Chapter – 12: Introduction to Three-Dimensional Geometry
EXERCISE 12.1
1. A point is on the x-axis. What are its y coordinate and z-coordinates?
Solution: –
If a point is on the x-axis, then the coordinates of y and z are 0.
So the point is (x, 0, 0).
2. A point is in the XZ-plane. What can you say about its y-coordinate?
Solution: –
If a point is in XZ plane, then its y-co-ordinate is 0.
3. Name the octants in which the following points lie:
(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (2, – 4, –7).
Solution: –
Here is the table which represents the octants:
| Octants | I | II | III | IV | V | VI | VII | VIII |
| x | + | – | – | + | + | – | – | + |
| y | + | + | – | – | + | + | – | – |
| z | + | + | + | + | – | – | – | – |
(i) (1, 2, 3)
Here x is positive, y is positive and z is positive.
So it lies in I octant.
(ii) (4, -2, 3)
Here x is positive, y is negative and z is positive.
So it lies in IV octant.
(iii) (4, -2, -5)
Here x is positive, y is negative and z is negative.
So it lies in VIII octant.
(iv) (4, 2, -5)
Here x is positive, y is positive and z is negative.
So it lies in V octant.
(v) (-4, 2, -5)
Here x is negative, y is positive and z is negative.
So it lies in VI octant.
(vi) (-4, 2, 5)
Here x is negative, y is positive and z is positive.
So it lies in II octant.
(vii) (-3, -1, 6)
Here x is negative, y is negative and z is positive.
So it lies in III octant.
(viii) (2, -4, -7)
Here x is positive, y is negative and z is negative.
So it lies in VIII octant.
4. Fill in the blanks:
(i) The x-axis and y-axis taken together determine a plane known as _______.
(ii) The coordinates of points in the XY-plane are of the form _______.
(iii) Coordinate planes divide the space into ______ octants.
Solution: –
(i) The x-axis and y-axis taken together determine a plane known as XY Plane.
(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).
(iii) Coordinate planes divide the space into eight octants.
EXERCISE 12.2
1. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, – 4) and (1, –3, 4)
(iv) (2, –1, 3) and (–2, 1, 3)
Solution: –
(i) (2, 3, 5) and (4, 3, 1)
Let P be (2, 3, 5) and Q be (4, 3, 1)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = 3, z1 = 5
x2 = 4, y2 = 3, z2 = 1
Distance PQ = √[(4 – 2)2 + (3 – 3)2 + (1 – 5)2]
= √[(2)2 + 02 + (-4)2]
= √[4 + 0 + 16]
= √20
= 2√5
∴ The required distance is 2√5 units.
(ii) (–3, 7, 2) and (2, 4, –1)
Let P be (– 3, 7, 2) and Q be (2, 4, – 1)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 3, y1 = 7, z1 = 2
x2 = 2, y2 = 4, z2 = – 1
Distance PQ = √[(2 – (-3))2 + (4 – 7)2 + (-1 – 2)2]
= √[(5)2 + (-3)2 + (-3)2]
= √[25 + 9 + 9]
= √43
∴ The required distance is √43 units.
(iii) (–1, 3, – 4) and (1, –3, 4)
Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 1, y1 = 3, z1 = – 4
x2 = 1, y2 = – 3, z2 = 4
Distance PQ = √[(1 – (-1))2 + (-3 – 3)2 + (4 – (-4))2]
= √[(2)2 + (-6)2 + (8)2]
= √[4 + 36 + 64]
= √104
= 2√26
∴ The required distance is 2√26 units.
(iv) (2, –1, 3) and (–2, 1, 3)
Let P be (2, – 1, 3) and Q be (– 2, 1, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 1, z1 = 3
x2 = – 2, y2 = 1, z2 = 3
Distance PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
∴ The required distance is 2√5 units.
2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
Solution: –
If three points are collinear, then they lie on a line.
Firstly let us calculate distance between the 3 points
i.e. PQ, QR and PR
Calculating PQ
P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 2, y1 = 3, z1 = 5
x2 = 1, y2 = 2, z2 = 3
Distance PQ = √[(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]
= √[(3)2 + (-1)2 + (-2)2]
= √[9 + 1 + 4]
= √14
Calculating QR
Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 1, y1 = 2, z1 = 3
x2 = 7, y2 = 0, z2 = – 1
Distance QR = √[(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]
= √[(6)2 + (-2)2 + (-4)2]
= √[36 + 4 + 16]
= √56
= 2√14
Calculating PR
P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 2, y1 = 3, z1 = 5
x2 = 7, y2 = 0, z2 = – 1
Distance PR = √[(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]
= √[(9)2 + (-3)2 + (-6)2]
= √[81 + 9 + 36]
= √126
= 3√14
Thus, PQ = √14, QR = 2√14 and PR = 3√14
So, PQ + QR = √14 + 2√14
= 3√14
= PR
∴ The points P, Q and R are collinear.
3. Verify the following:
(i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.
Solution: –
(i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
Let us consider the points be
P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6)
If any 2 sides are equal, hence it will be an isosceles triangle
So firstly let us calculate the distance of PQ, QR
Calculating PQ
P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 0, y1 = 7, z1 = – 10
x2 = 1, y2 = 6, z2 = – 6
Distance PQ = √[(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]
= √[(1)2 + (-1)2 + (4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 1, y1 = 6, z1 = – 6
x2 = 4, y2 = 9, z2 = – 6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Hence, PQ = QR
18 = 18
2 sides are equal
∴ PQR is an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.
Let the points be
P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)
Firstly let us calculate the distance of PQ, OR and PR
Calculating PQ
P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 0, y1 = 7, z1 = 10
x2 = – 1, y2 = 6, z2 = 6
Distance PQ = √[(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]
= √[(-1)2 + (-1)2 + (-4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 1, y1 = 6, z1 = – 6
x2 = 4, y2 = 9, z2 = – 6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Calculating PR
P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 0, y1 = 7, z1 = 10
x2 = – 4, y2 = 9, z2 = 6
Distance PR = √[(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]
= √[(-4)2 + (2)2 + (-4)2]
= √[16 + 4 + 16]
= √36
Now,
PQ2 + QR2 = 18 + 18
= 36
= PR2
By using converse of Pythagoras theorem,
∴ The given vertices P, Q & R are the vertices of a right – angled triangle at Q.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.
Let the points be: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)
ABCD can be vertices of parallelogram only if opposite sides are equal.
i.e. AB = CD and BC = AD
Firstly let us calculate the distance
Calculating AB
A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)
By using the formula,
Distance AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 1, y1 = 2, z1 = 1
x2 = 1, y2 = – 2, z2 = 5
Distance AB = √[(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]
= √[(2)2 + (-4)2 + (4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating BC
B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)
By using the formula,
Distance BC = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 1, y1 = – 2, z1 = 5
x2 = 4, y2 = – 7, z2 = 8
Distance BC = √[(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]
= √[(3)2 + (-5)2 + (3)2]
= √[9 + 25 + 9]
= √43
Calculating CD
C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)
By using the formula,
Distance CD = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 4, y1 = – 7, z1 = 8
x2 = 2, y2 = – 3, z2 = 4
Distance CD = √[(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]
= √[(-2)2 + (4)2 + (-4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating DA
D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)
By using the formula,
Distance DA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 3, z1 = 4
x2 = – 1, y2 = 2, z2 = 1
Distance DA = √[(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]
= √[(-3)2 + (5)2 + (-3)2]
= √[9 + 25 + 9]
= √43
Since AB = CD and BC = DA (given)
So, In ABCD both pairs of opposite sides are equal.
∴ ABCD is a parallelogram.
4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Solution: –
Let A (1, 2, 3) & B (3, 2, – 1)
Let point P be (x, y, z)
Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1)
i.e. PA = PB
Firstly let us calculate
Calculating PA
P ≡ (x, y, z) and A ≡ (1, 2, 3)
By using the formula,
Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = x, y1 = y, z1 = z
x2 = 1, y2 = 2, z2 = 3
Distance PA = √[(1 – x)2 + (2 – y)2 + (3 – z)2]
Calculating PB
P ≡ (x, y, z) and B ≡ (3, 2, – 1)
By using the formula,
Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = x, y1 = y, z1 = z
x2 = 3, y2 = 2, z2 = – 1
Distance PB = √[(3 – x)2 + (2 – y)2 + (-1 – z)2]
Since PA = PB
Square on both the sides, we get
PA2 = PB2
(1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2
(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)
(9 + x2 – 6x) + (4 + y2 – 4y) + (1 + z2 + 2z)
– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
4x – 8z = 0
x – 2z = 0
∴ The required equation is x – 2z = 0
5. Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10.
Solution: –
Let A (4, 0, 0) & B (– 4, 0, 0)
Let the coordinates of point P be (x, y, z)
Calculating PA
P ≡ (x, y, z) and A ≡ (4, 0, 0)
By using the formula,
Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = x, y1 = y, z1 = z
x2 = 4, y2 = 0, z2 = 0
Distance PA = √[(4– x)2 + (0 – y)2 + (0 – z)2]
Calculating PB
P ≡ (x, y, z) and B ≡ (– 4, 0, 0)
By using the formula,
Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = x, y1 = y, z1 = z
x2 = – 4, y2 = 0, z2 = 0
Distance PB = √[(-4– x)2 + (0 – y)2 + (0 – z)2]
Now it is given that:
PA + PB = 10
PA = 10 – PB
Square on both the sides, we get
PA2 = (10 – PB)2
PA2 = 100 + PB2 – 20 PB
(4 – x)2 + (0 – y)2 + (0 – z)2
100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB
(16 + x2 – 8x) + (y2) + (z2)
100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB
20 PB = 16x + 100
5 PB = (4x + 25)
Square on both the sides again, we get
25 PB2 = 16x2 + 200x + 625
25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625
25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625
25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625
9x2 + 25y2 + 25z2 – 225 = 0
∴ The required equation is 9x2 + 25y2 + 25z2 – 225 = 0
EXERCISE 12.3
1. Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2: 3 internally, (ii) 2: 3 externally.
Solution: –
Let the line segment joining the points P (-2, 3, 5) and Q (1, -4, 6) be PQ.
(i) 2: 3 internally
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:
Upon comparing we have
x1 = -2, y1 = 3, z1 = 5;
x2 = 1, y2 = -4, z2 = 6 and
m = 2, n = 3
So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 internally is given by:
Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-4/5, 1/5, 27/5)
(ii) 2: 3 externally
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) externally in the ratio m: n is given by:
Upon comparing we have
x1 = -2, y1 = 3, z1 = 5;
x2 = 1, y2 = -4, z2 = 6 and
m = 2, n = 3
So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2: 3 externally is given by:
∴ The co-ordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).
2. Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.
Solution: –
Let us consider Q divides PR 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:
Upon comparing we have,
x1 = 3, y1 = 2, z1 = -4;
x2 = 9, y2 = 8, z2 = -10 and
m = k, n = 1
So, we have
9k + 3 = 5 (k+1)
9k + 3 = 5k + 5
9k – 5k = 5 – 3
4k = 2
k = 2/4
= ½
Hence, the ratio in which Q divides PR is 1: 2.
3. Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).
Solution: –
Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.
We know that any point on the YZ-plane is of the form (0, y, z).
So now, let R (0, y, z) divides the line segment PQ in the ratio k: 1.
Then,
Upon comparing we have,
x1 = -2, y1 = 4, z1 = 7;
x2 = 3, y2 = -5, z2 = 8 and
m = k, n = 1
By using the 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 we have,
3k – 2 = 0
3k = 2
k = 2/3
Hence, the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8) is 2:3.
4. Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2) are collinear.
Solution: –
Let the point P divides AB in the ratio k: 1.
Upon comparing we have,
x1 = 2, y1 = -3, z1 = 4;
x2 = -1, y2 = 2, z2 = 1 and
m = k, n = 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 we have,
Now, we check if for some value of k, the point coincides with the point C.
Put (-k+2)/(k+1) = 0
-k + 2 = 0
k = 2
When k = 2, then (2k-3)/(k+1) = (2(2)-3)/(2+1)
= (4-3)/3
= 1/3
And, (k+4)/(k+1) = (2+4)/(2+1)
= 6/3
= 2
∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2: 1 and is same as P.
Hence, A, B, C are collinear.
5. Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).
Solution: –
Let A (x1, y1, z1) and B (x2, y2, z2) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).
A divides the line segment PQ in the ratio 1: 2.
Upon comparing we have,
x1 = 4, y1 = 2, z1 = -6;
x2 = 10, y2 = -16, z2 = 6 and
m = 1, n = 2
By using the 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 we have,
Similarly, we know that B divides the line segment PQ in the ratio 2: 1.
Upon comparing we have,
x1 = 4, y1 = 2, z1 = -6;
x2 = 10, y2 = -16, z2 = 6 and
m = 2, n = 1
By using the 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 we have,
∴ The coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).
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