Exercise Miscellaneous - Introduction to Three Dimensional Geometry NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download

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Q1: 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.
Ans: Given:
ABCD is a parallelogram with vertices A (3, -1, 2), B (1, 2, -4), and 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 know that the diagonals of a parallelogram bisect each other, so the midpoints of AC and BD are equal, i.e., Midpoint of AC = Midpoint of BD ……….(1)
Now, by the 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 are D (1, -2, 8).

Q2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
Ans: 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 the midpoint formula, the coordinates of the midpoint 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 the distance formula, 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.

Q3:  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.
Ans:

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


Q4: 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.
Ans: Given:
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 the distance formula,
We know that the distance between two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is given by

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 – 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) = (k² – 109)/2
Hence, the required equation is (x² + y² + z² – 2x – 7y + 2z) = (k² – 109)/2.