NCERT Solutions Class 11th Maths Chapter 12- Introduction to Three Dimensional Geometry Miscellaneous Exercise

NCERT Solutions Class 11th Maths Chapter 12- Introduction to Three Dimensional Geometry Miscellaneous Exercise

NCERT Solutions Class 11th Maths Chapter 12- Introduction to Three Dimensional Geometry Miscellaneous Exercise

?Chapter – 12?

✍Introduction to Three Dimensional Geometry✍

?Miscellaneous Exercise?

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: Given:
ABCD is a parallelogram, with vertices A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2).
Where, x₁ = 3, y₁ = -1, z₁ = 2;
x₂ = 1, y₂ = 2, z₂ = -4;
x₃ = -1, y₃ = 1, z₃ = 2

Let the coordinates of the fourth vertex be D (x, y, z).
We also know that the diagonals of a parallelogram bisect each other, so the mid points of AC and BD are equal, i.e. Midpoint of AC = Midpoint of BD ……….(1)
Now, by Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) are [(x₁+x₂)/², (y₁+y₂)/², (z₁+z₂)/²]
So we have,

= (2/2, 0/2, 4/2)
= (1, 0, 2)

1 + x = 2, 2 + y = 0, -4 + z = 4
x = 1, y = -2, z = 8
Hence, the coordinates of the fourth vertex is D (1, -2, 8).

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: Given:
The vertices of the triangle are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
x₁ = 0, y₁ = 0, z₁ = 6;
x₂ = 0, y₂ = 4, z₂ = 0;
x₃ = 6, y₃ = 0, z₃ = 0

So, let the medians of this triangle 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.
By Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) are [(x₁+x₂)/², (y₁+y₂)/², (z₁+z₂)/²]
So we have,

By Distance Formula, we know that the distance between two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is given by

∴ The lengths of the medians of the given triangle are 7, √34 and 7.

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: Given:

The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).
Where,
x₁ = 2a, y₁ = 2, z₁ = 6;
x₂ = -4, y₂ = 3b, z₂ = -10;
x₃ = 8, y₃ = 14, z₃ = 2c
We know that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃), are [(x₁+x₂+x₃)/³, (y₁+y₂+y₃)/³, (z₁+z₂+z₃)/³]
So, the coordinates of the centroid of the triangle PQR are

2a + 4 = 0, 3b + 16 = 0, 2c – 4 = 0
a = -2, b = -16/3, c = 2
∴ The values of a, b and c are a = -2, b = -16/3, c = 2

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: Let the point on y-axis be A (0, y, 0).
Then, it is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is 5√2.
Now, by using distance formula,
We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by
Distance of PQ = √[(x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z1)²]
So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by
Distance of AP = √[(x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)²]
= √[(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
∴ The points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.

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: Given:
The coordinates of the points P (2, -3, 4) and Q (8, 0, 10).
x₁ = 2, y₁ = -3, z₁ = 4;
x₂ = 8, y₂ = 0, z₂ = 10
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.
By using Section Formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m: n is given by:

So, the coordinates of the point R are given by

So, we have

8k + 2 = 4 (k + 1)
8k + 2 = 4k + 4
8k – 4k = 4 – 2
4k = 2
k = 2/4
= 1/2
Now let us substitute the values, we get

= 6
∴ The coordinates of the required point are (4, -2, 6).

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: Given:
The points A (3, 4, 5) and B (-1, 3, -7)
x1 = 3, y1 = 4, z1 = 5;
x2 = -1, y2 = 3, z2 = -7;
PA2 + PB2 = k2 ……….(1)
Let the point be P (x, y, z).
Now by using distance formula,

We know that the distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is given by

So,

And

Now, substituting these values in (1), we have
[(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 – 10₂)] + [(1 + x₂ + 2x) + (9 + y₂– 6y) + (49 + z₂ + 14₂)] = 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) = (k2 – 109)/2
Hence, the required equation is (x₂ + y₂ + z₂ – 2x – 7y + 2z) = (k2 – 109)/2