Class 11 NCERT Solutions- Chapter 12 Introduction to three dimensional Geometry – Miscellaneous Exercise on Chapter 12

Question 1: Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4), and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Solution: 

ABCD is a parallelogram, with vertices A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2) and D (x, y, z).

Using the property:

The diagonals of a parallelogram bisect each other, 

Midpoint of AC = Midpoint of BD = Point O

Now, by using Midpoint section formula

Coordinates of O for the line segment joining (x₁,y₁,z₁) and (x₂,y₂,z₂) = 

So, Coordinates of O for the line segment joining AC = 

= 

= (1, 0, 2) ……………………….(1)

and, Coordinates of O for the line segment joining BD =  ………….(2)

Using the Eq(1) and Eq(2), we get

 = 1

x = 1

 = 0

y = -2

 = 2

z = 8

Hence, the coordinates of the fourth vertex is D (1, -2, 8).

Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0), and (6, 0, 0).

Solution: 

The vertices of the triangle are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

So, let the medians be AD, BE and CF corresponding to the vertices A, B and C respectively.

D, E and F are the midpoints of the sides BC, AC and AB respectively.

Coordinates of mid-point for the line segment joining (x₁,y₁,z₁) and (x₂,y₂,z₂) = 

So, Coordinates of D for the line segment joining BC = 

Coordinates of D = (3, 2, 0)

and, Coordinates of E for the line segment joining AC = 

Coordinates of E = (3, 0, 3)

and, Coordinates of F for the line segment joining AB = 

Coordinates of F = (0, 2, 3)

By using the distance formula for two points, P(x₁,y₁,z₁) and Q(x₂,y₂,z₂)

PQ =

So, AD = 

AD =

and, BE = 

BE = 

and, CF = 

CF =  = 7

Hence, the lengths of the medians are 7, √34 and 7.

Question 3: If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10), and R(8, 14, 2c), then find the values of a, b and c.

Solution: 

The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

Coordinates of centroid(0, 0, 0) of the triangle having vertices (x₁,y₁,z₁), (x₂,y₂,z₂) and (x₃,y₃,z₃) = 

(0, 0, 0) =

(0, 0, 0) = 

So,  = 0

a = -2

and,  = 0

b =

and,  = 0

c = 2

Hence, the values of a, b and c are a = -2, b =  and c = 2

Question 4: Find the coordinates of a point on y-axis which are at a distance of 5√2 from the point P (3, –2, 5).

Solution: 

Point on y-axis = A (0, y, 0).

Distance between the points A (0, y, 0) and P (3, -2, 5) = 5√2.

Now, by using distance formula,

Distance of PQ = 

So, the distance between the points A (0, y, 0) and P (3, -2, 5) will be

Distance of AP = √[(3-0)² + (-2-y)² + (5-0)²]

= √[3² + (-2-y)² + 5²]

= √[(-2-y)² + 9 + 25]

5√2 = √[(-2-y)² + 34]

Squaring on both the sides, we get

(-2 -y)² + 34 = 25 × 2

(-2 -y)² = 50 – 34

4 + y² + (2 × -2 × -y) = 16

y² + 4y + 4 -16 = 0

y² + 4y – 12 = 0

y² + 6y – 2y – 12 = 0

y (y + 6) – 2 (y + 6) = 0

(y + 6) (y – 2) = 0

y = -6, y = 2

Hence, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.

Question 5: A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

[Hint: Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by ]

Solution: 

Let the coordinates of the required point be (4, y, z).

So now, let the point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) in the ratio k: 1.

Coordinates of the point which divides PQ in the ratio k : 1 = 

So, we have

 = (4, y, z)

 = 4

8k + 2 = 4 (k + 1)

8k + 2 = 4k + 4

8k – 4k = 4 – 2

4k = 2

k = 

k =

Now, substituting the value we get,

y =  = -2

z =  = 6

Hence, the coordinates of the required point are (4, -2, 6).

Question 6: If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA² + PB² = k², where k is a constant. 

Solution: 

The points A (3, 4, 5) and B (-1, 3, -7)

Let the point be P (x, y, z).

Now by using distance formula,

Distance of point (x₁, y₁, z₁) and (x₂, y₂, z₂) = 

So, the distance between the points A (3, 4, 5) and P (x,y,z)) will be

Distance of PA = √[(3-x)² + (4-y)² + (5-z)²]

Distance of PB = √[(-1-x)² + (3-y)² + (-7-z)²]

As, PA² + PB² = k²

[(3 – x)² + (4 – y)² + (5 – z)²] + [(-1 – x)² + (3 – y)² + (-7 – z)²] = k²

[(9 + x² – 6x) + (16 + y² – 8y) + (25 + z² – 10z)] + [(1 + x² + 2x) + (9 + y² – 6y) + (49 + z² + 14z)] = k²

9 + x² – 6x + 16 + y² – 8y + 25 + z² – 10z + 1 + x² + 2x + 9 + y² – 6y + 49 + z² + 14z = k²

2x² + 2y² + 2z² – 4x – 14y + 4z + 109 = k²

2x² + 2y² + 2z² – 4x – 14y + 4z = k² – 109

2 (x² + y² + z² – 2x – 7y + 2z) = k² – 109

(x² + y² + z² – 2x – 7y + 2z) = 

Hence, the required equation is (x² + y² + z² – 2x – 7y + 2z) =