RD Chapter 28 Introduction to 3D coordinate geometry Ex 28.2 Solutions
Question - 1 : - Find the distance between the following pairs of points:
(i) P(1, -1, 0) and Q (2, 1, 2)
(ii) A(3, 2, -1) and B (-1, -1, -1)
Answer - 1 : -
(i) P(1,-1, 0) and Q (2, 1, 2)
Given:
The points P(1, -1, 0) and Q (2, 1, 2)
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) isgiven by,
∴ TheDistance between P and Q is 3 units.
(ii) A (3,2, -1) and B (-1, -1, -1)
Given:
The points A (3, 2, -1) and B (-1, -1, -1)
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) isgiven by,
= 5
∴ TheDistance between A and B is 5 units.
Question - 2 : - Find the distance between the points P and Q having coordinates (-2, 3, 1) and (2, 1, 2).
Answer - 2 : -
Given:
The points (-2, 3, 1) and (2, 1, 2)
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) isgiven by,
∴ TheDistance between the given two points is √21units.
Question - 3 : - Using distance formula prove that the following points are collinear:
(i) A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
(ii) P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)
(iii) A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
Answer - 3 : -
(i) A(4,-3, -1), B(5, -7, 6) and C(3, 1, -8)
Given:
The points A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)
Points A, B and C are collinear if AB + BC = AC or AB + AC = BCor AC + BC = AB
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
∴Thepoints A, B and C are collinear.
(ii) P (0,7, -7), Q (1, 4, -5) and R (-1, 10, -9)
Given:
The points P (0, 7, -7), Q (1, 4, -5) and R (-1, 10, -9)
Points P, Q and R are collinear if PQ + QR = PR or PQ + PR = QRor PR + QR = PQ
By using the formula,
Distance between any two points (a, b, c) and (m, n, o) is givenby,
∴Thepoints P, Q and R are collinear.
(iii) A(3,-5, 1), B(-1, 0, 8) and C(7, -10, -6)
Given:
The points A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)
Points A, B and C are collinear if AB + BC = AC or AB + AC = BCor AC + BC = AB
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
∴Thepoints A, B and C are collinear.
Question - 4 : - Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1).
Answer - 4 : -
Given:
The points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)
(i) xy-plane
We know z = 0 in xy-plane.
So let P(x, y, 0) be any point in xy-plane
According to the question:
PA = PB = PC
PA2 = PB2 = PC2
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) is given by,
We know PA2 = PB2
So, (x – 1)2+ (y + 1)2 = (x – 2)2 +(y – 1)2 + 4
x2+ 1 – 2x + y2 + 1 + 2y = x2+4 – 4x + y2 + 1 – 2y + 4
– 2x + 2 + 2y = 9 – 4x – 2y
– 2x + 2 + 2y – 9 + 4x + 2y = 0
2x + 4y – 7 = 0
2x = – 4y + 7……………………(1)
Since, PA2 = PC2
So, (x – 1)2+ (y + 1)2 = (x – 3)2 +(y – 2)2 + 1
x2+ 1 – 2x + y2 + 1 + 2y = x2+9 – 6x + y2 + 4 – 4y + 1
– 2x + 2 + 2y = 14 – 6x – 4y
– 2x + 2 + 2y – 14 + 6x + 4y = 0
4x + 6y – 12 = 0
2(2x + 3y – 6) = 0
Now substitute the value of 2x (obtained in equation (1)), weget
7 – 4y + 3y – 6 = 0
– y + 1 = 0
y = 1
By substituting the value of y back in equation (1) we get,
2x = 7 – 4y
2x = 7 – 4(1)
2x = 3
x = 3/2
∴Thepoint P (3/2, 1, 0) in xy-plane is equidistant from A, B and C.
(ii) yz-plane
We know x = 0 in yz-plane.
Let Q(0, y, z) any point in yz-plane
According to the question:
QA = QB = QC
QA2 = QB2 = QC2
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
We know, QA2 = QB2
So, 1 + z2+ (y + 1)2 = (z – 2)2 +(y – 1)2 + 4
z2+ 1 + y2 + 1 + 2y = z2+4 – 4z + y2 + 1 – 2y + 4
2 + 2y = 9 – 4z – 2y
2 + 2y – 9 + 4z + 2y = 0
4y + 4z – 7 = 0
4z = –4y + 7
z = [–4y + 7]/4 …. (1)
Since, QA2 = QC2
So, 1 + z2+ (y + 1)2 = (z + 1)2 +(y – 2)2 + 9
2+ 1 + y2 + 1 + 2y = z2+ 1 + 2z + y2 +4 – 4y + 9
2 + 2y = 14 + 2z – 4y
2 + 2y – 14 – 2z + 4y = 0
–2z + 6y – 12 = 0
2(–z + 3y – 6) = 0
Now, substitute the value of z [obtained from (1)] we get
12y + 4y – 7 – 24 = 0
16y – 31 = 0
y = 31/16
Substitute the value of y back in equation (1), we get
∴Thepoint Q (0, 31/16, -3/16) in yz-plane is equidistant from A, B and C.
(iii) zx-plane
We know y = 0 in xz-plane.
Let R(x, 0, z) any point in xz-plane
According to the question:
RA = RB = RC
RA2 = RB2 = RC2
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
We know, RA2 = RB2
So, 1 + z2+ (x – 1)2 = (z – 2)2 +(x – 2)2 + 1
z2+ 1 + x2 + 1 – 2x = z2+4 – 4z + x2 + 4 – 4x + 1
2 – 2x = 9 – 4z – 4x
2 + 4z – 9 + 4x – 2x = 0
2x + 4z – 7 = 0
2x = –4z + 7……………………………(1)
Since, RA2 = RC2
So, 1 + z2+ (x – 1)2 = (z + 1)2 +(x – 3)2 + 4
z2+ 1 + x2 + 1 – 2x = z2+1 + 2z + x2 + 9 – 6x + 4
2 – 2x = 14 + 2z – 6x
2 – 2x – 14 – 2z + 6x = 0
–2z + 4x – 12 = 0
2(2x) = 12 + 2z
Substitute the value of 2x [obtained from equation (1)] we get,
2(–4z + 7) = 12 + 2z
–8z + 14 = 12 + 2z
14 – 12 = 8z + 2z
10z = 2
z = 2/10
= 1/5
Now, substitute the value of z back in equation (1), we get
2x = -4z + 7
∴Thepoint R (31/10, 0, 1/5) in xz-plane is equidistant from A, B and C.
Answer - 5 : -
Given:
The points (1, 5, 7) and (5, 1, -4)
We know x = 0 and y = 0 on z-axis
Let R(0, 0, z) any point on z-axis
According to the question:
RA = RB
RA2 = RB2
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) is given by,
We know, RA2 = RB2
26+ (z – 7)2 = (z + 4)2 + 26
z2+ 49 – 14z + 26 = z2+ 16 + 8z + 26
49 – 14z = 16 + 8z
49 – 16 = 14z + 8z
22z = 33
z = 33/22
= 3/2
∴Thepoint R (0, 0, 3/2) on z-axis is equidistant from (1, 5, 7) and (5, 1,-4).
Answer - 6 : -
Given:
The points (3, 1, 2) and (5, 5, 2)
We know x = 0 and z = 0 on y-axis
Let R(0, y, 0) any point on the y-axis
According to the question:
RA = RB
RA2 = RB2
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) is given by,
So,
We know, RA2 = RB2
13+ (y – 1)2 = (y – 5)2 + 29
y2+ 1 – 2y + 13 = y2+ 25 – 10y + 29
10y – 2y = 54 – 14
8y = 40
y = 40/8
= 5
∴Thepoint R (0, 5, 0) on y-axis is equidistant from (3, 1, 2) and (5, 5, 2).
Answer - 7 : -
Given:
The point (1, 2, 3)
Distance = √21
We know x = 0 and y = 0 on z-axis
Let R(0, 0, z) any point on z-axis
According to question:
RA = √21
RA2 = 21
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
We know, RA2 = 21
5 + (z – 3)2 = 21
z2+ 9 – 6z + 5 = 21
z2 – 6z = 21 – 14
z2– 6z – 7 = 0
z2– 7z + z – 7 = 0
z(z– 7) + 1(z – 7) = 0
(z– 7) (z + 1) = 0
(z– 7) = 0 or (z + 1) = 0
z= 7 or z = -1
∴The points(0, 0, 7) and (0, 0, -1) on z-axis is equidistant from (1, 2, 3).
Question - 8 : - Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.
Answer - 8 : -
Given:
The points (1, 2, 3), (2, 3, 1) and (3, 1, 2)
An equilateral triangle is a triangle whose all sides are equal.
So let us prove AB = BC = AC
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) is given by,
It is clear that,
AB = BC = AC
Δ ABC is a equilateral triangle
Hence Proved.
Answer - 9 : -
Given:
The points (0, 7, 10), (-1, 6, 6) and (-4, 9, 6)
Isosceles right-angled triangle is a triangle whose two sidesare equal and also satisfies Pythagoras Theorem.
By using the formula,
The distance between any two points (a, b, c) and (m, n, o)is given by,
Since, AB = BC
So, AB2 + BC2
= (3√2)2 +(3√2)2
= 18 + 18
= 36
= AC2
We know that, AB = BC and AB2 + BC2 = AC2
So, Δ ABC is an isosceles-right angled triangle
Hence Proved.
Question - 10 : - Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of squares.
Answer - 10 : -
Given:
The points A (3, 3, 3), B (0, 6, 3), C (1, 7, 7) and D (4, 4, 7)
We know that all sides of a square are equal.
By using the formula,
The distance between any two points (a, b, c) and (m, n, o) is given by,
It is clear that,
AB = BC = CD = AD
Quadrilateral formed by ABCD is a square. [Since all sides are equal]
Hence Proved.