# 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 (x

_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2}) =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 = 8Hence, 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 (x

_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2}) =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(x

_{1},y_{1},z_{1}) and Q(x_{2},y_{2},z_{2})

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 (x

_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2}) and (x_{3},y_{3},z_{3}) =(0, 0, 0) =

(0, 0, 0) =

So, = 0

a = -2and, = 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}]= √[3

^{2}+ (-2-y)^{2}+ 5^{2}]= √[(-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 – 344 + y

^{2}+ (2 × -2 × -y) = 16y

^{2}+ 4y + 4 -16 = 0y

^{2}+ 4y – 12 = 0y

^{2}+ 6y – 2y – 12 = 0y (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 PA^{2} + PB^{2} = k^{2}, 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 (x

_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{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,

PA^{2}+ PB^{2}= k^{2}[(3 – x)

^{2}+ (4 – y)^{2}+ (5 – z)^{2}] + [(-1 – x)^{2}+ (3 – y)^{2}+ (-7 – z)^{2}] = k^{2}[(9 + x

^{2}– 6x) + (16 + y^{2}– 8y) + (25 + z^{2}– 10z)] + [(1 + x^{2}+ 2x) + (9 + y^{2}– 6y) + (49 + z^{2}+ 14z)] = k^{2}9 + x

^{2}– 6x + 16 + y^{2}– 8y + 25 + z^{2}– 10z + 1 + x^{2}+ 2x + 9 + y^{2}– 6y + 49 + z^{2}+ 14z = k^{2}2x

^{2}+ 2y^{2}+ 2z^{2}– 4x – 14y + 4z + 109 = k^{2}2x

^{2}+ 2y^{2}+ 2z^{2}– 4x – 14y + 4z = k^{2}– 1092 (x

^{2}+ y^{2}+ z^{2}– 2x – 7y + 2z) = k^{2}– 109(x

^{2}+ y^{2}+ z^{2}– 2x – 7y + 2z) =Hence,

the required equation is (x^{2}+ y^{2}+ z^{2}– 2x – 7y + 2z) =

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