## MP Board Class 11th Maths Important Questions Chapter 12 Introduction to Three Dimensional Geometry

### Introduction to Three Dimensional Geometry Important Questions

** Introduction to Three Dimensional Geometry Objective Type Questions**

(A) Choose the correct option :

Question 1.

Distance of a point (3,4, 5) from origin:

(a) 3

(b) 4

(c) 5

(d) 5\(\sqrt {2}\)

Answer:

(d) 5\(\sqrt {2}\)

Question 2.

The perpendicular distance of the point P (x, y, z) from X – axis is:

(a) \(\sqrt { { x }^{ 2 } + { z }^{ 2 } }\)

(b) \(\sqrt { { y }^{ 2 } + { z }^{ 2 } }\)

(c) \(\sqrt { { x }^{ 2 } + { y }^{ 2 } }\)

(d) \(\sqrt { { x }^{ 2 } + { y }^{ 2 } + { z }^{ 2 } }\)

Answer:

(b) \(\sqrt { { y }^{ 2 } + { z }^{ 2 } }\)

Question 3.

Distance of a point (3, 2, 5) from X – axis is:

(a) \(\sqrt {28}\)

(b)\(\sqrt {30}\)

(c) \(\sqrt {29}\)

(d) 3

Answer:

(c) \(\sqrt {29}\)

Question 4.

The YZ – plane divides the line segment joining the points (2, 3, 4) and (3, 5, – 4) in the ratio:

(a) 2 : 3 internally

(b) 1 : 2

(c) 2 : 3 externally

(d) 1 : 3

Answer:

(c) 2 : 3 externally

Question 5.

Distance between the points (1, 2,3) and (1,3, -2) is :

(a) \(\sqrt {- 26}\)

(b) \(\sqrt {26}\)

(c) 26

(d) ± \(\sqrt {24}\)

Answer:

(b) \(\sqrt {26}\)

Question 6.

Coordinate of a point on Z – axis, equidistant from the points (1, 5, 7) and (5, 1, – 4) is :

(a) (0, 0, \(\frac { 3 }{ 2 }\)

(b) (o, \(\frac { 3 }{ 2 }\), 0)

(c) (\(\frac { 3 }{ 2 }\), 0, 0)

(d) (4, – 4, – 11)

Answer:

(a) (0, 0, \(\frac { 3 }{ 2 }\)

Question 7.

Distance of a point (2, 1, 4) from Y – axis is:

(a) \(\sqrt {20}\)

(b) 1

(c) \(\sqrt {12}\)

(d) \(\sqrt {10}\)

Answer:

(a) \(\sqrt {20}\)

(B) Match the following :

Answer:

- (e)
- (d)
- (b)
- (a)
- (c)

(C) Fill in the blanks :

- If the points (- 1, 3, 2), (- 4, 2, – 2) and (5, 5, λ ) are collinear, then value of λ is …………….
- The perpendicular distance of point P(x, y, z) from YZ – plane is …………….
- XY – plane divides the line segment joining the points (- 3, 4, 8) and (5, – 6, 4) in the ratio …………….
- Distance between the points (1, – 3, 4) and (3, 11, – 6) is …………….
- Perpendicular distance of a point P (3,4, 5) from YZ – plane is …………….

Answer:

- 10
- x
- 2 : 1
- 10\(\sqrt {3}\)
- 3

(D) Write true / false :

- Distance of a point (4, 3, 5) from Y – axis is \(\sqrt {40}\).
- Distance of a point (5, 12, 13) from YZ – axis is \(\sqrt {313}\).
- The coordinate of a point where YZ – plane divides the join of the points (3, 5, – 7) and (- 2, 1, 8) are (0, \(\frac { 13 }{ 5 }\), 2).
- Coordinate of mid point of the line segment joining the points (- 3, 4, – 8) and (5, – 6, 4) are (1, – 1, 2).
- Points A(1, 2, 3), B(4, 0, 4) and C (- 2, 4, 2) are collinear.

Answer:

- False
- True
- True
- False
- True

(E) Write answer in one word / sentence :

- A point R lies on the line segment joining the points P (2, – 3, 4) and Q(8, 0, 10) whose x coordinate is 4, then the coordinate of R will be.
- In what ratio does the plane 2x + y – z = 3 divide the line segment joining the points A (2, 1, 3) and 5(4, – 2, 5)?
- If the origin is the centroid of the triangle ABC with vertices A (a, 1, 3), B (- 2, b, – 5), C (4, 7, c) then write the value of a, b and c.
- Find the locus of the point, which is equidistant to the points (3, 4, – 5) and (- 2, 1, 4).
- In which ratio the XY – plane divides the line joining the points (2, 4, 2) and (2, 5, – 4)?

Answer:

- (4, – 2, 6)
- 1 : 2 externally
- – 2, – 8, – 2
- 10x + 6y – 18z – 29 = 0
- 1 : 2 internally.

**Introduction to Three Dimensional Geometry Very Short Answer Type Questions**

Question 1.

A point is on X – axis. What are its y – co – ordinate and z – co – ordinate?

Answer:

y – co – ordinate and z – co – ordinate of point on X – axis is 0, 0 is (x, 0, 0).

Question 2.

A point is in the xz – plane. What can you say about is y – co – ordinate?

Answer:

On xz – plane the y – co – ordinate will be zero.

Question 3.

The X – axis and Y – axis taken together to form a plane, name of the plane is.

Answer:

The plane is known as xy – plane.

Question 4.

Co – ordinate of any point on xy – plane is.

Answer:

The co – ordinate of any point on xy – plane is (x, y, 0).

Question 5.

Length of perpendicular of point P(x, y, z) from X – axis is.

Answer:

Length of perpendicular = \(\sqrt { { y }^{ 2 } + { z }^{ 2 } }\)

Question 6.

Length of perpendicular of point P(x, y, z) from yz – plane is.

Answer:

Length of perpendicular of point P(x, y, z) from yz – plane is x.

Question 7.

Find the distance of point (3, 4, 5) from xz – plane.

Answer:

Distance of point (3, 4, 5) from xz – plane M(3, 0, 5) = \(\sqrt {0 + 16 + 0}\) = 4.

Question 8.

What is the centroid of ∆ABC whose vertices A(x_{1}, y_{1}, z_{1}), B(x_{2}, y_{2}, Z_{2}) and C(x_{3}, y_{3}, z_{3})?

Answer:

Centroid of ∆ABC is

Question 9.

Find distance between points (2,3, 5) and (4,3,1).

Solution:

Distance between two points A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, Z_{2}):

Question 10.

Find the distance between points (- 3, 7, 2) and (2, 4, – 1).

Solution:

Question 11.

Find the distance of the point (3, 2, 5) from X – axis.

Solution:

Point on X – axis (3, 0, 0)

Question 12.

Find the distance of the point (2, 1, 4) from Y – axis.

Solution:

Point on Y – axis (0, 1, 0)

Question 13.

The three dimensional planes divides the space in how many octants?

Answer:

Three dimensional plane divides the space in eight octants.

**Introduction to Three Dimensional Geometry Short Answer Type Questions**

Question 1.

Prove that the points (- 2, 3, 5) (1, 2, 3) and (7, 0, – 1) are collinear. (NCERT)

Solution:

Given point are A (- 2, 3, 5), B(1, 2, 3) and C (7, 0, – 1).

Hence points A, B, Care collinear.

Question 2.

Find the co – ordinates of the point which divides the line segment joining the points (- 2, 3, 5) and (1, – 4, 6) in the ratio 2 : 3 internally. (NCERT)

Solution:

Required co – ordinates are :

Question 3.

Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, – 10) are collinear. Find the ratio in which Q divides PR. (NCERT)

Solution:

⇒ 5k + 5 = 9k + 3

⇒ 4k = 2

⇒ k = \(\frac { 1 }{ 2 }\)

Hence, point Q divides PR internally in ratio 1 : 2.

Question 4.

Find the ratio in which the YZ – plane divides the line segment formed by joining the points (- 2, 4, 7) and (3, – 5, 8).

Solution:

Let the point R (0, y, z) on YZ – plane divides the points P (- 2, 4, 7) and Q (3, – 5, 8) in the ratio m : n.

x = \(\frac{m x_{2}+n x_{1}}{m+n}\)

⇒ 0 = \(\frac { m × 3 + n( – 2) }{ m + n }\)

⇒ 3m – 2n = 0

⇒ 3m = 2n

⇒ \(\frac { m }{ n }\) = \(\frac { 2 }{ 3 }\)

⇒ m : n = 2 : 3

YZ – plane divides the line PQ internally in ratio 2:3.

Question 5.

If the origin is the centroid of the ∆PQR with vertices P(2a, 2, 6), Q(- 4, 3b, – 10) and R (8, 14, 2c) then find the value of a, b and c. (NCERT)

Solution:

Centroid of APQR is :

x = \(\frac{x_{1} + x_{2} + x_{3}}{3}\), y = \(\frac{y_{1} + y_{2} + y_{3}}{3}\), z = \(\frac{z_{1} + z_{2} + z_{3}}{3}\)

Co – ordinate of centroid are (0, 0, 0),

Question 6.

Find the co – ordinates of a point on Y – axis which are at a distance of 5\(\sqrt {2}\) from the point P(3, – 2, 5).

Solution:

Let A (0, y, 0) be any point on Y – axis.

Given : PA = 5\(\sqrt {2}\)

PA^{2} = 50

⇒ (3 – 0)^{2} + (- 2 – y)^{2} + (5 – 0)^{2} = 50

⇒ 9 + 4 + y^{2} + 4y + 25 = 50

⇒ y^{2} + 4y + 38 – 50 = 0

⇒ y^{2} + 4y – 12 = 0

⇒ y^{2} + 6y – 2y – 12 = 0

⇒ y(y + 6) – 2(y + 6) = 0

⇒ (y – 2)(y + 6) = 0

⇒ y = 2, – 6

The co – ordinate on Y – axis is (0, 2, 0) or (0, – 6, 0).

Question 7.

A point R with x co – ordinate 4 lies on the line segment joining the points P (2, – 3, 4) and Q (8, 0, 10). Find the co – ordinate of the point R.

Solution:

Let the point R (x, y, z) divides PQ in ratio m : n.

Co – ordinate of R (4, – 2, 6).

Question 8.

Find the equation of set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1). (NCERT)

Solution:

Let point P (x, y, z) be equidistant from points A (1, 2, 3) and B (3, 2, – 1).

∴ PA = PB

⇒ PA^{2} = PB^{2}

⇒ (x – 1)^{2} + (y – 2)^{2} + (z – 3)^{2} = (x – 3)^{2} + (y – 2)^{2} + (z +1)^{2}

⇒ x^{2} – 2x + 1 + z^{2} – 6z + 9

– 2x – 6z + 10 = – 6x + 2z + 10

⇒ 4x – 8z = 0

⇒ x – 2z = 0.

This is the required equation.

Question 9.

In which ratio the point (1, 1, 1) divides the line joining the points (3, – 2, 4) and (- 1, 4, 2)?

Solution:

Let point R (1,1,1), divides the line joining the points P (3, – 2, 4) and Q(- 1, 4, 2) in ratio λ : 1.

By formula:

x = \(\frac{m x_{2}+n x_{1}}{m+n}\)

Here x_{1} = 3, x_{2} = – 1, x = 1, m = λ and n = 1.

∴ 1 = \(\frac { λ(- 1) + 1 × 3 }{ λ + 1}\)

⇒ λ + 1 = – λ + 3

⇒ 2λ = 2

∴ λ = 1

Hence, required ratio is 1 : 1.