NCERT Solutions: Exercise Miscellaneous - Introduction to Three Dimensional Geometry
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.
