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

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) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})

So, Coordinates of O for the line segment joining AC = (\frac{3+(-1)}{2}, \frac{-1+1}{2}, \frac{2+2}{2})

= (\frac{2}{2}, 0, \frac{4}{2})

= (1, 0, 2) ............................(1)

and, Coordinates of O for the line segment joining BD = (\frac{1+x}{2}, \frac{2+y}{2}, \frac{-4+z}{2})  .............(2)

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

\frac{1+x}{2}  = 1

x = 1

\frac{2+y}{2}  = 0

y = -2

\frac{z-4}{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) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2})

So, Coordinates of D for the line segment joining BC = (\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2})

Coordinates of D = (3, 2, 0)

and, Coordinates of E for the line segment joining AC = (\frac{6+0}{2}, \frac{0+0}{2}, \frac{0+6}{2})

Coordinates of E = (3, 0, 3)

and, Coordinates of F for the line segment joining AB = (\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2})

Coordinates of F = (0, 2, 3)

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

PQ = \mathbf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}}

So, AD = \sqrt{(0-3)^2+(0-2)^2+(6-0)^2}

AD = \sqrt{9+4+36} = 7

and, BE = \sqrt{(0-3)^2+(4-0)^2+(0-3)^2}

BE = \sqrt{9+16+9} = \sqrt{34}

and, CF = \sqrt{(6-0)^2+(0-2)^2+(0-3)^2}

CF = \sqrt{36+4+9}  = 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) = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})

(0, 0, 0) = (\frac{2a-4+8}{3}, \frac{2+3b+14}{3}, \frac{6-10+2c}{3})

(0, 0, 0) = (\frac{2a+4}{3}, \frac{3b+16}{3}, \frac{2c-4}{3})

So, \frac{2a+4}{3}  = 0

a = -2

and, \frac{3b+16}{3}  = 0

b = \mathbf{\frac{-16}{3}}

and, \frac{2c-4}{3}  = 0

c = 2

Hence, the values of a, b and c are a = -2, b = \mathbf{\frac{-16}{3}}  and c = 2

Question 4: 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) = \mathbf{\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}}

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) = \frac{(k^2 – 109)}{2}

Hence, the required equation is (x² + y² + z² – 2x – 7y + 2z) = \mathbf{\frac{(k^2 – 109)}{2}}