7) from the x-axis is

is a point on the line segment joining the points (3, 5, –1) and (6, 3, –2). If y-coordinate of point P is 2, then its

(d) –5

and the plane 6

1

is 2, then its x-coordinate

1

ANS: (c), as let P divides the join of (3, 5,

⇒ 3k + 5 = 2k + 2 ⇒ k = –3 ∴ x-coordinate is

55 Direction ratios of a line are 2, 3, –6. Then direction

(a) (b) (c

ANS: (c), as direction cosines of a line whose direction

As angle with the y-axis is obtuse. ∴ cos β < 0, Therefore direction ratios are

56 A line makes angle α, β, γ with x-axis, y-axis and

(a) 2 (b) 1 (c) –2 (d)

ANS: (d), as cos2 α + cos2 β + cos2 γ = 1 ⇒

⇒ cos 2α + cos 2β + cos 2γ = –1

57 The equations of y-axis in space are

divides the join of (3, 5, –1) and (6, 3, –2) in the ratio k : 1 ∴

coordinate is

direction cosines of a line making obtuse angle with the y-axis are

c) (d)

), as direction cosines of a line whose direction ratio are 2, 3, –6 are .

cos β < 0, Therefore direction ratios are .

axis and z-axis respectively then cos 2α + cos 2β + cos 2γ is equal to

) –1

⇒

axis are

1

is equal to 1

1

(a) x = 0, y = 0 (b) x = 0, z = 0

ANS: (b), as on the y-axis, x-coordinate and

58: If the direction cosines of a line are , then value

(a) k > 0 (b) 0 < k < 1 (c) k

ANS: (d), as 3 × = 1 ⇒ k = ±

59 Distance of plane , from

(a) 2 (b) 14 (c)

ANS: (c), as normal form of plane is

60 Distance between planes

(a) (b) (c)

(c) y = 0, z = 0 (d) y = 0

coordinate and z-coordinate are zeroes.

, then value of k is

k = (d) k = ±

, from origin is

(d) –

∴ distance =

and is

(d)

1

1

1

ANS: (c), as distance = =

61 Intercept cut by the plane 2x – y + 2z + 7 = 0 on the

(a) 2 (b) (c) –

ANS:

62 Direction ratios of a line passing through the points

(a) (1, 1, –1) (b) 1, 1, –1 (c) < 5, 3,

ANS: (b), as DR’s are 3 – 2, 2 – 1, – 1 – 0 i.e. 1, 1,

63 Angle between the lines with direction ratios 2, 1, 2 and 3, 2,

(a) cos–1(– 4) (b) cos–1 (

units

+ 7 = 0 on the x-axis is

(d) –2

Direction ratios of a line passing through the points (2, 1, 0) and (3, 2, –1) are

) < 5, 3, –1 > (d) none of these

0 i.e. 1, 1, –1

Angle between the lines with direction ratios 2, 1, 2 and 3, 2, – 6 is

(c) – (d) none of these

1

1

1

ANS:

64

Equation of the line passing through the point (2, 1, 3)

(a) (b)

ANS: (d), as line is

65 The ratio in which the line segment joining the points

(a) 5 : 4 internally (b) 4 : 5 externally

ANS: (c), if ratio is k : 1, then = 0 ⇒

66 Direction cosines of a unit vector perpendicular to the

(a) 6, –3, –2 (b)

Equation of the line passing through the point (2, 1, 3) and perpendicular to the lines

(c) (d) none of these

and a + 3b + 3c = 0 –3a + 2b + 5c = 0

The ratio in which the line segment joining the points (2, 4, 5) and (3, 5, – 4) is divided by YZ-plane is

(c) 2 : 3 externally (d) none of these

⇒ k = – ⇒ 2 : 3 externally.

Direction cosines of a unit vector perpendicular to the plane are

(c) (d) none of these

and is

) none of these

1

1

1

ANS:

67 A line makes equal angles with axes, direction cosines

(a) 1, 1, 1 (b)

ANS: (c)

68 Projection of a line segment joining the points (2,

(a) – (b) (c)

ANS: (c)

69 Direction ratios of the line

(a) 2, 6, 3 (b) –2, 6, 3

ANS: (c)

70 Distance between parallel planes 2x – y + 3z

A line makes equal angles with axes, direction cosines of line are

(c) (d)

segment joining the points (2, 0, 5) and (0, 3, 1) on the line whose direction ratios are

(d) 19

are

(c) 2, – 6, 3 (d) none of these

z – 4 = 0 and 6x – 3y + 9z + 13 = 0 is

1

1) on the line whose direction ratios are 2, 3, 6 is

1

1

1

(a) 17 (b)

ANS: (c)

71

If line is parallel to the plane

(a) 2 (b) –2 (c)

ANS: (b)

(c) (d)

is parallel to the plane px + 3y – z + 5 = 0, then the value of ‘p’ is

(d) none of these

1